Step 1: Understanding the Concept:

The angle \(\theta\) between two non-zero vectors \(\vec{a}\) and \(\vec{b}\) is obtuse if and only if \(\cos \theta < 0\).

Since the dot product is defined as \(\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta\), and the magnitudes \(|\vec{a}|\) and \(|\vec{b}|\) are always positive, the sign of \(\cos \theta\) is determined by the sign of the dot product \(\vec{a} \cdot \vec{b}\).

Therefore, the angle is obtuse if \(\vec{a} \cdot \vec{b} < 0\).


Step 2: Key Formula or Approach:

Calculate the dot product of the two given vectors and set the resulting expression to be less than zero.
\[ \vec{a} \cdot \vec{b} < 0 \]

Step 3: Detailed Explanation:

Given vectors are \(\vec{a} = 2x^2\hat{i} + 4x\hat{j} + \hat{k}\) and \(\vec{b} = 7\hat{i} - 2\hat{j} + x\hat{k}\).

First, we compute the dot product:
\[ \vec{a} \cdot \vec{b} = (2x^2)(7) + (4x)(-2) + (1)(x) \] \[ \vec{a} \cdot \vec{b} = 14x^2 - 8x + x \] \[ \vec{a} \cdot \vec{b} = 14x^2 - 7x \]
For the angle between the vectors to be obtuse, the dot product must be negative:
\[ 14x^2 - 7x < 0 \]
Factor out the common term \(7x\):
\[ 7x(2x - 1) < 0 \]
To solve this inequality, we find the critical points by setting the expression to zero: \(7x(2x - 1) = 0\).

This gives \(x = 0\) and \(2x - 1 = 0 \implies x = \frac{1}{2}\).

These critical points divide the number line into three intervals: \((-\infty, 0)\), \((0, \frac{1}{2})\), and \((\frac{1}{2}, \infty)\). We test a value from each interval to see where the inequality holds.


For \(x < 0\) (e.g., \(x = -1\)): \(7(-1)(2(-1)-1) = (-7)(-3) = 21 > 0\).
For \(0 < x < \frac{1}{2}\) (e.g., \(x = \frac{1}{4}\)): \(7(\frac{1}{4})(2(\frac{1}{4})-1) = \frac{7}{4}(\frac{1}{2}-1) = \frac{7}{4}(-\frac{1}{2}) = -\frac{7}{8} < 0\).
For \(x > \frac{1}{2}\) (e.g., \(x = 1\)): \(7(1)(2(1)-1) = 7(1) = 7 > 0\).

The inequality \(7x(2x - 1) < 0\) is satisfied only in the interval \(0 < x < \frac{1}{2}\).

Note that if the dot product is zero, the vectors are orthogonal (angle is \(90^\circ\)), not obtuse. An obtuse angle is strictly between \(90^\circ\) and \(180^\circ\).


Step 4: Final Answer:

The values of \(x\) for which the angle is obtuse are given by the interval \((0, \frac{1}{2})\).
Quick Tip: Remember the geometric interpretation of the dot product: \(\vec{a} \cdot \vec{b} > 0 \implies\) Angle is acute (\(0^\circ \le \theta < 90^\circ\)). \(\vec{a} \cdot \vec{b} < 0 \implies\) Angle is obtuse (\(90^\circ < \theta \le 180^\circ\)). \(\vec{a} \cdot \vec{b} = 0 \implies\) Angle is right (\(\theta = 90^\circ\)). This simple check can quickly solve many vector problems.

Step 1: Understanding the Concept:

If a line in 3D space makes angles \(\alpha\), \(\beta\), and \(\gamma\) with the positive x, y, and z axes respectively, then the cosines of these angles are called the direction cosines of the line, denoted by \(l, m, n\).
\(l = \cos \alpha\), \(m = \cos \beta\), \(n = \cos \gamma\).

These direction cosines are related by a fundamental identity.


Step 2: Key Formula or Approach:

The sum of the squares of the direction cosines of any line is always equal to 1.
\[ l^2 + m^2 + n^2 = 1 \]
or \[ \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 \]

Step 3: Detailed Explanation:

We are given the angles that the line makes with the x and y axes:

Angle with x-axis, \(\alpha = \frac{3\pi}{4}\).

Angle with y-axis, \(\beta = \frac{\pi}{4}\).

Angle with z-axis, \(\gamma = \theta\).

First, we find the corresponding direction cosines \(l\) and \(m\):
\[ l = \cos \alpha = \cos\left(\frac{3\pi}{4}\right) = \cos\left(\pi - \frac{\pi}{4}\right) = -\cos\left(\frac{\pi}{4}\right) = -\frac{1}{\sqrt{2}} \] \[ m = \cos \beta = \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]
The direction cosine with the z-axis is \(n = \cos \gamma = \cos \theta\).

Now, we apply the identity \(\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1\):
\[ \left(-\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 + \cos^2 \theta = 1 \] \[ \frac{1}{2} + \frac{1}{2} + \cos^2 \theta = 1 \] \[ 1 + \cos^2 \theta = 1 \] \[ \cos^2 \theta = 0 \] \[ \cos \theta = 0 \]
The principal value for \(\theta\) in the range \([0, \pi]\) for which \(\cos \theta = 0\) is \(\theta = \frac{\pi}{2}\).


Step 4: Final Answer:

The angle the line makes with the positive z-axis is \(\frac{\pi}{2}\).
Quick Tip: When dealing with direction cosines, remember the identity \(l^2+m^2+n^2=1\). This is one of the most fundamental and frequently used formulas in 3D geometry. Also, recall the values of cosine for standard angles like \(\frac{\pi}{4}, \frac{\pi}{3}, \frac{\pi}{2}, \frac{2\pi}{3}, \frac{3\pi}{4}\), etc.

Step 1: Understanding the Concept:

This problem involves the integration of a trigonometric function. The key is to manipulate the integrand into a form that can be easily integrated using standard formulas. Using the Pythagorean identity \(\sin^2 x + \cos^2 x = 1\) is a common strategy.


Step 2: Key Formula or Approach:

1. Replace 1 in the numerator with \(\sin^2 x + \cos^2 x\).

2. Split the integral into two parts.

3. Use the standard integration formulas: \(\int \sec^2 x \, dx = \tan x + C\) and \(\int \csc^2 x \, dx = -\cot x + C\).


Step 3: Detailed Explanation:

Let the given integral be \(I\).
\[ I = \int \frac{dx}{\sin^2 x \cos^2 x} \]
We can write the numerator as 1. Using the identity \(\sin^2 x + \cos^2 x = 1\), we have:
\[ I = \int \frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x} \, dx \]
Now, split the fraction into two parts:
\[ I = \int \left( \frac{\sin^2 x}{\sin^2 x \cos^2 x} + \frac{\cos^2 x}{\sin^2 x \cos^2 x} \right) \, dx \]
Cancel out the common terms in each fraction:
\[ I = \int \left( \frac{1}{\cos^2 x} + \frac{1}{\sin^2 x} \right) \, dx \]
Recall that \(\frac{1}{\cos^2 x} = \sec^2 x\) and \(\frac{1}{\sin^2 x} = \csc^2 x\).
\[ I = \int (\sec^2 x + \csc^2 x) \, dx \]
Now, integrate term by term:
\[ I = \int \sec^2 x \, dx + \int \csc^2 x \, dx \]
Using the standard integration formulas:
\[ I = (\tan x) + (-\cot x) + C \] \[ I = \tan x - \cot x + C \]
where C is the constant of integration.


Alternative Method:
\[ I = \int \frac{dx}{\sin^2 x \cos^2 x} = \int \frac{4 \, dx}{4 \sin^2 x \cos^2 x} = \int \frac{4 \, dx}{(2 \sin x \cos x)^2} \]
Using the identity \(\sin 2x = 2 \sin x \cos x\): \[ I = \int \frac{4 \, dx}{(\sin 2x)^2} = 4 \int \csc^2(2x) \, dx \]
Now, integrate using the formula \(\int \csc^2(au) \, du = -\frac{1}{a} \cot(au) + C\).
\[ I = 4 \left( -\frac{1}{2} \cot(2x) \right) + C = -2 \cot(2x) + C \]
We can transform this back to \(\tan x\) and \(\cot x\).
\[ -2 \cot(2x) = -2 \frac{\cos(2x)}{\sin(2x)} = -2 \frac{\cos^2 x - \sin^2 x}{2 \sin x \cos x} = - \frac{\cos^2 x - \sin^2 x}{\sin x \cos x} \] \[ = -\left(\frac{\cos^2 x}{\sin x \cos x} - \frac{\sin^2 x}{\sin x \cos x}\right) = -\left(\frac{\cos x}{\sin x} - \frac{\sin x}{\cos x}\right) = -(\cot x - \tan x) = \tan x - \cot x \]
So, \(I = \tan x - \cot x + C\). Both methods yield the same result.


Step 4: Final Answer:

The value of the integral is \(\tan x - \cot x + C\).
Quick Tip: When you see \(\sin^2 x \cos^2 x\) in the denominator, think of two main strategies: either using \(1 = \sin^2 x + \cos^2 x\) in the numerator or using the double angle formula \(\sin 2x = 2\sin x \cos x\). The first method is often more direct.

Step 1: Understanding the Concept:

This question tests the properties of determinants of skew-symmetric matrices. A matrix \(P\) is skew-symmetric if its transpose is equal to its negative, i.e., \(P^T = -P\). There is a specific property for the determinant of a skew-symmetric matrix based on its order (the number of rows or columns).


Step 2: Key Formula or Approach:

The key property is: The determinant of a skew-symmetric matrix of odd order is always zero.

We will use this property along with the determinant properties \(\det(A^T) = \det(A)\) and \(\det(kA) = k^n \det(A)\) for an \(n \times n\) matrix \(A\).


Step 3: Detailed Explanation:

Let \(P\) be a skew-symmetric matrix of order \(n=3\).

By definition of a skew-symmetric matrix, we have:
\[ P^T = -P \]
Taking the determinant on both sides:
\[ \det(P^T) = \det(-P) \]
We know two properties of determinants:
1. \(\det(A^T) = \det(A)\)
2. \(\det(kA) = k^n \det(A)\) where \(n\) is the order of the matrix \(A\).

Applying the first property to the left side:
\[ \det(P) = \det(-P) \]
Applying the second property to the right side, with \(k = -1\) and \(n = 3\) (since the order is 3):
\[ \det(-P) = (-1)^3 \det(P) = -1 \cdot \det(P) = -\det(P) \]
So, the equation becomes:
\[ \det(P) = -\det(P) \]
Adding \(\det(P)\) to both sides:
\[ 2 \det(P) = 0 \] \[ \det(P) = 0 \]
The problem states that \(\det(P) = \alpha\). Therefore, \(\alpha = 0\).

We are asked to find the value of \((2025)^\alpha\).

Substituting \(\alpha = 0\):
\[ (2025)^0 \]
Any non-zero number raised to the power of 0 is 1.
\[ (2025)^0 = 1 \]

Step 4: Final Answer:

The value of \((2025)^\alpha\) is 1.
Quick Tip: This is a standard result that's worth memorizing for competitive exams: The determinant of a skew-symmetric matrix of \textbf{odd} order is \textbf{zero}. The determinant of a skew-symmetric matrix of \textbf{even} order is a \textbf{perfect square}. Knowing this fact directly would solve the problem in seconds.

Step 1: Understanding the Concept:

This problem requires finding the principal value of an inverse trigonometric function. The principal value branch for \(\sin^{-1}(x)\) is the interval \([-\frac{\pi}{2}, \frac{\pi}{2}]\). The solution must lie within this range. We first need to simplify the argument inside the \(\sin^{-1}\) function.


Step 2: Key Formula or Approach:

1. Simplify the angle \(\frac{43\pi}{5}\) by finding a coterminal angle in a simpler range. Use the periodicity of the cosine function, \(\cos(2n\pi + x) = \cos(x)\).

2. Convert the cosine function to a sine function using the identity \(\cos(x) = \sin\left(\frac{\pi}{2} - x\right)\).

3. Use the property \(\sin^{-1}(\sin(y)) = y\), provided \(y\) is in the principal value range \([-\frac{\pi}{2}, \frac{\pi}{2}]\).


Step 3: Detailed Explanation:

Let's evaluate the expression step-by-step. First, simplify the argument of the cosine function.
\[ \frac{43\pi}{5} = \frac{40\pi + 3\pi}{5} = \frac{40\pi}{5} + \frac{3\pi}{5} = 8\pi + \frac{3\pi}{5} \]
Now, apply this to the cosine function. Since cosine has a period of \(2\pi\), we can remove multiples of \(2\pi\). \(8\pi\) is \(4 \times 2\pi\).
\[ \cos\left(\frac{43\pi}{5}\right) = \cos\left(8\pi + \frac{3\pi}{5}\right) = \cos\left(\frac{3\pi}{5}\right) \]
The expression now becomes:
\[ \sin^{-1}\left(\cos\left(\frac{3\pi}{5}\right)\right) \]
To proceed, we need to express \(\cos\left(\frac{3\pi}{5}\right)\) in terms of a sine function. We use the complementary angle identity \(\cos(x) = \sin\left(\frac{\pi}{2} - x\right)\).
\[ \cos\left(\frac{3\pi}{5}\right) = \sin\left(\frac{\pi}{2} - \frac{3\pi}{5}\right) \] \[ \frac{\pi}{2} - \frac{3\pi}{5} = \frac{5\pi - 6\pi}{10} = -\frac{\pi}{10} \]
So, \(\cos\left(\frac{3\pi}{5}\right) = \sin\left(-\frac{\pi}{10}\right)\).

Substituting this back into our expression:
\[ \sin^{-1}\left(\sin\left(-\frac{\pi}{10}\right)\right) \]
Now we must check if the angle \(-\frac{\pi}{10}\) lies within the principal value range of \(\sin^{-1}(x)\), which is \([-\frac{\pi}{2}, \frac{\pi}{2}]\).

The range is equivalent to \([-\frac{5\pi}{10}, \frac{5\pi}{10}]\).

Since \(-\frac{5\pi}{10} \le -\frac{\pi}{10} \le \frac{5\pi}{10}\), the angle is within the principal value branch.

Therefore, we can directly write:
\[ \sin^{-1}\left(\sin\left(-\frac{\pi}{10}\right)\right) = -\frac{\pi}{10} \]

Step 4: Final Answer:

The principal value is \(-\frac{\pi}{10}\).
Quick Tip: Always simplify the angle inside the trigonometric function first. Then, use identities like \(\cos(x) = \sin(\pi/2 - x)\) or \(\sin(x) = \cos(\pi/2 - x)\) to match the outer inverse function. Finally, ensure your result is within the defined principal value range before concluding the answer.

Step 1: Understanding the Concept:

This question tests the fundamental relationship between continuity and differentiability of functions.

Continuity: A function is continuous at a point if its graph has no breaks, jumps, or holes at that point.

Differentiability: A function is differentiable at a point if it has a well-defined, non-vertical tangent line at that point. This implies the graph is "smooth" and has no sharp corners or cusps.


Step 2: Key Theorem:

The core theorem connecting these two concepts is:

If a function is differentiable at a point, then it must be continuous at that point.

However, the converse is not true. A function can be continuous at a point but not differentiable at that point.


Step 3: Detailed Explanation:

Let's analyze the relationship based on the theorem.

1. Differentiability implies Continuity: Every function that belongs to the set of differentiable functions (Set B) must also belong to the set of continuous functions (Set A). In set theory terms, this means that B is a subset of A (\(B \subseteq A\)).

2. Continuity does not imply Differentiability: There exist functions that are continuous but not differentiable. A classic example is the absolute value function, \(f(x) = |x|\). It is continuous everywhere, including at \(x=0\), but it is not differentiable at \(x=0\) because of the sharp corner. This means there are elements in Set A (continuous functions) that are not in Set B (differentiable functions).

Therefore, Set B is a proper subset of Set A (\(B \subset A\)). This means all elements of B are in A, but A contains elements that are not in B.


Step 4: Venn Diagram Representation:

The relationship \(B \subset A\) is represented by a Venn diagram where the circle for Set B is drawn completely inside the circle for Set A. This corresponds to option (B).


Option (A) is incorrect because the sets are not disjoint; differentiable functions are a type of continuous function.
Option (C) is incorrect because it implies some differentiable functions are not continuous, which is false.
Option (D) is incorrect because there are continuous functions that are not differentiable.


Step 5: Final Answer:

The correct depiction is that the set of differentiable functions (B) is a proper subset of the set of continuous functions (A).
Quick Tip: Remember the mnemonic "Differentiability \(\implies\) Continuity". Think of it as a hierarchy: differentiability is a stronger condition than continuity. For a visual cue, remember the example \(f(x) = |x|\), which is continuous but has a "corner" where it's not differentiable.

Step 1: Understanding the Concept:

The area of a region bounded by a curve can be calculated by integration. If the area is bounded by the y-axis, it is often easier to integrate with respect to \(y\). This involves expressing \(x\) as a function of \(y\) and using the y-coordinates as the limits of integration.


Step 2: Key Formula or Approach:

The area of a region bounded by the curve \(x = g(y)\), the y-axis (\(x=0\)), and the horizontal lines \(y=c\) and \(y=d\) is given by: \[ Area = \int_{c}^{d} g(y) \, dy \]

Step 3: Detailed Explanation:

1. Identify the Boundaries: The shaded region is bounded by:

The curve \(y = x^2\).
The y-axis (which is the line \(x=0\)).
The bottom boundary is the x-axis, which corresponds to \(y=0\).
The top boundary is a horizontal line corresponding to the value of \(y\) when \(x=2\). When \(x=2\), \(y = (2)^2 = 4\). So the top boundary is the line \(y=4\).

2. Set up the Integral: Since the region is bounded by the y-axis, we should integrate with respect to \(y\). To do this, we need to express \(x\) in terms of \(y\).
From the equation \(y = x^2\), we get \(x = \pm \sqrt{y}\).
Since the shaded region is in the first quadrant where \(x \ge 0\), we take the positive root: \(x = \sqrt{y}\).
3. Determine the Limits of Integration: We are integrating with respect to \(y\), so we need the y-values that define the region's vertical extent. The region starts at \(y=0\) and ends at \(y=4\).
So, the limits of integration are from \(c=0\) to \(d=4\).
4. Write the Integral:
Using the formula, the area is:
\[ Area = \int_{0}^{4} x \, dy = \int_{0}^{4} \sqrt{y} \, dy \]
This matches option (B).


Analysis of other options:

(A) \(\int_{0}^{2} x^2 \, dx\): This represents the area under the curve \(y=x^2\) and above the x-axis, between \(x=0\) and \(x=2\). This is the unshaded region in the rectangle formed by \(x=0, x=2, y=0, y=4\).
(C) \(\int_{0}^{4} x^2 \, dx\): This integral is incorrectly set up, mixing limits for \(y\) with an integrand and differential for \(x\).
(D) \(\int_{0}^{2} \sqrt{y} \, dy\): The limits of integration are incorrect for an integral with respect to \(y\). The limits should be from 0 to 4.


Step 4: Final Answer:

The integral representing the area of the shaded region is \(\int_{0}^{4} \sqrt{y} \, dy\).
Quick Tip: When finding the area, always check the boundaries of the region. If the region is defined by the y-axis, it's a strong hint to integrate with respect to \(y\). This means you need to express \(x\) as \(g(y)\) and use y-values for the limits of integration.

Step 1: Understanding the Concept:

This problem involves the simplification of a matrix expression. The key concept is to apply the rules of matrix addition and scalar multiplication, keeping in mind that matrix multiplication is generally not commutative (i.e., \(AB \ne BA\)).


Step 2: Key Formula or Approach:

Use the distributive property of matrix multiplication over addition: \(k(A+B) = kA + kB\), where \(k\) is a scalar. Then, combine like terms.


Step 3: Detailed Explanation:

Let's simplify the given expression: \[ 4 AB + 3(AB + BA) - 4 BA \]
First, apply the distributive property to the term \(3(AB + BA)\): \[ 3(AB + BA) = 3AB + 3BA \]
Now, substitute this back into the original expression: \[ 4 AB + (3AB + 3BA) - 4 BA \]
Remove the parentheses and rearrange the terms to group like terms together. We can group the 'AB' terms and the 'BA' terms. \[ (4 AB + 3 AB) + (3 BA - 4 BA) \]
Now, perform the addition and subtraction on the coefficients of the like terms: \[ (4+3)AB + (3-4)BA \] \[ 7AB + (-1)BA \] \[ 7AB - BA \]
The simplified expression is \(7AB - BA\).


Step 4: Comparing with the answers:


Abhay's answer: 6 AB
Bina's answer: 7 AB - BA
Chhaya's answer: 8 AB
Devesh's answer: 7 BA - AB

Comparing our result with the given answers, we see that Bina's answer is correct.

The conditions \(AB \ne BA\) and \(A^{-1} \ne B\) are given to ensure that no further simplification is possible. If \(AB=BA\), the expression would simplify to \(7AB - AB = 6AB\), which would make Abhay's answer correct. However, since they are not equal, we must leave the terms separate.


Step 5: Final Answer:

Bina answered the question correctly.
Quick Tip: Always be careful with matrix algebra. Unlike regular algebra with real numbers, matrix multiplication is not commutative. Treat \(AB\) and \(BA\) as distinct entities unless it is specified that the matrices commute.

Step 1: Understanding the Concept:


Order of a differential equation is the order of the highest derivative appearing in the equation.
Degree of a differential equation is the power of the highest order derivative after the equation has been made free from radicals and fractions in its derivatives.

First, we need to simplify the given differential equation to clearly identify the highest order derivative and its power.


Step 2: Key Formula or Approach:

Apply the chain rule for differentiation to expand the term \(\frac{d}{dx}\left(\left(\frac{dy}{dx}\right)^3\right)\).
The chain rule states \(\frac{d}{dx}(u^n) = n u^{n-1} \frac{du}{dx}\).


Step 3: Detailed Explanation:

The given differential equation is: \[ \frac{d}{dx}\left(\left(\frac{dy}{dx}\right)^3\right) = 0 \]
Let \(u = \frac{dy}{dx}\). The equation becomes \(\frac{d}{dx}(u^3) = 0\).
Using the chain rule: \[ 3u^2 \frac{du}{dx} = 0 \]
Now substitute back \(u = \frac{dy}{dx}\): \[ 3\left(\frac{dy}{dx}\right)^2 \frac{d}{dx}\left(\frac{dy}{dx}\right) = 0 \] \[ 3\left(\frac{dy}{dx}\right)^2 \left(\frac{d^2y}{dx^2}\right) = 0 \]
This is the simplified form of the differential equation.


Finding the Order (p):

The derivatives present in the equation are \(\frac{dy}{dx}\) (first order) and \(\frac{d^2y}{dx^2}\) (second order).
The highest order of the derivative is 2.
So, the order \(p = 2\).


Finding the Degree (q):

The highest order derivative is \(\frac{d^2y}{dx^2}\). The power to which this term is raised is 1.
The equation is already a polynomial in its derivatives, so we can directly find the degree.
The degree \(q = 1\).


Calculating \((p-q)^q\):

We need to find the value of \((p-q)^q\).
Substitute the values \(p=2\) and \(q=1\): \[ (2-1)^1 = (1)^1 = 1 \]

Step 4: Final Answer:

The value of \((p-q)^q\) is 1.
Quick Tip: To find the order and degree, you must first simplify the differential equation. Do not determine the order and degree from the initial expression. Differentiate any terms as needed to get an equation that explicitly shows all the derivative terms.

Step 1: Understanding the Concept:

A function \(f(x)\) is said to be increasing on an interval if its first derivative, \(f'(x)\), is non-negative on that interval. That is, \(f'(x) \ge 0\). If \(f'(x) > 0\), the function is strictly increasing.


Step 2: Key Formula or Approach:

1. Find the first derivative of the function, \(f'(x)\).
2. Set up the inequality \(f'(x) \ge 0\) to find the interval(s) where the function is increasing.
3. Solve the inequality for \(x\).


Step 3: Detailed Explanation:

The given function is: \[ f(x) = x^2 - 4x + 6 \]
First, we find the derivative of \(f(x)\) with respect to \(x\): \[ f'(x) = \frac{d}{dx}(x^2 - 4x + 6) \] \[ f'(x) = 2x - 4 \]
For the function to be increasing, we must have \(f'(x) \ge 0\). \[ 2x - 4 \ge 0 \]
Now, we solve this linear inequality for \(x\):
Add 4 to both sides: \[ 2x \ge 4 \]
Divide by 2: \[ x \ge 2 \]
This means the function is increasing for all values of \(x\) that are greater than or equal to 2.

In interval notation, this is represented as \([2, \infty)\).


Alternative Method (Analyzing the Parabola):

The function \(f(x) = x^2 - 4x + 6\) is a quadratic function, and its graph is an upward-opening parabola (since the coefficient of \(x^2\) is positive, i.e., 1 > 0).

The vertex of a parabola given by \(ax^2 + bx + c\) occurs at \(x = -\frac{b}{2a}\).
For our function, \(a=1, b=-4, c=6\).
The x-coordinate of the vertex is: \[ x = -\frac{-4}{2(1)} = \frac{4}{2} = 2 \]
An upward-opening parabola is decreasing to the left of its vertex and increasing to the right of its vertex.
Therefore, the function \(f(x)\) is increasing for \(x \ge 2\), which is the interval \([2, \infty)\).


Step 4: Final Answer:

The function is increasing in the interval \([2, \infty)\).
Quick Tip: For polynomial functions, the first derivative test is the most straightforward way to find intervals of increase or decrease. Find \(f'(x)\), find the critical points by solving \(f'(x)=0\), and then test the intervals between these points.

Step 1: Understanding the Concept:

A probability distribution for a discrete random variable \(X\) lists all the possible values of \(X\) along with their corresponding probabilities \(P(X)\). A fundamental property of any probability distribution is that the sum of all probabilities must be equal to 1.


Step 2: Key Formula or Approach:

The sum of all probabilities for all possible values of the random variable \(X\) is 1. \[ \sum P(X_i) = 1 \]

Step 3: Detailed Explanation:

From the given probability distribution table, the possible values of \(X\) are 0, 1, 2, and 3. The corresponding probabilities are:

\(P(X=0) = p\)
\(P(X=1) = p\)
\(P(X=2) = 0.3\)
\(P(X=3) = 2p\)

According to the property of probability distributions, the sum of these probabilities must be 1. \[ P(X=0) + P(X=1) + P(X=2) + P(X=3) = 1 \]
Substitute the given values: \[ p + p + 0.3 + 2p = 1 \]
Now, we solve this linear equation for \(p\).
Combine the terms with \(p\): \[ (1 + 1 + 2)p + 0.3 = 1 \] \[ 4p + 0.3 = 1 \]
Subtract 0.3 from both sides: \[ 4p = 1 - 0.3 \] \[ 4p = 0.7 \]
Divide by 4: \[ p = \frac{0.7}{4} \]
To express this as a fraction, we can write \(0.7\) as \(\frac{7}{10}\). \[ p = \frac{7/10}{4} = \frac{7}{10 \times 4} = \frac{7}{40} \]

Step 4: Final Answer:

The value of p is \(\frac{7}{40}\).
Quick Tip: For any question involving a probability distribution table, the very first thing to check is the condition that the sum of probabilities equals 1. This property is the key to solving for unknown variables in the distribution. Also, remember that each individual probability must be between 0 and 1, inclusive.

Step 1: Understanding the Concept:

The magnitude of the cross product of two vectors, \(\vec{a}\) and \(\vec{b}\), represents the area of the parallelogram formed by these two vectors as adjacent sides. The area of the triangle formed by these two vectors is exactly half the area of that parallelogram.


Step 2: Key Formula or Approach:

If \(\vec{PQ}\) and \(\vec{PR}\) are two vectors representing the adjacent sides of a triangle PQR, then the area of the triangle is given by: \[ Area(\triangle PQR) = \frac{1}{2} |\vec{PQ} \times \vec{PR}| \]
where \(|\vec{a} \times \vec{b}|\) is the magnitude of the cross product vector.
The magnitude of a vector \(x\hat{i} + y\hat{j} + z\hat{k}\) is \(\sqrt{x^2 + y^2 + z^2}\).


Step 3: Detailed Explanation:

We are given the cross product vector: \[ \vec{PQ} \times \vec{PR} = 4\hat{i} + 8\hat{j} - 8\hat{k} \]
First, we need to calculate the magnitude of this vector. \[ |\vec{PQ} \times \vec{PR}| = \sqrt{(4)^2 + (8)^2 + (-8)^2} \] \[ |\vec{PQ} \times \vec{PR}| = \sqrt{16 + 64 + 64} \] \[ |\vec{PQ} \times \vec{PR}| = \sqrt{144} \] \[ |\vec{PQ} \times \vec{PR}| = 12 \]
This magnitude, 12 sq units, represents the area of the parallelogram with adjacent sides \(\vec{PQ}\) and \(\vec{PR}\).

Now, we find the area of the triangle PQR using the formula: \[ Area(\triangle PQR) = \frac{1}{2} |\vec{PQ} \times \vec{PR}| \] \[ Area(\triangle PQR) = \frac{1}{2} \times 12 \] \[ Area(\triangle PQR) = 6 \]
The area of the triangle is 6 square units.


Step 4: Final Answer:

The area (\(\triangle\)PQR) is 6 sq units.
Quick Tip: Don't confuse the area of a parallelogram with the area of a triangle. The magnitude of the cross product \(|\vec{a} \times \vec{b}|\) gives the area of the parallelogram. For the triangle formed by the same vectors, you must divide by 2. This is a common mistake in exams.

Step 1: Understanding the Concept:

This question deals with conditional probability. P(A|B) denotes the probability of event A occurring given that event B has already occurred. E' represents the complement of event E. The condition P(F) \(\ne\) 0 is necessary for the conditional probability P(E'|F) to be defined. The given condition P(E) > 0 is extra information, while P(F) \(\ne\) 1 is also not strictly necessary for the derivation but might be given to define other terms like P(E|F').


Step 2: Key Formula or Approach:

1. The definition of conditional probability: \(P(A|B) = \frac{P(A \cap B)}{P(B)}\).

2. The relationship between an event and its complement within a given condition: For any event E, given an event F, the sum of the probabilities of E and its complement E' is 1. That is, \(P(E|F) + P(E'|F) = 1\).


Step 3: Detailed Explanation:

Method 1: Using the definition of conditional probability

By definition, the conditional probability of E' given F is: \[ P(E'|F) = \frac{P(E' \cap F)}{P(F)} \]
The event \(E' \cap F\) means that F occurs but E does not. This is the same as the part of F that is not in E. We can write the probability of F as the sum of the probabilities of its parts: the part that intersects with E and the part that does not. \[ P(F) = P(E \cap F) + P(E' \cap F) \]
From this, we can express \(P(E' \cap F)\) as: \[ P(E' \cap F) = P(F) - P(E \cap F) \]
Now, substitute this back into the formula for \(P(E'|F)\): \[ P(E'|F) = \frac{P(F) - P(E \cap F)}{P(F)} \]
Split the fraction: \[ P(E'|F) = \frac{P(F)}{P(F)} - \frac{P(E \cap F)}{P(F)} \] \[ P(E'|F) = 1 - \frac{P(E \cap F)}{P(F)} \]
We know that \(P(E|F) = \frac{P(E \cap F)}{P(F)}\). Therefore: \[ P(E'|F) = 1 - P(E|F) \]

Method 2: Using the property of complementary events

The sample space, given that F has occurred, is F itself. Within this new sample space, either E occurs or E does not occur (E'). There are no other possibilities. Therefore, the sum of their conditional probabilities must be 1. \[ P(E occurs given F) + P(E does not occur given F) = 1 \] \[ P(E|F) + P(E'|F) = 1 \]
Rearranging this equation to solve for \(P(E'|F)\) gives: \[ P(E'|F) = 1 - P(E|F) \]

Step 4: Final Answer:

The expression for P(E'|F) is \(1 - P(E|F)\).
Quick Tip: Remember that the rules for complementary events hold in the context of conditional probability as well. Just as \(P(A) + P(A') = 1\), it is also true that \(P(A|B) + P(A'|B) = 1\). This is a very useful identity.

Step 1: Understanding the Concept:

Let's recall the definitions of symmetric and skew-symmetric matrices.

Symmetric Matrix: A square matrix \(A\) is symmetric if its transpose is equal to the matrix itself. That is, \(A^T = A\). This means \(a_{ij} = a_{ji}\) for all \(i, j\).
Skew-Symmetric Matrix: A square matrix \(A\) is skew-symmetric if its transpose is equal to the negative of the matrix. That is, \(A^T = -A\). This means \(a_{ij} = -a_{ji}\) for all \(i, j\). Note that this implies all diagonal elements must be zero (\(a_{ii} = -a_{ii} \implies 2a_{ii} = 0 \implies a_{ii} = 0\)).


Step 2: Key Formula or Approach:

For a matrix \(A\) to be both symmetric and skew-symmetric, it must satisfy both conditions simultaneously:
1. \(A^T = A\) (Symmetric condition)
2. \(A^T = -A\) (Skew-symmetric condition)


Step 3: Detailed Explanation:

If a matrix \(A\) satisfies both conditions, we can equate the two expressions for \(A^T\): \[ A = -A \]
Now, add the matrix \(A\) to both sides of the equation: \[ A + A = -A + A \] \[ 2A = O \]
where \(O\) is the null matrix (a matrix of the same order as A with all elements equal to zero).

Dividing by the scalar 2, we get: \[ A = O \]
This shows that the only matrix that can be both symmetric and skew-symmetric is the null matrix.


Analysis of the options:

(A) Unit Matrix (I): The unit matrix is symmetric (\(I^T = I\)), but it is not skew-symmetric since \(I \ne -I\) (unless it's a 1x1 zero matrix, which is not a unit matrix).
(B) Diagonal Matrix (D): A diagonal matrix is always symmetric. For it to be skew-symmetric, all its diagonal elements must be zero (\(d_{ii} = -d_{ii} \implies d_{ii} = 0\)). This means it must be the null matrix. So, a general non-zero diagonal matrix is not both.
(C) Null Matrix (O): The null matrix satisfies \(O^T = O\) (symmetric) and \(O^T = -O\) (since \(-O\) is also the null matrix). Thus, the null matrix is both symmetric and skew-symmetric.
(D) Row Matrix: A row matrix is not a square matrix (unless it's 1x1), and the concepts of symmetric and skew-symmetric are typically defined for square matrices.


Step 4: Final Answer:

The null matrix is the only matrix that is both symmetric and skew-symmetric.
Quick Tip: This is a standard property tested in exams. Any square matrix \(A\) can be uniquely expressed as the sum of a symmetric matrix \(P = \frac{1}{2}(A+A^T)\) and a skew-symmetric matrix \(Q = \frac{1}{2}(A-A^T)\). The only matrix for which \(P=Q=A\) is the null matrix.

Step 1: Understanding the Concept:

The equation of a straight line in three-dimensional space can be represented in vector form or Cartesian form. We need a point on the line and a vector parallel to the line (the direction vector).


Step 2: Key Formula or Approach:

The vector equation of a line passing through a point with position vector \(\vec{a}\) and parallel to a vector \(\vec{b}\) is given by: \[ \vec{r} = \vec{a} + t\vec{b} \]
where \(t\) is a scalar parameter.

If the point is \((x_1, y_1, z_1)\) and the direction vector is \(\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}\), the parametric Cartesian equations are: \[ x = x_1 + t b_1 \] \[ y = y_1 + t b_2 \] \[ z = z_1 + t b_3 \]

Step 3: Detailed Explanation:

From the problem statement:

The line passes through the point \((x_1, y_1, z_1) = (4, -3, 7)\).
The line is parallel to the vector \(3\hat{i} + \hat{j} + 2\hat{k}\). This is our direction vector, \(\vec{b}\). The components of the direction vector (direction ratios) are \((b_1, b_2, b_3) = (3, 1, 2)\).

Now, we substitute these values into the parametric equations: \[ x = 4 + t(3) \implies x = 3t + 4 \] \[ y = -3 + t(1) \implies y = t - 3 \] \[ z = 7 + t(2) \implies z = 2t + 7 \]
These are the parametric equations of the required line.

We now compare this result with the given options:

(A) \(x = 4t + 3, y = -3t + 1, z = 7t + 2\): Incorrect direction ratios (4, -3, 7).
(B) \(x = 3t + 4, y = t + 3, z = 2t + 7\): The y-equation is \(y = t+3\), which does not pass through \(y=-3\).
(C) \(x = 3t + 4, y = t - 3, z = 2t + 7\): This matches our derived equations exactly.
(D) \(x = 3t + 4, y = -t + 3, z = 2t + 7\): Incorrect direction ratio for y (-1 instead of 1).


Step 4: Final Answer:

The correct equation of the line is \(x = 3t + 4, y = t - 3, z = 2t + 7\).
Quick Tip: To quickly find the equation of a line in parametric form, remember the structure: \(x = (x-coordinate of point) + t \times (x-component of direction vector)\), and similarly for y and z. Always double-check both the point and the direction vector components against the options.

Step 1: Understanding the Concept:

This problem tests the properties of matrix multiplication. Unlike multiplication of real numbers, matrix multiplication is not generally commutative, meaning \(AB\) is not always equal to \(BA\). The identity \(a^2 - b^2 = (a-b)(a+b)\) holds for real numbers because of commutativity. We need to find the condition under which this identity holds for matrices.


Step 2: Key Formula or Approach:

Expand the right-hand side (RHS) of the given matrix equation using the distributive property of matrix multiplication and then compare it with the left-hand side (LHS).


Step 3: Detailed Explanation:

The given equation is: \[ A^2 - B^2 = (A - B)(A + B) \]
Let's expand the RHS: \[ (A - B)(A + B) = A(A + B) - B(A + B) \]
Using the distributive property again: \[ = (A \cdot A + A \cdot B) - (B \cdot A + B \cdot B) \] \[ = A^2 + AB - BA - B^2 \]
Now, we set the given LHS equal to this expanded RHS: \[ A^2 - B^2 = A^2 + AB - BA - B^2 \]
We can subtract \(A^2\) and add \(B^2\) to both sides of the equation: \[ (A^2 - B^2) - A^2 + B^2 = (A^2 + AB - BA - B^2) - A^2 + B^2 \] \[ 0 = AB - BA \]
Adding \(BA\) to both sides gives: \[ BA = AB \]
This shows that the given identity holds if and only if the matrices A and B commute (i.e., their product is independent of the order of multiplication).


Step 4: Final Answer:

The condition that must always be correct is \(AB = BA\).
Quick Tip: Whenever you see algebraic identities applied to matrices, be suspicious. The failure of commutativity (\(AB \ne BA\)) is the most common reason why standard identities do not hold for matrices. Always expand expressions like \((A+B)^2\) or \((A-B)(A+B)\) carefully. \((A+B)^2 = A^2 + AB + BA + B^2\), not \(A^2 + 2AB + B^2\), unless \(AB=BA\).

Step 1: Understanding the Concept:

The given equations are the parametric equations of a line in 3D space. A point \((x_0, y_0, z_0)\) lies on this line if there exists a single, consistent value of the parameter \(\mu\) that satisfies all three equations when the coordinates of the point are substituted.


Step 2: Key Formula or Approach:

For each point given in the options, substitute its x, y, and z coordinates into the line's equations and solve for \(\mu\). If the value of \(\mu\) is the same for all three equations, the point is on the line.


Step 3: Detailed Explanation:

The equations of the line are:
1. \(x = 1 + 5\mu\)
2. \(y = -5 + \mu\)
3. \(z = -6 - 3\mu\)


Let's test each option:
(A) Point (1, -5, 6):

For x=1: \(1 = 1 + 5\mu \implies 5\mu = 0 \implies \mu = 0\)
For y=-5: \(-5 = -5 + \mu \implies \mu = 0\)
For z=6: \(6 = -6 - 3\mu \implies 12 = -3\mu \implies \mu = -4\)

The values of \(\mu\) (0, 0, -4) are not consistent. So, this point is not on the line.


(B) Point (1, 5, 6):

For x=1: \(1 = 1 + 5\mu \implies \mu = 0\)
For y=5: \(5 = -5 + \mu \implies \mu = 10\)

The values of \(\mu\) are already inconsistent. So, this point is not on the line.


(C) Point (1, -5, -6):

For x=1: \(1 = 1 + 5\mu \implies 5\mu = 0 \implies \mu = 0\)
For y=-5: \(-5 = -5 + \mu \implies \mu = 0\)
For z=-6: \(-6 = -6 - 3\mu \implies 0 = -3\mu \implies \mu = 0\)

The value \(\mu = 0\) is consistent for all three coordinates. Therefore, this point lies on the line.


(D) Point (-1, -5, 6):

For x=-1: \(-1 = 1 + 5\mu \implies -2 = 5\mu \implies \mu = -2/5\)
For y=-5: \(-5 = -5 + \mu \implies \mu = 0\)

The values of \(\mu\) are inconsistent. So, this point is not on the line.


Step 4: Final Answer:

The point \((1, -5, -6)\) lies on the given line.
Quick Tip: A quick way to check is to find the value of the parameter from the simplest equation first. For example, in this case, solving for \(\mu\) from \(y = -5 + \mu\) is very easy. Then, substitute that value of \(\mu\) into the other two equations to see if they hold true.

Step 1: Understanding the Concept:

This question is about interpreting an objective function in the context of a Linear Programming Problem (LPP). The objective function is a mathematical expression that represents the quantity to be optimized (maximized or minimized), such as profit, cost, or resource usage.


Step 2: Key Formula or Approach:

Analyze the components of the given objective function \(Z = 5x + 7y\) based on the problem description.

\(x\): number of units of product X sold.
\(y\): number of units of product Y sold.
5: profit per unit of product X.
7: profit per unit of product Y.


Step 3: Detailed Explanation:

Let's break down the objective function \(Z = 5x + 7y\):

The term \(5x\) represents the total profit from selling \(x\) units of product X (since each unit gives a profit of 5).
The term \(7y\) represents the total profit from selling \(y\) units of product Y (since each unit gives a profit of 7).
The sum \(5x + 7y\) therefore represents the total combined profit from selling \(x\) units of X and \(y\) units of Y.

In a business context, the goal associated with a profit function is typically to maximize it. So, the purpose of this objective function in an LPP would be to find the values of \(x\) and \(y\) (subject to certain constraints) that maximize the total profit \(Z\).


Now let's evaluate the given statements:

(A) ...maximizes the difference of the profit...: Incorrect. The function is a sum (\(5x+7y\)), not a difference (like \(5x-7y\) or \(7y-5x\)).
(B) ...measures the total production...: Incorrect. The total production would be represented by \(x+y\). The function \(Z\) represents total profit.
(C) ...maximizes the combined profit earned from selling X and Y.: Correct. This statement accurately describes the function and its goal in an LPP.
(D) ...ensures the company produces more of product X than product Y.: Incorrect. The objective function itself does not impose any constraints. A condition like \(x > y\) would be a constraint, not part of the objective function's definition.


Step 4: Final Answer:

The correct statement is that the objective function maximizes the combined profit earned from selling X and Y.
Quick Tip: In LPP word problems, carefully identify what each variable and coefficient represents. The objective function will always be a linear combination of the variables representing the quantity to be optimized (e.g., total profit, total cost). The constraints will be inequalities representing limitations (e.g., available resources, production capacity).

Step 1: Understanding the Concept:

This question requires understanding the definitions of a diagonal matrix and a scalar matrix.


Diagonal Matrix: A square matrix in which all the elements are zero except for the elements on the main diagonal.
Scalar Matrix: A special type of diagonal matrix in which all the main diagonal elements are equal to the same non-zero constant.


Step 2: Detailed Explanation:

Analysis of Reason (R):

The statement is: "If a diagonal matrix has all non-zero elements equal, it is known as a scalar matrix." This is the correct definition of a scalar matrix. For example, \(\begin{pmatrix} k & 0
0 & k \end{pmatrix}\) is a scalar matrix. So, Reason (R) is a true statement.


Analysis of Assertion (A):

The statement is: "A = diag [3 5 2] is a scalar matrix of order 3 \(\times\) 3."

The matrix A can be written as: \[ A = \begin{pmatrix} 3 & 0 & 0
0 & 5 & 0
0 & 0 & 2 \end{pmatrix} \]
This is a diagonal matrix of order 3 \(\times\) 3 because all its non-diagonal elements are zero.

However, for it to be a scalar matrix, all its diagonal elements must be equal. Here, the diagonal elements are 3, 5, and 2, which are not equal.

Therefore, A is a diagonal matrix, but it is not a scalar matrix.

So, Assertion (A) is a false statement.


Step 3: Final Conclusion:

Assertion (A) is false and Reason (R) is true. This corresponds to option (D).
Quick Tip: Remember the hierarchy of matrices: Scalar Matrix \(\subset\) Diagonal Matrix \(\subset\) Square Matrix. All scalar matrices are diagonal matrices, but not all diagonal matrices are scalar matrices. This distinction is key to solving such assertion-reason problems.

Step 1: Understanding the Concept:

This question relates to the fundamental principles of Linear Programming Problems (LPP).

Feasible Region: The set of all points that satisfy all the constraints of an LPP.
Optimal Solution: A point within the feasible region where the objective function reaches its maximum or minimum value.


Step 2: Detailed Explanation:

Analysis of Assertion (A):

The statement is: "Every point of the feasible region of a Linear Programming Problem is an optimal solution."

This is incorrect. The feasible region contains infinitely many points (if it's not empty). The optimal solution is a specific point or set of points within this region that optimizes the objective function. For example, if we want to maximize \(Z = x+y\) in a feasible region, only the point(s) furthest from the origin in the direction of the vector \((1,1)\) will be optimal, not every point in the region. So, Assertion (A) is false.


Analysis of Reason (R):

The statement is: "The optimal solution for a Linear Programming Problem exists only at one or more corner point(s) of the feasible region."

This is the statement of the Fundamental Theorem of Linear Programming. It states that if an optimal solution to an LPP exists, then it must occur at a vertex (corner point) of the feasible region. It is possible for the optimal solution to lie along an entire edge of the feasible region, but in such cases, the optimal value is still achieved at the corner points that define that edge. Thus, the search for the optimal solution can be restricted to the corner points. So, Reason (R) is true.


Step 3: Final Conclusion:

Assertion (A) is false and Reason (R) is true. This corresponds to option (D).
Quick Tip: The core idea of solving an LPP graphically is based on the Reason (R). You find the feasible region, identify all the corner points, and then evaluate the objective function at each of these points to find the optimal value. This theorem drastically simplifies the search for a solution.

Step 1: Understanding the Concept:

A function \(f(x)\) is defined as increasing on an interval if its first derivative, \(f'(x)\), is greater than or equal to zero for all \(x\) in that interval. For the function to be increasing on the entire set of real numbers R, we must have \(f'(x) \ge 0\) for all \(x \in R\).


Step 2: Key Formula or Approach:

1. Find the first derivative of the function, \(f'(x)\).

2. Set the condition \(f'(x) \ge 0\).

3. Solve the resulting inequality for the parameter 'a', ensuring it holds for all \(x \in R\).


Step 3: Detailed Explanation:

The given function is: \[ f(x) = \sin x - ax + b \]
First, we find the derivative of \(f(x)\) with respect to \(x\): \[ f'(x) = \frac{d}{dx}(\sin x - ax + b) \] \[ f'(x) = \cos x - a \]
For the function to be increasing on R, the condition is \(f'(x) \ge 0\) for all \(x \in R\). \[ \cos x - a \ge 0 \]
This can be rewritten as: \[ \cos x \ge a \]
This inequality must hold true for all real values of \(x\).

We know that the range of the cosine function is \([-1, 1]\). That is, \(-1 \le \cos x \le 1\).

The minimum value of \(\cos x\) is -1.

For the condition \(a \le \cos x\) to be true for every possible value of \(\cos x\), 'a' must be less than or equal to the smallest possible value of \(\cos x\).

Therefore, 'a' must satisfy: \[ a \le \min(\cos x) \] \[ a \le -1 \]
This ensures that no matter what value \(\cos x\) takes in its range, 'a' will always be less than or equal to it.


Step 4: Final Answer:

The values of 'a' for which the function is increasing on R are given by the interval \((-\infty, -1]\).
Quick Tip: When an inequality involving a trigonometric function must hold for all real numbers, find the range of the trigonometric function. Then, relate the unknown parameter to the maximum or minimum value of that range, as required by the inequality. For \(a \le g(x)\), you need \(a \le \min(g(x))\). For \(a \ge g(x)\), you need \(a \ge \max(g(x))\).

Step 1: Understanding the Concept:

This problem involves evaluating a definite integral with a parameter \(p\). A useful technique for such integrals is to establish a recurrence relation by relating the integral for \(p\) to the integral for \(p-1\).


Step 2: Key Formula or Approach:

Let \(I_p = \int_{0}^{\pi} \frac{\sin 2px}{\sin x} dx\).

We will use the trigonometric identity: \(\sin A - \sin B = 2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)\).

Consider the difference \(I_p - I_{p-1}\).


Step 3: Detailed Explanation:

Let \(I_p = \int_{0}^{\pi} \frac{\sin 2px}{\sin x} dx\).
Now, consider the difference between consecutive terms: \[ I_p - I_{p-1} = \int_{0}^{\pi} \frac{\sin 2px}{\sin x} dx - \int_{0}^{\pi} \frac{\sin 2(p-1)x}{\sin x} dx \] \[ I_p - I_{p-1} = \int_{0}^{\pi} \frac{\sin(2px) - \sin(2px-2x)}{\sin x} dx \]
Using the identity \(\sin A - \sin B = 2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)\) with \(A=2px\) and \(B=2px-2x\): \[ \sin(2px) - \sin(2px-2x) = 2\cos\left(\frac{2px + 2px - 2x}{2}\right)\sin\left(\frac{2px - (2px-2x)}{2}\right) \] \[ = 2\cos\left(\frac{4px - 2x}{2}\right)\sin\left(\frac{2x}{2}\right) \] \[ = 2\cos((2p-1)x)\sin(x) \]
Substitute this back into the integral for the difference: \[ I_p - I_{p-1} = \int_{0}^{\pi} \frac{2\cos((2p-1)x)\sin x}{\sin x} dx \] \[ I_p - I_{p-1} = 2 \int_{0}^{\pi} \cos((2p-1)x) dx \]
Now, evaluate this integral: \[ = 2 \left[ \frac{\sin((2p-1)x)}{2p-1} \right]_{0}^{\pi} \] \[ = \frac{2}{2p-1} [\sin((2p-1)\pi) - \sin(0)] \]
Since \(p \in N\), \(2p-1\) is an integer. The sine of any integer multiple of \(\pi\) is 0. \[ = \frac{2}{2p-1} [0 - 0] = 0 \]
So, we have established that \(I_p - I_{p-1} = 0\), which means \(I_p = I_{p-1}\).

This implies that the value of the integral is the same for all natural numbers \(p\). \[ I_p = I_{p-1} = I_{p-2} = \dots = I_1 \]
We can find the value by calculating \(I_1\): \[ I_1 = \int_{0}^{\pi} \frac{\sin(2 \cdot 1 \cdot x)}{\sin x} dx = \int_{0}^{\pi} \frac{\sin 2x}{\sin x} dx \]
Using the double angle identity \(\sin 2x = 2\sin x \cos x\): \[ I_1 = \int_{0}^{\pi} \frac{2\sin x \cos x}{\sin x} dx = \int_{0}^{\pi} 2\cos x dx \] \[ = [2\sin x]_{0}^{\pi} = 2\sin(\pi) - 2\sin(0) = 2(0) - 2(0) = 0 \]
Since \(I_p = I_1\), we have \(I_p = 0\).


Step 4: Final Answer:

The value of the integral is 0.
Quick Tip: When an integral contains a parameter \(n\) or \(p\), look for a way to create a recurrence relation. This often involves considering \(I_n - I_{n-1}\) or a similar expression and using trigonometric identities or integration by parts to simplify it.

Step 1: Understanding the Concept:

We are given an equation that implicitly defines \(y\) as a function of \(x\). We need to find the derivative \(\frac{dy}{dx}\). The easiest way to handle an exponent that involves variables is to use logarithms to simplify the expression before differentiating.


Step 2: Key Formula or Approach:

1. Take the natural logarithm of both sides of the equation to eliminate the exponential function.
2. Rearrange the resulting equation.
3. Use implicit differentiation with respect to \(x\).
4. Solve for \(\frac{dy}{dx}\).


Step 3: Detailed Explanation:

The given equation is: \[ x = e^{\frac{x}{y}} \]
Taking the natural logarithm (\(\log\)) on both sides: \[ \log x = \log(e^{\frac{x}{y}}) \]
Using the property \(\log(e^a) = a\): \[ \log x = \frac{x}{y} \]
To make differentiation easier, we can rearrange this equation by multiplying both sides by \(y\): \[ y \log x = x \]
Now, we differentiate both sides of this equation with respect to \(x\). We use the product rule on the left side, \( \frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx} \). \[ \frac{d}{dx}(y \log x) = \frac{d}{dx}(x) \] \[ y \cdot \frac{d}{dx}(\log x) + \log x \cdot \frac{d}{dx}(y) = 1 \] \[ y \cdot \left(\frac{1}{x}\right) + \log x \cdot \frac{dy}{dx} = 1 \] \[ \frac{y}{x} + \log x \frac{dy}{dx} = 1 \]
Now, we isolate the term with \(\frac{dy}{dx}\): \[ \log x \frac{dy}{dx} = 1 - \frac{y}{x} \] \[ \log x \frac{dy}{dx} = \frac{x - y}{x} \]
Finally, divide by \(\log x\) to solve for \(\frac{dy}{dx}\): \[ \frac{dy}{dx} = \frac{x-y}{x \log x} \]
This is the required expression. Hence proved.


Step 4: Final Answer:

By using logarithmic and implicit differentiation, we have proved that \(\frac{dy}{dx} = \frac{x-y}{x \log x}\).
Quick Tip: Logarithmic differentiation is a powerful tool. Use it whenever you see a function of the form \(f(x)^{g(x)}\), a complex product/quotient, or variables in an exponent as in this problem. It often simplifies the expression significantly before you need to differentiate.

Step 1: Understanding the Concept:

For a function to be differentiable at a point \(x=a\), it must first be continuous at that point. If it is continuous, its Left-Hand Derivative (LHD) and Right-Hand Derivative (RHD) at \(x=a\) must exist and be equal. A crucial prerequisite for differentiability is continuity.


Step 2: Key Formula or Approach:

1. Check for continuity at \(x=-2\) by comparing the Left-Hand Limit (LHL), Right-Hand Limit (RHL), and the value of the function at the point.

LHL = \(\lim_{h \to 0^+} f(a-h)\)
RHL = \(\lim_{h \to 0^+} f(a+h)\)
\(f(a)\)

For continuity, LHL = RHL = \(f(a)\).
2. If the function is not continuous, it is not differentiable. No further checks are needed.


Step 3: Detailed Explanation:

The function is defined as: \[ f(x) = \begin{cases} 2x-3 & , -3 \le x \le -2
x+1 & , -2 < x \le 0 \end{cases} \]
We need to check differentiability at \(x=-2\). First, let's check for continuity at \(x=-2\).


Checking Continuity at x = -2:

Left-Hand Limit (LHL):
For \(x \to -2^-\), we use the definition \(f(x) = 2x-3\).
\[ LHL = \lim_{x \to -2^-} f(x) = \lim_{x \to -2^-} (2x-3) = 2(-2) - 3 = -4 - 3 = -7 \]
Right-Hand Limit (RHL):
For \(x \to -2^+\), we use the definition \(f(x) = x+1\).
\[ RHL = \lim_{x \to -2^+} f(x) = \lim_{x \to -2^+} (x+1) = -2 + 1 = -1 \]

Since the Left-Hand Limit (\(-7\)) is not equal to the Right-Hand Limit (\(-1\)), the limit \(\lim_{x \to -2} f(x)\) does not exist.

Therefore, the function \(f(x)\) is not continuous at \(x=-2\).


Conclusion on Differentiability:
A fundamental theorem in calculus states that if a function is differentiable at a point, it must be continuous at that point. The contrapositive of this theorem is that if a function is not continuous at a point, it cannot be differentiable at that point.

Since \(f(x)\) is not continuous at \(x=-2\), it is not differentiable at \(x=-2\).


Step 4: Final Answer:

The function \(f(x)\) is not differentiable at \(x = -2\) because it is not continuous at \(x = -2\).
Quick Tip: Always check for continuity before checking for differentiability for piecewise functions. If the function is discontinuous at the point in question, you can immediately conclude it's not differentiable and save the time you would have spent calculating the LHD and RHD.

Step 1: Understanding the Concept:

To express a vector \(\vec{r}\) as a linear combination of two other non-collinear vectors \(\vec{p}\) and \(\vec{q}\), all three vectors must be coplanar. If they are coplanar, we can set up a vector equation and solve for the scalar coefficients \(\lambda\) and \(\mu\) by equating the components (\(\hat{i}, \hat{j}, \hat{k}\)) on both sides. This results in a system of linear equations.


Step 2: Key Formula or Approach:

1. Set up the vector equation: \(\vec{r} = \lambda\vec{p} + \mu\vec{q}\).
2. Substitute the component forms of the vectors.
3. Group terms by \(\hat{i}, \hat{j},\) and \(\hat{k}\).
4. Equate the corresponding components to form a system of three linear equations in two variables, \(\lambda\) and \(\mu\).
5. Solve any two of the equations to find \(\lambda\) and \(\mu\).
6. Verify the solution using the third equation.


Step 3: Detailed Explanation:

We are given the vectors: \(\vec{p} = 2\hat{i} - 3\hat{j} - \hat{k}\)
\(\vec{q} = -3\hat{i} + 4\hat{j} + \hat{k}\)
\(\vec{r} = \hat{i} + \hat{j} + 2\hat{k}\)


Set up the equation \(\vec{r} = \lambda\vec{p} + \mu\vec{q}\): \[ \hat{i} + \hat{j} + 2\hat{k} = \lambda(2\hat{i} - 3\hat{j} - \hat{k}) + \mu(-3\hat{i} + 4\hat{j} + \hat{k}) \]
Distribute the scalars \(\lambda\) and \(\mu\): \[ \hat{i} + \hat{j} + 2\hat{k} = (2\lambda\hat{i} - 3\lambda\hat{j} - \lambda\hat{k}) + (-3\mu\hat{i} + 4\mu\hat{j} + \mu\hat{k}) \]
Group the components: \[ 1\hat{i} + 1\hat{j} + 2\hat{k} = (2\lambda - 3\mu)\hat{i} + (-3\lambda + 4\mu)\hat{j} + (-\lambda + \mu)\hat{k} \]
Equating the coefficients of \(\hat{i}, \hat{j},\) and \(\hat{k}\) on both sides, we get a system of equations:
\begin{align
2\lambda - 3\mu &= 1 \quad &(1)

-3\lambda + 4\mu &= 1 \quad &(2)

-\lambda + \mu &= 2 \quad &(3)
\end{align
Let's solve equations (1) and (3). From equation (3), we can express \(\mu\) in terms of \(\lambda\): \[ \mu = 2 + \lambda \]
Substitute this expression for \(\mu\) into equation (1): \[ 2\lambda - 3(2 + \lambda) = 1 \] \[ 2\lambda - 6 - 3\lambda = 1 \] \[ -\lambda - 6 = 1 \] \[ -\lambda = 7 \] \[ \lambda = -7 \]
Now find \(\mu\) using \(\mu = 2 + \lambda\): \[ \mu = 2 + (-7) = -5 \]
So, our potential solution is \(\lambda = -7\) and \(\mu = -5\).


Now we must verify this solution with the remaining equation (2): \[ -3\lambda + 4\mu = 1 \] \[ -3(-7) + 4(-5) = 21 - 20 = 1 \]
Since \(1 = 1\), the solution is correct.


Step 4: Final Answer:

The vector \(\vec{r}\) can be expressed as \(\vec{r} = -7\vec{p} - 5\vec{q}\).

The values of the scalars are \(\lambda = -7\) and \(\mu = -5\).
Quick Tip: When you get a system of three equations with two unknowns, it's called an overdetermined system. It will only have a solution if the equations are consistent (which implies the vectors are coplanar). Always use two equations to solve for the variables and then substitute the values into the third equation to check for consistency.

Step 1: Understanding the Concept:

If a vector makes angles \(\alpha, \beta, \gamma\) with the positive x, y, and z axes respectively, then its direction cosines are \(l = \cos\alpha\), \(m = \cos\beta\), and \(n = \cos\gamma\). These are related by the identity \(l^2 + m^2 + n^2 = 1\). A vector \(\vec{a}\) can be expressed using its magnitude and direction cosines as \(\vec{a} = |\vec{a}|(l\hat{i} + m\hat{j} + n\hat{k})\).


Step 2: Key Formula or Approach:

1. Use the given condition that the angles are equal, i.e., \(\alpha = \beta = \gamma\), which implies \(l=m=n\).
2. Substitute this into the identity \(l^2+m^2+n^2=1\) to find the values of the direction cosines.
3. Use the formula \(\vec{a} = |\vec{a}|(l\hat{i} + m\hat{j} + n\hat{k})\) to find the vector.


Step 3: Detailed Explanation:

Let the vector \(\vec{a}\) make an angle \(\theta\) with each of the three axes. So, \(\alpha = \beta = \gamma = \theta\).

The direction cosines are: \[ l = \cos\theta, \quad m = \cos\theta, \quad n = \cos\theta \]
Using the property of direction cosines: \[ l^2 + m^2 + n^2 = 1 \] \[ \cos^2\theta + \cos^2\theta + \cos^2\theta = 1 \] \[ 3\cos^2\theta = 1 \] \[ \cos^2\theta = \frac{1}{3} \] \[ \cos\theta = \pm\frac{1}{\sqrt{3}} \]
So, the direction cosines are \(l = m = n = \pm\frac{1}{\sqrt{3}}\).

The magnitude of the vector is given as \(|\vec{a}| = 5\sqrt{3}\).

The vector \(\vec{a}\) is given by \(\vec{a} = |\vec{a}|(l\hat{i} + m\hat{j} + n\hat{k})\).

Substituting the values: \[ \vec{a} = 5\sqrt{3} \left( \pm\frac{1}{\sqrt{3}}\hat{i} + \pm\frac{1}{\sqrt{3}}\hat{j} + \pm\frac{1}{\sqrt{3}}\hat{k} \right) \] \[ \vec{a} = \pm 5\sqrt{3} \frac{1}{\sqrt{3}} (\hat{i} + \hat{j} + \hat{k}) \] \[ \vec{a} = \pm 5(\hat{i} + \hat{j} + \hat{k}) \]
This can be written as \(5\hat{i} + 5\hat{j} + 5\hat{k}\) or \(-5\hat{i} - 5\hat{j} - 5\hat{k}\).


Step 4: Final Answer:

The required vector \(\vec{a}\) is \(\pm(5\hat{i} + 5\hat{j} + 5\hat{k})\).
Quick Tip: A vector equally inclined to the coordinate axes will always be of the form \(k(\hat{i} + \hat{j} + \hat{k})\). The value of \(k\) can be found using the magnitude of the vector. The magnitude of \(\hat{i} + \hat{j} + \hat{k}\) is \(\sqrt{1^2+1^2+1^2} = \sqrt{3}\). So, the unit vector is \(\frac{1}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k})\).

Step 1: Understanding the Concept:

The phrase "in QP produced" means that we extend the line segment from Q through P. So the order of the collinear points is Q, P, R. This means that P divides the line segment QR internally. We can use the section formula for position vectors to find the position vector of R.


Step 2: Key Formula or Approach:

1. Let the position vectors of P, Q, and R be \(\vec{p} = \vec{\alpha}\), \(\vec{q} = \vec{\beta}\), and \(\vec{r}\) respectively.
2. Use the given vector relation \(\vec{QR} = \frac{3}{2} \vec{QP}\) to establish a relationship between the position vectors.
3. Solve the resulting equation for \(\vec{r}\).

Alternatively, determine the ratio in which P divides QR and use the section formula.


Step 3: Detailed Explanation:

Let O be the origin. We are given: \(\vec{OP} = \vec{\alpha}\) \(\vec{OQ} = \vec{\beta}\)
Let the position vector of R be \(\vec{OR} = \vec{r}\).


The vectors \(\vec{QR}\) and \(\vec{QP}\) can be expressed in terms of position vectors: \(\vec{QR} = \vec{OR} - \vec{OQ} = \vec{r} - \vec{\beta}\)
\(\vec{QP} = \vec{OP} - \vec{OQ} = \vec{\alpha} - \vec{\beta}\)


We are given the condition \(QR = \frac{3}{2} QP\). Since R is on the line QP produced, the vectors \(\vec{QR}\) and \(\vec{QP}\) are in the same direction. So we can write the vector equation: \[ \vec{QR} = \frac{3}{2} \vec{QP} \]
Substitute the expressions in terms of position vectors: \[ \vec{r} - \vec{\beta} = \frac{3}{2} (\vec{\alpha} - \vec{\beta}) \]
Now, solve for \(\vec{r}\): \[ \vec{r} = \vec{\beta} + \frac{3}{2} (\vec{\alpha} - \vec{\beta}) \] \[ \vec{r} = \vec{\beta} + \frac{3}{2}\vec{\alpha} - \frac{3}{2}\vec{\beta} \] \[ \vec{r} = \frac{3}{2}\vec{\alpha} + \left(1 - \frac{3}{2}\right)\vec{\beta} \] \[ \vec{r} = \frac{3}{2}\vec{\alpha} - \frac{1}{2}\vec{\beta} \] \[ \vec{r} = \frac{3\vec{\alpha} - \vec{\beta}}{2} \]

Alternative Method (Section Formula):

The order of points is Q, P, R. So P divides QR internally.
We have \(QR = \frac{3}{2} QP\).
We can write \(QR = QP + PR\).
So, \(QP + PR = \frac{3}{2} QP\). \[ PR = \frac{3}{2} QP - QP = \frac{1}{2} QP \]
This gives the ratio of the lengths \(QP : PR = QP : \frac{1}{2} QP = 2 : 1\).
So, P divides the segment QR in the ratio 2:1.
Using the section formula for internal division: \[ \vec{OP} = \frac{2 \cdot \vec{OR} + 1 \cdot \vec{OQ}}{2+1} \] \[ \vec{\alpha} = \frac{2\vec{r} + \vec{\beta}}{3} \]
Now, solve for \(\vec{r}\): \[ 3\vec{\alpha} = 2\vec{r} + \vec{\beta} \] \[ 2\vec{r} = 3\vec{\alpha} - \vec{\beta} \] \[ \vec{r} = \frac{3\vec{\alpha} - \vec{\beta}}{2} \]
Both methods yield the same result.


Step 4: Final Answer:

The position vector of point R is \(\frac{3\vec{\alpha} - \vec{\beta}}{2}\).
Quick Tip: Carefully interpret phrases like "in AB produced" or "in BA produced". "AB produced" means the order of points is A, B, C. "BA produced" means the order is B, A, C. A diagram can be very helpful to correctly apply vector addition or the section formula.

Step 1: Understanding the Concept:

This problem involves second-order derivatives and requires proving a given differential equation. The best strategy is often to simplify the original function, find the first derivative, manipulate the resulting equation to avoid complex quotients, and then differentiate again.

Note: There appears to be a typo in the question's target equation. We will derive the correct differential equation satisfied by the function.


Step 2: Key Formula or Approach:

1. Simplify the expression for \(y\).
2. Differentiate \(y\) to find \(y_1 = \frac{dy{dx}\).
3. Rearrange the expression for \(y_1\) to form an equation without fractions, e.g., \(g(x) y_1 = h(x)\).
4. Differentiate this equation again using the product rule to find a relationship involving \(y_2\).


Step 3: Detailed Explanation:

First, simplify the function \(y\). \[ \left(\sqrt{x} + \frac{1}{\sqrt{x}}\right)^2 = (\sqrt{x})^2 + 2(\sqrt{x})\left(\frac{1}{\sqrt{x}}\right) + \left(\frac{1}{\sqrt{x}}\right)^2 = x + 2 + \frac{1}{x} \] \[ = \frac{x^2 + 2x + 1}{x} = \frac{(x+1)^2}{x} \]
So, the function becomes: \[ y = \log\left(\frac{(x+1)^2}{x}\right) \]
Using properties of logarithms: \[ y = \log((x+1)^2) - \log(x) \] \[ y = 2\log(x+1) - \log(x) \]
Now, differentiate with respect to \(x\) to find \(y_1\): \[ y_1 = \frac{dy}{dx} = 2 \cdot \frac{1}{x+1} - \frac{1}{x} \] \[ y_1 = \frac{2x - (x+1)}{x(x+1)} = \frac{x-1}{x(x+1)} \]
To avoid using the quotient rule for the second derivative, we can rearrange this equation: \[ x(x+1)y_1 = x-1 \] \[ (x^2+x)y_1 = x-1 \]
Now, differentiate this equation with respect to \(x\) using the product rule on the left side: \[ \frac{d}{dx}((x^2+x)y_1) = \frac{d}{dx}(x-1) \] \[ \frac{d}{dx}(x^2+x) \cdot y_1 + (x^2+x) \cdot \frac{d}{dx}(y_1) = 1 \] \[ (2x+1)y_1 + (x^2+x)y_2 = 1 \] \[ x(x+1)y_2 + (2x+1)y_1 = 1 \]
This is the correct differential equation satisfied by the function. The equation given in the question, \(x(x+1)^2 y_2 + (x+1)^2 y_1 = 2\), seems to be incorrect.


Step 4: Final Answer:

The derivation shows that \(y\) satisfies the differential equation \(x(x+1)y_2 + (2x+1)y_1 = 1\). The equation stated in the problem appears to contain a misprint.
Quick Tip: For "show that" problems involving second derivatives, after finding the first derivative \(y_1\), it's almost always easier to cross-multiply to eliminate denominators and then differentiate the resulting expression implicitly. This avoids complex quotient rule calculations for \(y_2\).

Step 1: Understanding the Concept:

This problem requires finding the derivative of an implicit function. While implicit differentiation can be used directly, it often leads to a complex expression. A better approach is to first simplify the given relation algebraically to express \(y\) as an explicit function of \(x\), and then differentiate.


Step 2: Key Formula or Approach:

1. Isolate one term of the equation.
2. Square both sides to remove the square roots.
3. Rearrange and factor the resulting polynomial equation.
4. Use the condition \(x \ne y\) to simplify the equation.
5. Solve for \(y\) explicitly in terms of \(x\).
6. Differentiate the explicit function \(y(x)\) using the quotient rule.


Step 3: Detailed Explanation:

We are given the equation: \[ x\sqrt{1+y} + y\sqrt{1+x} = 0 \]
Rearrange the equation: \[ x\sqrt{1+y} = -y\sqrt{1+x} \]
Square both sides: \[ (x\sqrt{1+y})^2 = (-y\sqrt{1+x})^2 \] \[ x^2(1+y) = y^2(1+x) \]
Expand both sides: \[ x^2 + x^2y = y^2 + y^2x \]
Move all terms to one side and group them: \[ (x^2 - y^2) + (x^2y - y^2x) = 0 \]
Factor each group: \[ (x-y)(x+y) + xy(x-y) = 0 \]
Factor out the common term \((x-y)\): \[ (x-y)[(x+y) + xy] = 0 \]
Since we are given that \(x \ne y\), the term \((x-y)\) cannot be zero. Therefore, the other factor must be zero: \[ (x+y) + xy = 0 \]
Now, we can solve for \(y\) explicitly: \[ y + xy = -x \] \[ y(1+x) = -x \] \[ y = -\frac{x}{1+x} \]
Now, differentiate \(y\) with respect to \(x\) using the quotient rule, \(\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}\): \[ \frac{dy}{dx} = -\left[ \frac{(1)(1+x) - (x)(1)}{(1+x)^2} \right] \] \[ \frac{dy}{dx} = -\left[ \frac{1+x - x}{(1+x)^2} \right] \] \[ \frac{dy}{dx} = -\left[ \frac{1}{(1+x)^2} \right] \] \[ \frac{dy}{dx} = \frac{-1}{(1+x)^2} \]
Hence proved.


Step 4: Final Answer:

By algebraic manipulation and differentiation, we have proved that \(\frac{dy}{dx} = \frac{-1}{(1+x)^2}\).
Quick Tip: When dealing with implicit equations involving square roots, try to simplify the algebra first by squaring. This often transforms the problem into a simpler polynomial relationship from which you can find the derivative more easily, either implicitly or by solving for the function explicitly.

Step 1: Understanding the Concept:


One-one (Injective): A function \(f\) is one-one if every distinct element in the domain maps to a distinct element in the codomain. Mathematically, \(f(x_1) = f(x_2) \implies x_1 = x_2\).
Onto (Surjective): A function \(f\) is onto if its range is equal to its codomain. This means that for every element \(y\) in the codomain, there exists at least one element \(x\) in the domain such that \(f(x) = y\).

The domain and codomain are both the set of natural numbers, \(N = \{1, 2, 3, \dots\}\).


Step 2: Key Formula or Approach:

1. To prove one-one, assume \(f(x_1) = f(x_2)\) and show that this implies \(x_1 = x_2\).
2. To prove not onto, show that there is at least one element in the codomain \(N\) that is not the image of any element in the domain \(N\). This can be done by analyzing the range of the function.


Step 3: Detailed Explanation:

Part 1: Proving the function is one-one.

Let \(x_1, x_2 \in N\) be two elements in the domain such that \(f(x_1) = f(x_2)\). \[ ax_1 + b = ax_2 + b \]
Subtracting \(b\) from both sides: \[ ax_1 = ax_2 \]
Since \(a \in N\), we know \(a \ge 1\), so \(a \ne 0\). We can divide both sides by \(a\): \[ x_1 = x_2 \]
Since \(f(x_1) = f(x_2)\) implies \(x_1 = x_2\), the function \(f(x)\) is one-one.


Part 2: Proving the function is not onto.

For the function to be onto, for any \(y \in N\) (codomain), there must be an \(x \in N\) (domain) such that \(f(x) = y\).
Let's find the range of \(f(x)\). The domain is \(N = \{1, 2, 3, \dots\}\).
The possible values of \(f(x)\) are:

\(f(1) = a(1) + b = a+b\)
\(f(2) = a(2) + b = 2a+b\)
\(f(3) = a(3) + b = 3a+b\)
and so on...

The range of the function is the set \(\{a+b, 2a+b, 3a+b, \dots\}\).

We are given that \(a, b \in N\), which means \(a \ge 1\) and \(b \ge 1\).
The smallest value in the range is \(a+b\). Since \(a \ge 1\) and \(b \ge 1\), the minimum value of \(a+b\) is \(1+1=2\).
So, the range starts from a value greater than or equal to 2.
Consider the natural number \(1\), which is in the codomain \(N\). Does there exist an \(x \in N\) such that \(f(x)=1\)? \[ ax + b = 1 \] \[ ax = 1-b \]
Since \(b \in N\), \(b \ge 1\). This implies \(1-b \le 0\).
Since \(a \in N\), \(a \ge 1\).
So, \(x = \frac{1-b}{a} \le 0\).
An \(x\) value that is less than or equal to zero cannot be a natural number (\(x \notin N\)).
Therefore, the element \(1\) in the codomain has no pre-image in the domain.
This shows that the range is not equal to the codomain (\(Range \ne N\)). Hence, the function is not onto.


Step 4: Final Answer:

The function \(f(x) = ax+b\) is one-one because \(f(x_1) = f(x_2)\) leads to \(x_1 = x_2\). It is not onto because elements in the codomain N, such as 1, 2, ..., (a+b-1), have no pre-image in the domain N.
Quick Tip: To check if a function is onto, a good strategy is to find its range. If the range is a proper subset of the codomain, the function is not onto. For linear functions on \(N\), the range will be an arithmetic progression, which is usually a sparse subset of \(N\).

Step 1: Understanding the Concept:

The feasible region in a Linear Programming Problem (LPP) is the set of points satisfying all the given constraints. The constraints are linear inequalities. The boundary of the feasible region is formed by lines corresponding to these inequalities. We need to find the equations of these boundary lines and determine the correct inequality sign (\(\le\) or \(\ge\)) for each.


Step 2: Key Formula or Approach:

1. Identify the corner points of the convex feasible region.
2. The region is in the first quadrant, which gives the non-negativity constraints: \(x \ge 0, y \ge 0\).
3. For each line segment forming the boundary, find the equation of the line passing through the two corner points that define it. The two-point form of a line is \(y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)\).
4. For each line, determine the inequality by testing a point inside the feasible region (like the origin (0,0) if applicable).


Step 3: Detailed Explanation:

The corner points of the feasible region (the shaded polygon OABCD) are O(0,0), A(0, 30), B(15, 25), C(30, 10), and D(35, 0).


Constraint 1 \& 2 (Non-negativity):

Since the entire feasible region lies in the first quadrant, we have: \[ x \ge 0 \] \[ y \ge 0 \]

Constraint 3 (Line through A and B):

The line passes through A(0, 30) and B(15, 25).
Slope \(m_{AB} = \frac{25 - 30}{15 - 0} = \frac{-5}{15} = -\frac{1}{3}\).
The y-intercept is 30.
Equation of the line is \(y = -\frac{1}{3}x + 30 \implies 3y = -x + 90 \implies x + 3y = 90\).
To find the inequality, we test the origin (0,0), which is in the feasible region. \(0 + 3(0) = 0\). Since \(0 \le 90\), the constraint is: \[ x + 3y \le 90 \]

Constraint 4 (Line through B and C):

The line passes through B(15, 25) and C(30, 10).
Slope \(m_{BC} = \frac{10 - 25}{30 - 15} = \frac{-15}{15} = -1\).
Equation of the line using point C(30, 10): \(y - 10 = -1(x - 30) \implies y - 10 = -x + 30 \implies x + y = 40\).
Test the origin (0,0): \(0 + 0 = 0\). Since \(0 \le 40\), the constraint is: \[ x + y \le 40 \]

Constraint 5 (Line through C and D):

The line passes through C(30, 10) and D(35, 0).
Slope \(m_{CD} = \frac{0 - 10}{35 - 30} = \frac{-10}{5} = -2\).
Equation of the line using point D(35, 0): \(y - 0 = -2(x - 35) \implies y = -2x + 70 \implies 2x + y = 70\).
Test the origin (0,0): \(2(0) + 0 = 0\). Since \(0 \le 70\), the constraint is: \[ 2x + y \le 70 \]

Step 4: Final Answer:

The constraints for the given linear programming problem are:

\(x \ge 0\)
\(y \ge 0\)
\(x + 3y \le 90\)
\(x + y \le 40\)
\(2x + y \le 70\) Quick Tip: When deriving constraints from a graph, always identify the corner points first. Use pairs of adjacent corner points to find the equation of each boundary line. The origin (0,0) is the easiest point to test for determining the inequality direction, as long as the boundary line does not pass through it.

Step 1: Understanding the Concept:

The given differential equation is a first-order differential equation. We can check if it's separable, meaning we can rearrange it so that all terms involving \(y\) are on one side with \(dy\) and all terms involving \(x\) are on the other side with \(dx\). Once separated, we can integrate both sides to find the general solution and then use the initial condition to find the particular solution.


Step 2: Key Formula or Approach:

1. Rearrange the equation to the form \(g(y)dy = f(x)dx\).

2. Integrate both sides: \(\int g(y)dy = \int f(x)dx + C\).

3. Use the given initial condition \(y(x_0) = y_0\) to find the value of the constant of integration C.

4. Substitute C back into the general solution to get the particular solution.


Step 3: Detailed Explanation:

The given differential equation is: \[ 2(y + 3) - xy \frac{dy}{dx} = 0 \]
Rearrange to separate the variables: \[ 2(y + 3) = xy \frac{dy}{dx} \]
Divide by \(x(y+3)\) and multiply by \(dx\). Note that we assume \(x \ne 0\) and \(y \ne -3\). \[ \frac{2}{x} dx = \frac{y}{y+3} dy \]
Now, we integrate both sides. The right-hand side can be simplified by algebraic manipulation: \[ \frac{y}{y+3} = \frac{(y+3) - 3}{y+3} = 1 - \frac{3}{y+3} \]
So the equation becomes: \[ \int \frac{2}{x} dx = \int \left(1 - \frac{3}{y+3}\right) dy \]
Integrating both sides: \[ 2 \ln|x| = y - 3 \ln|y+3| + C \]
This is the general solution. Now, we use the initial condition \(y(1) = -2\) to find C.
Substitute \(x=1\) and \(y=-2\): \[ 2 \ln|1| = -2 - 3 \ln|-2+3| + C \] \[ 2(0) = -2 - 3 \ln|1| + C \] \[ 0 = -2 - 3(0) + C \] \[ C = 2 \]
Substitute the value of C back into the general solution to get the particular solution: \[ 2 \ln|x| = y - 3 \ln|y+3| + 2 \]

Step 4: Final Answer:

The particular solution to the differential equation is \(2 \ln|x| = y - 3 \ln|y+3| + 2\).
Quick Tip: When integrating a rational function like \(\frac{y}{y+3}\), if the degree of the numerator is equal to or greater than the degree of the denominator, perform polynomial division or algebraic manipulation (as shown) to simplify the integrand first.

Step 1: Understanding the Concept:

The given differential equation is a first-order linear differential equation. It can be written in the standard form \(\frac{dy}{dx} + P(x)y = Q(x)\). Such equations can be solved by finding an integrating factor (I.F.).


Step 2: Key Formula or Approach:

1. Write the equation in the standard form \(\frac{dy}{dx} + P(x)y = Q(x)\).

2. Calculate the integrating factor: I.F. = \(e^{\int P(x)dx}\).

3. The solution is given by the formula: \(y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C\).


Step 3: Detailed Explanation:

The given equation is: \[ (1 + x^2)\frac{dy}{dx} + 2xy = 4x^2 \]
First, divide the entire equation by \((1 + x^2)\) to bring it to the standard form: \[ \frac{dy}{dx} + \frac{2x}{1+x^2}y = \frac{4x^2}{1+x^2} \]
This is a linear differential equation with: \[ P(x) = \frac{2x}{1+x^2} \] \[ Q(x) = \frac{4x^2}{1+x^2} \]
Next, we find the integrating factor (I.F.): \[ I.F. = e^{\int P(x)dx} = e^{\int \frac{2x}{1+x^2}dx} \]
To evaluate the integral, let \(u = 1+x^2\), so \(du = 2x dx\). \[ \int \frac{2x}{1+x^2}dx = \int \frac{1}{u}du = \ln|u| = \ln(1+x^2) \quad (since 1+x^2 > 0) \] \[ I.F. = e^{\ln(1+x^2)} = 1+x^2 \]
Now, we use the solution formula: \[ y \cdot (1+x^2) = \int \frac{4x^2}{1+x^2} \cdot (1+x^2) dx + C \] \[ y(1+x^2) = \int 4x^2 dx + C \]
Integrate the right side: \[ y(1+x^2) = 4 \frac{x^3}{3} + C \]

Step 4: Final Answer:

The general solution to the differential equation is \(y(1+x^2) = \frac{4x^3}{3} + C\).
Quick Tip: Recognizing the form of a differential equation is the most crucial step. Notice that the term \(2xy\) is related to the derivative of \(y(1+x^2)\). The left side of the original equation, \((1 + x^2)\frac{dy}{dx} + 2xy\), is exactly the result of the product rule for \(\frac{d}{dx}(y(1+x^2))\). This provides a quick way to see the integrating factor.

Step 1: Understanding the Concept:

This problem describes a sequence of Bernoulli trials, which can be modeled by a binomial distribution. A Bernoulli trial is an experiment with exactly two outcomes: "success" and "failure". Here, a "success" is getting the number 2, and "failure" is getting any other number. The die is thrown a fixed number of times (\(n=2\)), and the trials are independent. The mean of a binomial distribution is given by the product of the number of trials and the probability of success.


Step 2: Key Formula or Approach:

For a binomial distribution:

Let \(n\) be the number of trials.
Let \(p\) be the probability of success in a single trial.
The mean (or expected value) of the number of successes, \(E(X)\), is given by the formula: \(E(X) = np\).


Step 3: Detailed Explanation:

Let X be the random variable representing the number of times the number 2 appears when the die is thrown twice.
The trials are independent, and the probability of success is constant for each trial. This is a binomial experiment.

The number of trials is \(n = 2\).
A "success" is defined as the event of getting the number 2 on the die.
The probability of success in a single trial is given as \(p = P(2) = \frac{3}{10}\).

We need to find the mean of the number of times number 2 appears, which is the mean of the random variable X.
Using the formula for the mean of a binomial distribution: \[ E(X) = np \]
Substitute the values of \(n\) and \(p\): \[ E(X) = 2 \times \frac{3}{10} \] \[ E(X) = \frac{6}{10} = \frac{3}{5} \]
The information about the probabilities of other numbers being equal is not needed to calculate the mean.


Step 4: Final Answer:

The mean of the number of times number 2 appears is \(\frac{3}{5}\) or 0.6.
Quick Tip: In probability questions, identify all the information given, but also recognize what is essential. Here, the key was to identify the problem as a binomial distribution and extract the necessary parameters, \(n\) and \(p\). The probabilities of the other outcomes were extraneous for finding the mean.

Step 1: Understanding the Concept:


Mutually Exclusive Events: Two events are mutually exclusive if they cannot occur at the same time. Mathematically, their intersection is empty: \(A \cap B = \emptyset\), which means \(P(A \cap B) = 0\).
Independent Events: Two events are independent if the occurrence of one does not affect the probability of the other. Mathematically, \(P(A \cap B) = P(A) \times P(B)\).

We need to find the probabilities of events A, B, and their intersection to check these conditions.


Step 2: Key Formula or Approach:

1. Determine the total number of outcomes in the sample space, \(n(S)\).
2. List the outcomes for event A, B, and \(A \cap B\), and find their cardinalities \(n(A), n(B), n(A \cap B)\).
3. Calculate the probabilities: \(P(A) = n(A)/n(S)\), \(P(B) = n(B)/n(S)\), and \(P(A \cap B) = n(A \cap B)/n(S)\).
4. Check the conditions for mutual exclusivity and independence.


Step 3: Detailed Explanation:

The sample space S for throwing two dice consists of \(6 \times 6 = 36\) equally likely outcomes. So, \(n(S) = 36\).


Event A: The sum is 9.
The outcomes for event A are: A = \(\{(3, 6), (4, 5), (5, 4), (6, 3)\}\).
The number of outcomes in A is \(n(A) = 4\).
The probability of A is \(P(A) = \frac{n(A)}{n(S)} = \frac{4}{36} = \frac{1}{9}\).


Event B: The first die is not 3.
The outcomes where the first die is 3 are \(\{(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)\}\). There are 6 such outcomes.
So, the number of outcomes in B is \(n(B) = n(S) - 6 = 36 - 6 = 30\).
The probability of B is \(P(B) = \frac{n(B)}{n(S)} = \frac{30}{36} = \frac{5}{6}\).


Intersection of A and B (\(A \cap B\)): \(A \cap B\) consists of outcomes that are in both A and B. These are outcomes where the sum is 9 AND the first die is not 3.
Looking at the set A = \(\{(3, 6), (4, 5), (5, 4), (6, 3)\}\), we remove the outcomes where the first die is 3.
So, \(A \cap B = \{(4, 5), (5, 4), (6, 3)\}\).
The number of outcomes in \(A \cap B\) is \(n(A \cap B) = 3\).
The probability of the intersection is \(P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{3}{36} = \frac{1}{12}\).


Check for Mutual Exclusivity:
Events are mutually exclusive if \(A \cap B = \emptyset\).
Here, \(A \cap B\) is not empty, so \(P(A \cap B) = \frac{1}{12} \ne 0\).
Therefore, events A and B are not mutually exclusive.


Check for Independence:
Events are independent if \(P(A \cap B) = P(A) \times P(B)\).
Let's check this condition: \[ P(A) \times P(B) = \frac{1}{9} \times \frac{5}{6} = \frac{5}{54} \]
We found that \(P(A \cap B) = \frac{1}{12}\).
Since \(\frac{1}{12} \ne \frac{5}{54}\), the condition for independence is not met.
Therefore, events A and B are not independent.


Step 4: Final Answer:

The events A and B are neither mutually exclusive nor independent.
Quick Tip: For problems with a small, discrete sample space like rolling dice, it's often best to list the actual outcomes for each event. This makes finding the intersection straightforward and reduces the chance of errors in counting.

Step 1: Understanding the Concept:

This problem requires the application of a key property of definite integrals, often called the "King's Property": \(\int_{0}^{a} h(x) dx = \int_{0}^{a} h(a-x) dx\). We will apply this property to the given integral and use the specified properties of functions f and g to simplify the expression.


Step 2: Key Formula or Approach:

1. Let \(I = \int_{0}^{a} f(x) g(x) dx\).
2. Apply the property \(\int_{0}^{a} h(x) dx = \int_{0}^{a} h(a-x) dx\) by substituting \(x\) with \((a-x)\) in the integrand.
3. Use the given relations: \(f(a-x) = f(x)\) and \(g(a-x) = a - g(x)\).
4. Add the original integral expression (from step 1) and the transformed integral expression (from step 3).
5. Solve the resulting equation for \(I\).


Step 3: Detailed Explanation:

Let the given integral be \(I\). \[ I = \int_{0}^{a} f(x) g(x) dx \quad \dots (1) \]
Using the property \(\int_{0}^{a} h(x) dx = \int_{0}^{a} h(a-x) dx\), we replace \(x\) with \((a-x)\): \[ I = \int_{0}^{a} f(a-x) g(a-x) dx \]
Now, we use the given properties of the functions \(f\) and \(g\):

\(f(a-x) = f(x)\)
\(g(x) + g(a-x) = a \implies g(a-x) = a - g(x)\)

Substitute these into the expression for \(I\): \[ I = \int_{0}^{a} f(x) (a - g(x)) dx \]
Distribute \(f(x)\) inside the integral: \[ I = \int_{0}^{a} [a f(x) - f(x)g(x)] dx \]
Split the integral into two parts: \[ I = \int_{0}^{a} a f(x) dx - \int_{0}^{a} f(x)g(x) dx \quad \dots (2) \]
We can see that the second term on the right-hand side is the original integral \(I\). \[ I = a \int_{0}^{a} f(x) dx - I \]
Now, add \(I\) to both sides of the equation: \[ 2I = a \int_{0}^{a} f(x) dx \]
Finally, divide by 2 to solve for \(I\): \[ I = \frac{a}{2} \int_{0}^{a} f(x) dx \]
Substituting back the original expression for \(I\): \[ \int_{0}^{a} f(x) g(x) dx = \frac{a}{2} \int_{0}^{a} f(x) dx \]
Hence proved.


Step 4: Final Answer:

By applying the properties of definite integrals and the given functional relations, we have shown that \(\int_{0}^{a} f(x) g(x) dx = \frac{a}{2} \int_{0}^{a} f(x) dx\).
Quick Tip: This "King's property" is one of the most powerful tools for definite integrals. When you see an integral from 0 to 'a', and the integrand involves functions with arguments like \((a-x)\), a good first step is to apply this property. The strategy of adding the original integral \(I\) to the transformed integral \(I\) often leads to a significant simplification.

Step 1: Understanding the Concept:

The shortest distance between two skew lines, given by \(\vec{r} = \vec{a}_1 + \lambda\vec{b}_1\) and \(\vec{r} = \vec{a}_2 + \mu\vec{b}_2\), is the length of the line segment that is perpendicular to both lines. It can be calculated using a standard vector formula.


Step 2: Key Formula or Approach:

The shortest distance \(d\) between two skew lines is given by the formula: \[ d = \left| \frac{(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)}{|\vec{b}_1 \times \vec{b}_2|} \right| \]
First, we need to extract the vectors \(\vec{a}_1, \vec{a}_2, \vec{b}_1,\) and \(\vec{b}_2\) from the Cartesian equations of the lines.


Step 3: Detailed Explanation:

The equations of the lines are:
Line 1: \(\frac{x - (-1)}{2} = \frac{y-1}{1} = \frac{z-9}{-3}\)
Line 2: \(\frac{x-3}{2} = \frac{y - (-15)}{-7} = \frac{z-9}{5}\)


From these equations, we can identify the position vectors of points on the lines and the direction vectors:
For Line 1: \(\vec{a}_1 = -\hat{i} + \hat{j} + 9\hat{k}\) \(\vec{b}_1 = 2\hat{i} + \hat{j} - 3\hat{k}\)


For Line 2: \(\vec{a}_2 = 3\hat{i} - 15\hat{j} + 9\hat{k}\) \(\vec{b}_2 = 2\hat{i} - 7\hat{j} + 5\hat{k}\)


Now, we calculate the required vector components for the formula:
1. \(\vec{a}_2 - \vec{a}_1\): \[ \vec{a}_2 - \vec{a}_1 = (3 - (-1))\hat{i} + (-15 - 1)\hat{j} + (9 - 9)\hat{k} = 4\hat{i} - 16\hat{j} + 0\hat{k} \]
2. \(\vec{b}_1 \times \vec{b}_2\): \[ \vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
2 & 1 & -3
2 & -7 & 5 \end{vmatrix} = \hat{i}(5 - 21) - \hat{j}(10 - (-6)) + \hat{k}(-14 - 2) \] \[ = -16\hat{i} - 16\hat{j} - 16\hat{k} \]
3. \((\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)\): \[ = (4\hat{i} - 16\hat{j}) \cdot (-16\hat{i} - 16\hat{j} - 16\hat{k}) = (4)(-16) + (-16)(-16) + (0)(-16) \] \[ = -64 + 256 = 192 \]
4. \(|\vec{b}_1 \times \vec{b}_2|\): \[ = \sqrt{(-16)^2 + (-16)^2 + (-16)^2} = \sqrt{256 + 256 + 256} = \sqrt{3 \times 256} = 16\sqrt{3} \]

Finally, substitute these values into the shortest distance formula: \[ d = \left| \frac{192}{16\sqrt{3}} \right| = \frac{12}{\sqrt{3}} = \frac{12\sqrt{3}}{3} = 4\sqrt{3} \]

Step 4: Final Answer:

The shortest distance between the two lines is \(4\sqrt{3}\) units.
Quick Tip: Before applying the skew lines formula, quickly check if the lines are parallel. If their direction vectors \(\vec{b}_1\) and \(\vec{b}_2\) are proportional, the lines are parallel, and a different formula for distance is used. Here, \(\vec{b}_1\) and \(\vec{b}_2\) are clearly not proportional, so they are skew or intersecting.

Step 1: Understanding the Concept:

The foot of the perpendicular from a point A to a line \(l\) is the point M on the line such that the vector \(\vec{AM}\) is perpendicular to the line's direction vector. The image A' is a point such that M is the midpoint of the line segment AA'.


Step 2: Key Formula or Approach:

1. Find the coordinates of a general point M on the line in terms of \(\lambda\).
2. Determine the direction vector \(\vec{AM}\).
3. Use the perpendicularity condition \(\vec{AM} \cdot \vec{b} = 0\), where \(\vec{b}\) is the direction vector of the line, to solve for \(\lambda\).
4. Substitute \(\lambda\) back to find the coordinates of M (the foot of the perpendicular).
5. Use the midpoint formula \(M = \frac{A+A'}{2}\) to find the coordinates of the image A'.
6. Find the equation of the line AA' using points A and A'.


Step 3: Detailed Explanation:

The point is A(2, 1, 2). The line \(l\) is \(\vec{r} = (4+\lambda)\hat{i} + (2-\lambda)\hat{j} + (2-\lambda)\hat{k}\).

The direction vector of the line is \(\vec{b} = \hat{i} - \hat{j} - \hat{k}\).


1. Find the foot of the perpendicular (M):

Let M be a general point on the line \(l\). Its coordinates are M\((4+\lambda, 2-\lambda, 2-\lambda)\).

The direction vector \(\vec{AM}\) is: \[ \vec{AM} = \vec{OM} - \vec{OA} = (4+\lambda - 2)\hat{i} + (2-\lambda - 1)\hat{j} + (2-\lambda - 2)\hat{k} \] \[ \vec{AM} = (2+\lambda)\hat{i} + (1-\lambda)\hat{j} + (-\lambda)\hat{k} \]
Since \(\vec{AM}\) is perpendicular to the line \(l\), \(\vec{AM} \cdot \vec{b} = 0\). \[ ((2+\lambda)\hat{i} + (1-\lambda)\hat{j} - \lambda\hat{k}) \cdot (\hat{i} - \hat{j} - \hat{k}) = 0 \] \[ (2+\lambda)(1) + (1-\lambda)(-1) + (-\lambda)(-1) = 0 \] \[ 2 + \lambda - 1 + \lambda + \lambda = 0 \] \[ 1 + 3\lambda = 0 \implies \lambda = -\frac{1}{3} \]
Now, substitute \(\lambda = -1/3\) into the coordinates of M: \[ M = \left(4-\frac{1}{3}, 2-(-\frac{1}{3}), 2-(-\frac{1}{3})\right) = \left(\frac{11}{3}, \frac{7}{3}, \frac{7}{3}\right) \]
The foot of the perpendicular is M\((\frac{11}{3}, \frac{7}{3}, \frac{7}{3})\).


2. Find the image (A'):

Let the image be A'(\(x', y', z'\)). M is the midpoint of AA'. \[ (\frac{11}{3}, \frac{7}{3}, \frac{7}{3}) = \left(\frac{2+x'}{2}, \frac{1+y'}{2}, \frac{2+z'}{2}\right) \]
Equating components: \[ \frac{2+x'}{2} = \frac{11}{3} \implies 6+3x' = 22 \implies 3x' = 16 \implies x' = \frac{16}{3} \] \[ \frac{1+y'}{2} = \frac{7}{3} \implies 3+3y' = 14 \implies 3y' = 11 \implies y' = \frac{11}{3} \] \[ \frac{2+z'}{2} = \frac{7}{3} \implies 6+3z' = 14 \implies 3z' = 8 \implies z' = \frac{8}{3} \]
The image is A'\((\frac{16}{3}, \frac{11}{3}, \frac{8}{3})\).


3. Find the equation of the line AA':

The line passes through A(2, 1, 2) and is parallel to the vector \(\vec{AM}\).
Using \(\lambda = -1/3\) in \(\vec{AM}\): \[ \vec{AM} = (2 - \frac{1}{3})\hat{i} + (1 - (-\frac{1}{3}))\hat{j} - (-\frac{1}{3})\hat{k} = \frac{5}{3}\hat{i} + \frac{4}{3}\hat{j} + \frac{1}{3}\hat{k} \]
The direction ratios of AA' are proportional to (5, 4, 1).
The equation of the line AA' in Cartesian form is: \[ \frac{x-2}{5} = \frac{y-1}{4} = \frac{z-2}{1} \]

Step 4: Final Answer:

- The foot of the perpendicular is M\((\frac{11}{3}, \frac{7}{3}, \frac{7}{3})\).

- The image of A is A'\((\frac{16}{3}, \frac{11}{3}, \frac{8}{3})\).

- The equation of the line AA' is \(\frac{x-2}{5} = \frac{y-1}{4} = \frac{z-2}{1}\).
Quick Tip: The vector \(\vec{AM}\) is the direction vector for the line passing through A and A'. So once you find \(\vec{AM}\), you can immediately write down the equation for the line AA' without needing to calculate the coordinates of A'.

Step 1: Understanding the Concept:

The integrand is a rational function where the denominator is a product of a linear factor and an irreducible quadratic factor. This type of integral can be solved using the method of partial fraction decomposition.


Step 2: Key Formula or Approach:

1. Decompose the fraction into partial fractions:
\[ \frac{5x}{(x+1)(x^2+9)} = \frac{A}{x+1} + \frac{Bx+C}{x^2+9} \]
2. Solve for the constants A, B, and C.
3. Integrate the resulting simpler fractions term by term. The standard forms will be \(\int \frac{1}{u}du = \ln|u|\), \(\int \frac{x}{x^2+a^2}dx = \frac{1}{2}\ln(x^2+a^2)\), and \(\int \frac{1}{x^2+a^2}dx = \frac{1}{a}\tan^{-1}(\frac{x}{a})\).


Step 3: Detailed Explanation:

Let's decompose the integrand: \[ \frac{5x}{(x+1)(x^2+9)} = \frac{A}{x+1} + \frac{Bx+C}{x^2+9} \]
Multiply both sides by \((x+1)(x^2+9)\): \[ 5x = A(x^2+9) + (Bx+C)(x+1) \]
To find the constants:
- Let \(x = -1\): \(5(-1) = A((-1)^2+9) + 0 \implies -5 = A(10) \implies A = -\frac{1}{2}\).
- Expand and equate coefficients:
\[ 5x = Ax^2 + 9A + Bx^2 + Bx + Cx + C \]
\[ 5x = (A+B)x^2 + (B+C)x + (9A+C) \]
- Equating coefficients of \(x^2\): \(A+B = 0 \implies B = -A = \frac{1}{2}\).
- Equating constant terms: \(9A+C = 0 \implies C = -9A = -9(-\frac{1}{2}) = \frac{9}{2}\).
- (Check with coefficient of x): \(B+C = \frac{1}{2} + \frac{9}{2} = \frac{10}{2} = 5\). This matches.

So, the partial fraction decomposition is: \[ \frac{5x}{(x+1)(x^2+9)} = \frac{-1/2}{x+1} + \frac{\frac{1}{2}x + \frac{9}{2}}{x^2+9} \]
Now, we integrate: \[ I = \int \left( \frac{-1/2}{x+1} + \frac{\frac{1}{2}x}{x^2+9} + \frac{9/2}{x^2+9} \right) dx \] \[ I = -\frac{1}{2}\int \frac{1}{x+1}dx + \frac{1}{2}\int \frac{x}{x^2+9}dx + \frac{9}{2}\int \frac{1}{x^2+9}dx \]
Let's evaluate each integral:
1. \(\int \frac{1}{x+1}dx = \ln|x+1|\)
2. \(\int \frac{x}{x^2+9}dx\): Let \(u=x^2+9, du=2xdx\). The integral is \(\frac{1}{2}\int \frac{du}{u} = \frac{1}{2}\ln|u| = \frac{1}{2}\ln(x^2+9)\).
3. \(\int \frac{1}{x^2+9}dx = \int \frac{1}{x^2+3^2}dx = \frac{1}{3}\tan^{-1}\left(\frac{x}{3}\right)\)

Combining the results: \[ I = -\frac{1}{2}\ln|x+1| + \frac{1}{2}\left(\frac{1}{2}\ln(x^2+9)\right) + \frac{9}{2}\left(\frac{1}{3}\tan^{-1}\left(\frac{x}{3}\right)\right) + C \] \[ I = -\frac{1}{2}\ln|x+1| + \frac{1}{4}\ln(x^2+9) + \frac{3}{2}\tan^{-1}\left(\frac{x}{3}\right) + C \]

Step 4: Final Answer:

The integral is \(-\frac{1}{2}\ln|x+1| + \frac{1}{4}\ln(x^2+9) + \frac{3}{2}\tan^{-1}\left(\frac{x}{3}\right) + C\).
Quick Tip: When solving for coefficients in partial fractions, using the substitution method (plugging in roots like x=-1) is fast for linear factors. For quadratic factors, equating coefficients is reliable. Always perform a quick check, like we did for the coefficient of 'x', to ensure your constants are correct before integrating.

Step 1: Understanding the Concept:

This problem connects matrix multiplication with solving a system of linear equations. If the product of two matrices, A and B, is a scalar multiple of the identity matrix (\(AB = kI\)), then the inverse of A is related to B (\(A^{-1} = \frac{1}{k}B\)). This inverse can then be used to solve a system of linear equations \(CX = D\) if the coefficient matrix C is equal to A.


Step 2: Key Formula or Approach:

1. Calculate the matrix product \(P = AB\).
2. Write the system of linear equations in the matrix form \(CX = D\).
3. Compare the coefficient matrix C with the given matrix A.
4. If \(AB = kI\) and \(C=B\), then from \(B^{-1} = \frac{1}{k}A\), the solution is \(X = B^{-1}D = \frac{1}{k}AD\). (Note: Let's check which matrix is the coefficient matrix).
5. Calculate the product AD and multiply by the scalar \(\frac{1{k}\) to find the solution vector \(X\).


Step 3: Detailed Explanation:

Part 1: Find AB \[ AB = \begin{bmatrix} -4 & 4 & 4
-7 & 1 & 3
5 & -3 & -1 \end{bmatrix} \begin{bmatrix} 1 & -1 & 1
1 & -2 & -2
2 & 1 & 3 \end{bmatrix} \] \[ AB = \begin{bmatrix} (-4+4+8) & (4-8+4) & (-4-8+12)
(-7+1+6) & (7-2+3) & (-7-2+9)
(5-3-2) & (-5+6-1) & (5+6-3) \end{bmatrix} \] \[ AB = \begin{bmatrix} 8 & 0 & 0
0 & 8 & 0
0 & 0 & 8 \end{bmatrix} = 8 \begin{bmatrix} 1 & 0 & 0
0 & 1 & 0
0 & 0 & 1 \end{bmatrix} = 8I \]

Part 2: Solve the system of equations
The given system is:
\begin{align x - y + z &= 4
x - 2y - 2z &= 9
2x + y + 3z &= 1 \end{align
In matrix form \(CX = D\), this is: \[ \begin{bmatrix} 1 & -1 & 1
1 & -2 & -2
2 & 1 & 3 \end{bmatrix} \begin{bmatrix} x
y
z \end{bmatrix} = \begin{bmatrix} 4
9
1 \end{bmatrix} \]
We observe that the coefficient matrix C is the same as the given matrix B. So, we have the equation \(BX = D\).
The solution is \(X = B^{-1}D\).

From our product calculation, \(AB = 8I\).
Multiplying by \(B^{-1}\) from the right: \(ABB^{-1} = 8IB^{-1} \implies A = 8B^{-1}\).
Therefore, \(B^{-1} = \frac{1}{8}A\).

Now we can find the solution: \[ X = B^{-1}D = \frac{1}{8}AD \] \[ \begin{bmatrix} x
y
z \end{bmatrix} = \frac{1}{8} \begin{bmatrix} -4 & 4 & 4
-7 & 1 & 3
5 & -3 & -1 \end{bmatrix} \begin{bmatrix} 4
9
1 \end{bmatrix} \] \[ \begin{bmatrix} x
y
z \end{bmatrix} = \frac{1}{8} \begin{bmatrix} (-4)(4) + (4)(9) + (4)(1)
(-7)(4) + (1)(9) + (3)(1)
(5)(4) + (-3)(9) + (-1)(1) \end{bmatrix} \] \[ \begin{bmatrix} x
y
z \end{bmatrix} = \frac{1}{8} \begin{bmatrix} -16 + 36 + 4
-28 + 9 + 3
20 - 27 - 1 \end{bmatrix} = \frac{1}{8} \begin{bmatrix} 24
-16
-8 \end{bmatrix} \] \[ \begin{bmatrix} x
y
z \end{bmatrix} = \begin{bmatrix} 3
-2
-1 \end{bmatrix} \]

Step 4: Final Answer:

The product \(AB = 8I\).
The solution to the system of equations is \(x = 3, y = -2, z = -1\).
Quick Tip: In this type of "Hence, solve..." problem, the coefficient matrix of the system is always one of the matrices given (or its transpose). After calculating the product AB, you can immediately write down the inverse of one matrix in terms of the other, which is the key to the solution.

Step 1: Understanding the Concept:

This problem requires first finding the inverse of a matrix A using the adjugate method, and then using this inverse to solve a system of linear equations. The system's coefficient matrix will be related to A.


Step 2: Key Formula or Approach:

1. Find \(A^{-1}\) using the formula \(A^{-1} = \frac{1}{\det(A)}adj(A)\).
- Calculate \(\det(A)\).
- Find the matrix of cofactors.
- Find the adjugate (transpose of the cofactor matrix).
2. Express the system of equations in matrix form \(CX = D\).
3. Relate C to A. Find the solution using \(X = C^{-1}D\).


Step 3: Detailed Explanation:

Part 1: Find \(A^{-1}\)
Given \(A = \begin{bmatrix} 1 & 2 & 0
-2 & -1 & -2
0 & -1 & 1 \end{bmatrix}\).
- Determinant:
\(\det(A) = 1(-1 - 2) - 2(-2 - 0) + 0 = -3 - 2(-2) = -3 + 4 = 1\).
- Cofactors:
\(C_{11} = (-1-2) = -3\)
\(C_{12} = -(-2-0) = 2\)
\(C_{13} = (2-0) = 2\)
\(C_{21} = -(2-0) = -2\)
\(C_{22} = (1-0) = 1\)
\(C_{23} = -(-1-0) = 1\)
\(C_{31} = (-4-0) = -4\)
\(C_{32} = -(-2-0) = 2\)
\(C_{33} = (-1-(-4)) = 3\)
- Cofactor Matrix and Adjugate:
Cofactor Matrix = \(\begin{bmatrix} -3 & 2 & 2
-2 & 1 & 1
-4 & 2 & 3 \end{bmatrix}\).
\(adj(A) = (Cofactor Matrix)^T = \begin{bmatrix} -3 & -2 & -4
2 & 1 & 2
2 & 1 & 3 \end{bmatrix}\).
- Inverse:
\(A^{-1} = \frac{1}{\det(A)}adj(A) = \frac{1}{1} \begin{bmatrix} -3 & -2 & -4
2 & 1 & 2
2 & 1 & 3 \end{bmatrix} = \begin{bmatrix} -3 & -2 & -4
2 & 1 & 2
2 & 1 & 3 \end{bmatrix}\).


Part 2: Solve the system of equations
The system is:
\begin{align x - 2y + 0z &= 10
2x - y - z &= 8
0x - 2y + z &= 7 \end{align
In matrix form \(CX = D\), this is: \[ \begin{bmatrix} 1 & -2 & 0
2 & -1 & -1
0 & -2 & 1 \end{bmatrix} \begin{bmatrix} x
y
z \end{bmatrix} = \begin{bmatrix} 10
8
7 \end{bmatrix} \]
Let's compare the coefficient matrix C with A. We see that \(C = A^T\).
We know that \((A^T)^{-1} = (A^{-1})^T\).
So, \(C^{-1} = (A^{-1})^T = \left(\begin{bmatrix} -3 & -2 & -4
2 & 1 & 2
2 & 1 & 3 \end{bmatrix}\right)^T = \begin{bmatrix} -3 & 2 & 2
-2 & 1 & 1
-4 & 2 & 3 \end{bmatrix}\).
The solution is \(X = C^{-1}D\): \[ \begin{bmatrix} x
y
z \end{bmatrix} = \begin{bmatrix} -3 & 2 & 2
-2 & 1 & 1
-4 & 2 & 3 \end{bmatrix} \begin{bmatrix} 10
8
7 \end{bmatrix} \] \[ \begin{bmatrix} x
y
z \end{bmatrix} = \begin{bmatrix} (-3)(10) + (2)(8) + (2)(7)
(-2)(10) + (1)(8) + (1)(7)
(-4)(10) + (2)(8) + (3)(7) \end{bmatrix} \] \[ \begin{bmatrix} x
y
z \end{bmatrix} = \begin{bmatrix} -30 + 16 + 14
-20 + 8 + 7
-40 + 16 + 21 \end{bmatrix} = \begin{bmatrix} 0
-5
-3 \end{bmatrix} \]

Step 4: Final Answer:
\(A^{-1} = \begin{bmatrix} -3 & -2 & -4
2 & 1 & 2
2 & 1 & 3 \end{bmatrix}\).
The solution to the system is \(x = 0, y = -5, z = -3\).
Quick Tip: When solving a system \(CX=D\) after finding \(A^{-1}\), always check if \(C=A\) or \(C=A^T\). This is a common pattern in exam questions. Remember the property \((A^T)^{-1} = (A^{-1})^T\), which allows you to find the inverse of the coefficient matrix easily if you have already calculated \(A^{-1}\).

Step 1: Understanding the Concept:

The area of a region bounded by a curve \(y=f(x)\), the x-axis, and vertical lines \(x=a\) and \(x=b\) is given by the definite integral \(\int_{a}^{b} f(x) dx\). However, if the function takes on negative values in the interval, the integral will yield a negative result for that part of the area. Since area must be positive, we must integrate the absolute value of the function, which means splitting the integral at the x-intercepts.


Step 2: Key Formula or Approach:

1. Find the x-intercept of the line \(y = 5x+2\) by setting \(y=0\).
2. Check if the x-intercept lies within the given interval \([-2, 2]\).
3. If it does, split the integral into two parts: one for the area below the x-axis and one for the area above.
Area = \(\int_{a}^{c} |f(x)| dx + \int_{c}^{b} |f(x)| dx\), where c is the x-intercept.
4. Evaluate the definite integrals.


Step 3: Detailed Explanation:

The function is \(y = 5x+2\). The interval is \([-2, 2]\).

First, find the x-intercept: \[ 5x + 2 = 0 \implies 5x = -2 \implies x = -\frac{2}{5} = -0.4 \]
Since \(-2 < -0.4 < 2\), the x-intercept is within the interval of integration.
The function is negative on the interval \([-2, -0.4)\) and positive on the interval \((-0.4, 2]\).
So, we must split the area calculation: \[ Area = \int_{-2}^{2} |5x+2| dx = \int_{-2}^{-2/5} -(5x+2) dx + \int_{-2/5}^{2} (5x+2) dx \]

Part 1: Integral from -2 to -2/5 \[ \int_{-2}^{-2/5} (-5x-2) dx = \left[ -\frac{5x^2}{2} - 2x \right]_{-2}^{-2/5} \] \[ = \left( -\frac{5(-2/5)^2}{2} - 2(-\frac{2}{5}) \right) - \left( -\frac{5(-2)^2}{2} - 2(-2) \right) \] \[ = \left( -\frac{5(4/25)}{2} + \frac{4}{5} \right) - \left( -\frac{5(4)}{2} + 4 \right) \] \[ = \left( -\frac{2}{5} + \frac{4}{5} \right) - (-10 + 4) = \frac{2}{5} - (-6) = \frac{2}{5} + 6 = \frac{32}{5} \]

Part 2: Integral from -2/5 to 2 \[ \int_{-2/5}^{2} (5x+2) dx = \left[ \frac{5x^2}{2} + 2x \right]_{-2/5}^{2} \] \[ = \left( \frac{5(2)^2}{2} + 2(2) \right) - \left( \frac{5(-2/5)^2}{2} + 2(-\frac{2}{5}) \right) \] \[ = \left( \frac{20}{2} + 4 \right) - \left( \frac{5(4/25)}{2} - \frac{4}{5} \right) \] \[ = (10 + 4) - \left( \frac{2}{5} - \frac{4}{5} \right) = 14 - (-\frac{2}{5}) = 14 + \frac{2}{5} = \frac{72}{5} \]

Total Area: \[ Total Area = \frac{32}{5} + \frac{72}{5} = \frac{104}{5} = 20.8 \]

Step 4: Final Answer:

The area of the required region is \(\frac{104}{5}\) or 20.8 square units.
Quick Tip: For linear functions, the region is composed of triangles. You can use the formula for the area of a triangle (\(\frac{1}{2} \times base \times height\)) to quickly find or verify your answer. Triangle 1 (below x-axis): Base from \(x=-2\) to \(x=-2/5\), length = \(8/5\). Height at \(x=-2\) is \(y=-8\). Area = \(\frac{1}{2} \cdot \frac{8}{5} \cdot 8 = \frac{32}{5}\). Triangle 2 (above x-axis): Base from \(x=-2/5\) to \(x=2\), length = \(12/5\). Height at \(x=2\) is \(y=12\). Area = \(\frac{1}{2} \cdot \frac{12}{5} \cdot 12 = \frac{72}{5}\). Total Area = \(\frac{32}{5} + \frac{72}{5} = \frac{104}{5}\).

Step 1: Understanding the Concept:

This is a problem of total probability. We want to find the total probability of an event (a car being electric) which can happen through several mutually exclusive events (being manufactured by Amber, Bonzi, or Comet).


Step 2: Key Formula or Approach:

Let A, B, C be the events that a car is manufactured by Amber, Bonzi, and Comet, respectively.
Let E be the event that a car is electric.
The Law of Total Probability states: \[ P(E) = P(A)P(E|A) + P(B)P(E|B) + P(C)P(E|C) \]
where \(P(E|A)\) is the conditional probability that a car is electric given it was made by Amber.


Step 3: Detailed Explanation:

From the problem statement, we extract the probabilities:

Production shares (prior probabilities):

\(P(A) = 60% = 0.6\)
\(P(B) = 30% = 0.3\)
\(P(C) = 10% = 0.1\)

Conditional probabilities of being electric:

\(P(E|A) = 20% = 0.2\)
\(P(E|B) = 10% = 0.1\)
\(P(E|C) = 5% = 0.05\)


Now, apply the Law of Total Probability: \[ P(E) = (0.6)(0.2) + (0.3)(0.1) + (0.1)(0.05) \] \[ P(E) = 0.12 + 0.03 + 0.005 \] \[ P(E) = 0.155 \]

Step 4: Final Answer:

The probability that a randomly selected car is an electric car is 0.155 or 15.5%.
Quick Tip: For total probability and Bayes' theorem problems, it's very helpful to first list all the prior probabilities (like market shares) and all the conditional probabilities (like defect rates or feature percentages). This organizes the information and makes it easier to plug into the formulas.

Step 1: Understanding the Concept:

The events "a car is electric" (E) and "a car is a petrol car" (P) are complementary events, as the cars run on either petrol or battery. The sum of the probabilities of complementary events is 1.


Step 2: Key Formula or Approach:

Let P be the event that a car is a petrol car.
Since a car is either electric or petrol, \(P(P) + P(E) = 1\).
Therefore, \(P(P) = 1 - P(E)\).


Step 3: Detailed Explanation:

From part (a), we calculated the probability that a car is electric: \[ P(E) = 0.155 \]
The probability that a car is a petrol car is the complement of this event. \[ P(P) = 1 - P(E) = 1 - 0.155 = 0.845 \]
Alternative Method (Total Probability):

Let P be the event a car is petrol. \(P(P|A) = 1 - P(E|A) = 1 - 0.2 = 0.8\) \(P(P|B) = 1 - P(E|B) = 1 - 0.1 = 0.9\) \(P(P|C) = 1 - P(E|C) = 1 - 0.05 = 0.95\)
Using the Law of Total Probability for P: \[ P(P) = P(A)P(P|A) + P(B)P(P|B) + P(C)P(P|C) \] \[ P(P) = (0.6)(0.8) + (0.3)(0.9) + (0.1)(0.95) \] \[ P(P) = 0.48 + 0.27 + 0.095 = 0.845 \]

Step 4: Final Answer:

The probability that a randomly selected car is a petrol car is 0.845 or 84.5%.
Quick Tip: Using the complement rule is often much faster if you have already calculated the probability of the complementary event. Both methods are valid, but efficiency is key in exams.

Step 1: Understanding the Concept:

This question asks for a conditional probability, specifically a posterior probability. We are given that an event (the car is electric) has occurred, and we need to find the probability that it was caused by a specific source (manufactured by Comet). This is a direct application of Bayes' Theorem.


Step 2: Key Formula or Approach:

Bayes' Theorem states: \[ P(C|E) = \frac{P(C)P(E|C)}{P(E)} \]
We need the values for \(P(C)\), \(P(E|C)\), and \(P(E)\).


Step 3: Detailed Explanation:

From the previous parts and the problem statement, we have:

\(P(C)\): The probability that a car is manufactured by Comet = 0.1.
\(P(E|C)\): The probability that a car is electric given it was made by Comet = 0.05.
\(P(E)\): The total probability that a car is electric (calculated in part (a)) = 0.155.

The term \(P(C)P(E|C)\) is the probability that a car is both made by Comet AND is electric. \[ P(C \cap E) = P(C)P(E|C) = (0.1)(0.05) = 0.005 \]
Now, apply Bayes' Theorem: \[ P(C|E) = \frac{P(C \cap E)}{P(E)} = \frac{0.005}{0.155} \]
To simplify the fraction, multiply the numerator and denominator by 1000: \[ P(C|E) = \frac{5}{155} = \frac{1}{31} \]

Step 4: Final Answer:

The probability that the electric car was manufactured by Comet is \(\frac{1}{31}\).
Quick Tip: Bayes' Theorem formula is essentially \(P(Cause | Effect) = P(Cause and Effect) / P(Effect)\). The numerator is the probability of the specific path (e.g., Comet and Electric), and the denominator is the sum of probabilities of all paths leading to the effect (Total Probability of Electric).

Step 1: Understanding the Concept:

We need to find the probability \(P(A \cup B | E)\). Since the events A (made by Amber) and B (made by Bonzi) are mutually exclusive (a car cannot be made by both), the probability of their union given E is the sum of their individual probabilities given E: \(P(A \cup B | E) = P(A|E) + P(B|E)\). An alternative, and often simpler, approach is to use the complement rule.


Step 2: Key Formula or Approach:

The event "manufactured by Amber or Bonzi" is the complement of the event "manufactured by Comet", given that the car is electric.
Therefore, \(P(A \cup B | E) = 1 - P(C|E)\).


Step 3: Detailed Explanation:

In the context of a car already known to be electric, there are only three possibilities for its manufacturer: Amber, Bonzi, or Comet. These are mutually exclusive and exhaustive events.
Therefore, the sum of their probabilities must be 1: \[ P(A|E) + P(B|E) + P(C|E) = 1 \]
We want to find \(P(A \cup B | E)\), which is equal to \(P(A|E) + P(B|E)\).
From the equation above, we can see that: \[ P(A|E) + P(B|E) = 1 - P(C|E) \]
From the previous part (b)(ii), we calculated: \[ P(C|E) = \frac{1}{31} \]
Substituting this value: \[ P(A \cup B | E) = 1 - \frac{1}{31} = \frac{31-1}{31} = \frac{30}{31} \]

Step 4: Final Answer:

The probability that the electric car was manufactured by Amber or Bonzi is \(\frac{30}{31}\).
Quick Tip: When asked for the probability of a union of events ("A or B") in a conditional probability problem, always check if it's easier to calculate the probability of the complement. If there are many events in the union but only one or two outside it, the complement approach is much faster.

Step 1: Understanding the Concept:

To find the intervals where a function is increasing or decreasing, we need to analyze the sign of its first derivative, \(f'(x)\).

If \(f'(x) > 0\), the function is increasing.
If \(f'(x) < 0\), the function is decreasing.

The points where \(f'(x) = 0\) or is undefined are the critical points, which are the boundaries of these intervals.


Step 2: Key Formula or Approach:

1. Find the first derivative \(f'(x)\).
2. Find the critical points by solving \(f'(x) = 0\) in the given interval \(x \in [0, \pi]\).
3. Test the sign of \(f'(x)\) in the sub-intervals created by the critical points.


Step 3: Detailed Explanation:

The function is \(f(x) = e^x \sin x\).
1. Find the first derivative:
Using the product rule \((uv)' = u'v + uv'\): \[ f'(x) = \frac{d}{dx}(e^x)\sin x + e^x \frac{d}{dx}(\sin x) \] \[ f'(x) = e^x \sin x + e^x \cos x = e^x(\sin x + \cos x) \]

2. Find the critical points:
Set \(f'(x) = 0\). Since \(e^x\) is always positive, we only need to solve \(\sin x + \cos x = 0\). \[ \sin x = -\cos x \] \[ \tan x = -1 \]
In the interval \([0, \pi]\), the value of \(x\) for which \(\tan x = -1\) is in the second quadrant.
The reference angle for \(\tan \theta = 1\) is \(\frac{\pi}{4}\). In the second quadrant, the angle is \(\pi - \frac{\pi}{4} = \frac{3\pi}{4}\).
So, the only critical point in \([0, \pi]\) is \(x = \frac{3\pi}{4}\).

3. Test the intervals:
The critical point divides the interval \([0, \pi]\) into two sub-intervals: \([0, \frac{3\pi}{4})\) and \((\frac{3\pi}{4}, \pi]\).

Interval \([0, \frac{3\pi}{4})\): Let's pick a test point, e.g., \(x = \frac{\pi}{2}\).
\[ f'(\frac{\pi}{2}) = e^{\pi/2}(\sin\frac{\pi}{2} + \cos\frac{\pi}{2}) = e^{\pi/2}(1 + 0) = e^{\pi/2} > 0 \]
Since \(f'(x) > 0\), the function is increasing on \([0, \frac{3\pi}{4})\).
Interval \((\frac{3\pi}{4}, \pi]\): Let's pick a test point, e.g., \(x = \pi\).
\[ f'(\pi) = e^{\pi}(\sin\pi + \cos\pi) = e^{\pi}(0 - 1) = -e^{\pi} < 0 \]
Since \(f'(x) < 0\), the function is decreasing on \((\frac{3\pi}{4}, \pi]\).


Step 4: Final Answer:

- The function \(f(x)\) is increasing on the interval \([0, \frac{3\pi}{4}]\).
- The function \(f(x)\) is decreasing on the interval \([\frac{3\pi}{4}, \pi]\).
Quick Tip: To solve \(\sin x + \cos x = 0\), you can also use the form \(R\sin(x+\alpha)\). Here, \(\sin x + \cos x = \sqrt{2}(\frac{1}{\sqrt{2}}\sin x + \frac{1}{\sqrt{2}}\cos x) = \sqrt{2}\sin(x+\frac{\pi}{4})\). So \(\sqrt{2}\sin(x+\frac{\pi}{4}) = 0\), which means \(x+\frac{\pi}{4} = k\pi\). For \(x \in [0, \pi]\), \(k=1\) gives \(x+\frac{\pi}{4}=\pi \implies x=\frac{3\pi}{4}\).

Step 1: Understanding the Concept:

We can classify critical points using either the First Derivative Test or the Second Derivative Test.

First Derivative Test: Check the sign of \(f'(x)\) on either side of the critical point. If the sign changes from + to -, it's a local maximum. If it changes from - to +, it's a local minimum. If the sign doesn't change, it could be a point of inflection.
Second Derivative Test: Find \(f''(x)\). At a critical point \(c\) where \(f'(c)=0\): if \(f''(c) < 0\), it's a local maximum. If \(f''(c) > 0\), it's a local minimum. If \(f''(c) = 0\), the test is inconclusive.


Step 2: Key Formula or Approach:

1. We have already found the critical point \(x = \frac{3\pi}{4}\) from part (i).
2. We can use the results of the First Derivative Test from part (i).
3. Alternatively, we can calculate the second derivative \(f''(x)\) and evaluate its sign at the critical point.


Step 3: Detailed Explanation:

From part (i), we found the only critical point in the interval \([0, \pi]\) is \(x = \frac{3\pi}{4}\).


Method 1: First Derivative Test

In part (i), we analyzed the sign of \(f'(x)\) around \(x = \frac{3\pi}{4}\):

For \(x < \frac{3\pi}{4}\) (in the interval \([0, \frac{3\pi}{4})\)), we found \(f'(x) > 0\).
For \(x > \frac{3\pi}{4}\) (in the interval \((\frac{3\pi}{4}, \pi]\)), we found \(f'(x) < 0\).

Since the sign of the first derivative changes from positive (+) to negative (-), the point \(x = \frac{3\pi}{4}\) is a point of local maximum.


Method 2: Second Derivative Test

First, we need to find the second derivative, \(f''(x)\).
We have \(f'(x) = e^x(\sin x + \cos x)\).
Using the product rule again: \[ f''(x) = \frac{d}{dx}(e^x)(\sin x + \cos x) + e^x \frac{d}{dx}(\sin x + \cos x) \] \[ f''(x) = e^x(\sin x + \cos x) + e^x(\cos x - \sin x) \] \[ f''(x) = e^x\sin x + e^x\cos x + e^x\cos x - e^x\sin x \] \[ f''(x) = 2e^x \cos x \]
Now, evaluate \(f''(x)\) at the critical point \(x = \frac{3\pi}{4}\): \[ f''\left(\frac{3\pi}{4}\right) = 2e^{3\pi/4} \cos\left(\frac{3\pi}{4}\right) \]
We know that \(\cos\left(\frac{3\pi}{4}\right) = -\frac{1}{\sqrt{2}}\) (since it's in the second quadrant). \[ f''\left(\frac{3\pi}{4}\right) = 2e^{3\pi/4} \left(-\frac{1}{\sqrt{2}}\right) = -\sqrt{2} e^{3\pi/4} \]
Since \(e^{3\pi/4}\) is positive, the value of \(f''\left(\frac{3\pi}{4}\right)\) is negative.
According to the Second Derivative Test, if \(f''(c) < 0\), then \(x=c\) is a point of local maximum.


A point of inflection occurs where \(f''(x) = 0\) and the concavity changes. Let's solve \(f''(x) = 0\): \[ 2e^x \cos x = 0 \implies \cos x = 0 \]
In the interval \([0, \pi]\), \(\cos x = 0\) at \(x = \frac{\pi}{2}\). This is a potential point of inflection, not a local extremum, as \(f'(\frac{\pi}{2}) \ne 0\). The question asks to classify the critical points only.


Step 4: Final Answer:

The critical point \(x = \frac{3\pi}{4}\) is a point of local maximum.
Quick Tip: The First Derivative Test is often sufficient and can be faster if you've already determined the intervals of increase and decrease. The Second Derivative Test is a good alternative if the second derivative is easy to compute.

Step 1: Understanding the Concept:

A relation from a set S to a set J is defined as any subset of the Cartesian product \(S \times J\). The total number of possible relations is equal to the total number of possible subsets of \(S \times J\).


Step 2: Key Formula or Approach:

If a set A has \(m\) elements and a set B has \(n\) elements, then:
1. The number of elements in the Cartesian product \(A \times B\) is \(m \times n\).
2. The number of subsets of a set with \(k\) elements is \(2^k\).
Therefore, the total number of relations from A to B is \(2^{m \times n}\).


Step 3: Detailed Explanation:

We are given the sets:

S = \(\{S_1, S_2, S_3, S_4\}\), so the number of elements in S is \(n(S) = 4\).
J = \(\{J_1, J_2, J_3\}\), so the number of elements in J is \(n(J) = 3\).

First, find the number of elements in the Cartesian product \(S \times J\): \[ n(S \times J) = n(S) \times n(J) = 4 \times 3 = 12 \]
The number of possible relations from S to J is the number of possible subsets of \(S \times J\). \[ Total number of relations = 2^{n(S \times J)} = 2^{12} \]
Calculating the value: \[ 2^{12} = (2^{10}) \times (2^2) = 1024 \times 4 = 4096 \]

Step 4: Final Answer:

The total number of possible relations from S to J is \(2^{12}\) or 4096.
Quick Tip: Remember the distinction: The number of elements in the Cartesian product \(A \times B\) is \(|A| \times |B|\), while the number of relations (which are subsets) is \(2^{|A| \times |B|}\).

Step 1: Understanding the Concept:

A function is bijective if it is both one-one (injective) and onto (surjective).

One-one (Injective): Different elements in the domain must have different images in the codomain.
Onto (Surjective): Every element in the codomain must be an image of at least one element from the domain. In other words, the range of the function must be equal to its codomain.


Step 2: Key Formula or Approach:

1. Check if the function is one-one by seeing if any two different inputs map to the same output.
2. Check if the function is onto by comparing the range of the function with the codomain.
3. Conclude whether it is bijective.


Step 3: Detailed Explanation:

The function is \(f: S \to J\), where \(S = \{S_1, S_2, S_3, S_4\}\) and \(J = \{J_1, J_2, J_3\}\).
The function is given as \(f = \{(S_1, J_1), (S_2, J_2), (S_3, J_2), (S_4, J_3)\}\).


Checking for One-one (Injective):

We examine the outputs for each input:

\(f(S_1) = J_1\)
\(f(S_2) = J_2\)
\(f(S_3) = J_2\)
\(f(S_4) = J_3\)

Here, two different elements from the domain, \(S_2\) and \(S_3\), are mapped to the same element \(J_2\) in the codomain.
Since \(f(S_2) = f(S_3)\) but \(S_2 \ne S_3\), the function is not one-one.


Checking for Onto (Surjective):

The codomain of the function is the set \(J = \{J_1, J_2, J_3\}\).
The range of the function is the set of all images (outputs).
Range(f) = \(\{f(S_1), f(S_2), f(S_3), f(S_4)\}\) = \(\{J_1, J_2, J_2, J_3\}\) = \(\{J_1, J_2, J_3\}\).
Since the Range(f) is equal to the Codomain J, the function is onto.


Conclusion on Bijectivity:

A function must be both one-one and onto to be bijective. Since this function is not one-one, it is not bijective.


Step 4: Final Answer:

The function is not bijective because it is not one-one.
Quick Tip: A quick check for a function \(f: A \to B\) to be one-one is that \(|A| \le |B|\). For it to be onto, \(|A| \ge |B|\). For it to be bijective, it must be both, so a necessary (but not sufficient) condition is \(|A| = |B|\). Here, \(|S|=4\) and \(|J|=3\), so a one-one function is impossible.

Step 1: Understanding the Concept:

A function \(f: S \to J\) is one-one (injective) if every element in the domain S is mapped to a unique element in the codomain J. This means no two elements in S can have the same image in J.


Step 2: Key Formula or Approach:

The number of one-one functions from a set A with \(m\) elements to a set B with \(n\) elements is given by the formula for permutations, \(^n P_m = \frac{n!}{(n-m)!}\). This is only possible if \(n \ge m\). If \(m > n\), it is impossible to create a one-one function, so the number of such functions is 0. This is based on the Pigeonhole Principle.


Step 3: Detailed Explanation:

We are defining a function from set S to set J.

The domain is S, with \(m = n(S) = 4\) elements.
The codomain is J, with \(n = n(J) = 3\) elements.

For a function to be one-one, we need to map the 4 distinct elements of S to 4 distinct elements in J. However, the codomain J only has 3 elements available.
According to the Pigeonhole Principle, if we try to map 4 items (pigeons) into 3 boxes (pigeonholes), at least one box must contain more than one item. In the context of functions, this means at least two elements from S must be mapped to the same element in J.
This violates the condition for a one-one function.
Mathematically, since \(m = 4 > n = 3\), it is impossible to define a one-one function from S to J.


Step 4: Final Answer:

The number of one-one functions that can be there from set S to set J is 0.
Quick Tip: Before starting calculations for the number of one-one functions from A to B, always compare the sizes of the sets. If the domain is larger than the codomain (\(|A| > |B|\)), the answer is immediately zero.

Step 1: Understanding the Concept:

We need to modify a given relation \(R_1\) on a set S to satisfy two properties:

Reflexive: A relation R on a set S is reflexive if for every element \(x \in S\), the pair \((x, x)\) is in R.
Not Symmetric: A relation R is not symmetric if there exists at least one pair \((x, y) \in R\) such that the reverse pair \((y, x)\) is not in R.

We must add the minimum number of pairs to achieve this.


Step 2: Key Formula or Approach:

1. To make the relation reflexive, identify all elements in set S and add the pairs \((x,x)\) for each element \(x\) that are not already present.
2. After making it reflexive, check if the resulting relation is symmetric. If it is not, we are done. If it is, we need to consider if the existing non-reflexive pairs can be used to break symmetry.


Step 3: Detailed Explanation:

The set is S = \(\{S_1, S_2, S_3, S_4\}\).
The initial relation is \(R_1 = \{(S_1, S_2), (S_2, S_4)\}\).


1. Making the relation reflexive:
For \(R_1\) to be reflexive on S, it must contain the pairs \((S_1, S_1), (S_2, S_2), (S_3, S_3),\) and \((S_4, S_4)\).
None of these are in the original \(R_1\). Therefore, to satisfy the reflexive property, we must add all four of these pairs.
The minimum pairs we must add are: \(\{(S_1, S_1), (S_2, S_2), (S_3, S_3), (S_4, S_4)\}\).


2. Checking the "not symmetric" condition:
Let's see what the new relation, let's call it \(R'_{1}\), looks like after adding these pairs: \[ R'_{1} = \{(S_1, S_2), (S_2, S_4), (S_1, S_1), (S_2, S_2), (S_3, S_3), (S_4, S_4)\} \]
Now we check if this relation \(R'_{1}\) is symmetric. A relation is symmetric if for every \((x, y) \in R\), we also have \((y, x) \in R\).
Let's check the non-reflexive pairs in \(R'_{1}\):

The pair \((S_1, S_2)\) is in \(R'_{1}\). For symmetry, \((S_2, S_1)\) must also be in \(R'_{1}\). It is not.
The pair \((S_2, S_4)\) is in \(R'_{1}\). For symmetry, \((S_4, S_2)\) must also be in \(R'_{1}\). It is not.

Since we have found at least one pair (\((S_1, S_2)\)) whose reverse is not in the relation, the relation \(R'_{1}\) is not symmetric.
Since adding the four reflexive pairs makes the relation reflexive AND it is already not symmetric, these are the only pairs we need to add.


Step 4: Final Answer:

The minimum ordered pairs that must be included are \((S_1, S_1), (S_2, S_2), (S_3, S_3),\) and \((S_4, S_4)\).
Quick Tip: To make a relation reflexive, you must add all pairs \((x,x)\) that are missing. To make a relation symmetric, you must add the reverse of every non-symmetric pair. To make it not symmetric, you just need to ensure there is at least one pair \((x,y)\) without its reverse \((y,x)\).