Step 1: Understanding the Question:


The question asks about the property of the sum of two skew-symmetric matrices of the same order.


Step 2: Key Formula or Approach:


A matrix 'M' is defined as skew-symmetric if its transpose is equal to its negative, i.e., \(M' = -M\).

We will use the property of transposes: \((A + B)' = A' + B'\).


Step 3: Detailed Explanation:


Let A and B be two skew-symmetric matrices of the same order.

By definition, we have \(A' = -A\) and \(B' = -B\).

Let C be their sum, so \(C = A + B\).

To check the property of C, we find its transpose:
\[ C' = (A + B)' \]

Using the transpose property:
\[ C' = A' + B' \]

Substitute the skew-symmetric conditions:
\[ C' = (-A) + (-B) = -(A + B) \]

Since \(C = A + B\), we get:
\[ C' = -C \]

This is the definition of a skew-symmetric matrix.


Step 4: Final Answer:


The sum of two skew-symmetric matrices is always a skew-symmetric matrix.
Quick Tip: Remember the basic algebra of special matrices:
- Sum of symmetric matrices is symmetric.
- Sum of skew-symmetric matrices is skew-symmetric.
- Scalar multiple of a symmetric/skew-symmetric matrix is also symmetric/skew-symmetric.

Step 1: Understanding the Question:


We need to identify the type of the given matrix A by checking its properties.


Step 2: Key Formula or Approach:


We will check the definitions of the given options:

- Symmetric matrix: \(A' = A\).

- Skew-symmetric matrix: \(A' = -A\). The diagonal elements must be zero.

- Diagonal matrix: All non-diagonal elements are zero.

- Null matrix: All elements are zero.


Step 3: Detailed Explanation:


The given matrix is \(A = \begin{bmatrix} 0 & -3 & 8
3 & 0 & 5
-8 & -5 & 0 \end{bmatrix}\).

First, let's find the transpose of A, denoted by A':
\[ A' = \begin{bmatrix} 0 & 3 & -8
-3 & 0 & -5
8 & 5 & 0 \end{bmatrix} \]

Now, let's find the negative of A, denoted by -A:
\[ -A = -\begin{bmatrix} 0 & -3 & 8
3 & 0 & 5
-8 & -5 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 3 & -8
-3 & 0 & -5
8 & 5 & 0 \end{bmatrix} \]

By comparing A' and -A, we see that \(A' = -A\).

This satisfies the condition for a skew-symmetric matrix.

Also, all diagonal elements are zero, which is another property of skew-symmetric matrices.


Step 4: Final Answer:


The matrix A is a skew-symmetric matrix.
Quick Tip: A quick check for a skew-symmetric matrix involves two steps:
1. Are all diagonal elements zero?
2. Is every off-diagonal element \(a_{ij}\) equal to the negative of its counterpart \(a_{ji}\)?
If both are true, the matrix is skew-symmetric.

Step 1: Understanding the Question:


We need to identify the function that corresponds to the given graph by analyzing its key features.


Step 2: Detailed Explanation:


Let's analyze the properties of the graphed function:

- Domain: The graph extends indefinitely to the left and right along the x-axis. The domain is all real numbers (\(-\infty, \infty\)).

- Range: The function's values are strictly between 0 and \(\pi\). The graph has horizontal asymptotes at y = 0 and y = \(\pi\). The range is (0, \(\pi\)).

- Behavior: The function is strictly decreasing over its entire domain. As x increases, y decreases.

- Key Point: The graph passes through the point (0, \(\pi/2\)).


Now let's check these features against the given options:

- (A) y = cot x: Has a range of (\(-\infty, \infty\)) and vertical asymptotes. This does not match.

- (B) y = cot\(^{-1}\) x: The principal value range of the inverse cotangent function is (0, \(\pi\)). Its domain is (\(-\infty, \infty\)), it is a decreasing function, and cot\(^{-1}\)(0) = \(\pi/2\). This perfectly matches all features of the graph.

- (C) y = tan x: Has a range of (\(-\infty, \infty\)) and vertical asymptotes. This does not match.

- (D) y = tan\(^{-1}\) x: Has a range of (\(-\pi/2, \pi/2\)). This does not match.


Step 3: Final Answer:


The graph shown depicts the function y = cot\(^{-1}\) x.
Quick Tip: Memorizing the graphs of the six inverse trigonometric functions is crucial. Pay close attention to their specific domains and principal value ranges.
- Range of sin\(^{-1}\)x is [-\(\pi/2\), \(\pi/2\)].
- Range of cos\(^{-1}\)x is [0, \(\pi\)].
- Range of tan\(^{-1}\)x is (-\(\pi/2\), \(\pi/2\)).
- Range of cot\(^{-1}\)x is (0, \(\pi\)).

Step 1: Understanding the Question:


We are given the order of matrix A and the fact that two matrix products involving A and the transpose of B are defined. We need to find the order of matrix B.


Step 2: Key Formula or Approach:


The rule for matrix multiplication states that for a product XY to be defined, the number of columns in X must equal the number of rows in Y.

If a matrix B has an order of p \(\times\) q, its transpose B' has an order of q \(\times\) p.


Step 3: Detailed Explanation:


Let the order of matrix A be n \(\times\) m.

Let the order of matrix B be p \(\times\) q.

This means the order of the transpose B' is q \(\times\) p.


Condition 1: AB' is defined.

The product is (A)\(_{n \times m}\) (B')\(_{q \times p}\).

For this to be defined, the number of columns of A must equal the number of rows of B'.

So, \(m = q\).


Condition 2: B'A is defined.

The product is (B')\(_{q \times p}\) (A)\(_{n \times m}\).

For this to be defined, the number of columns of B' must equal the number of rows of A.

So, \(p = n\).


Step 4: Final Answer:


We found that p = n and q = m.

Since the order of B was assumed to be p \(\times\) q, the order of B is n \(\times\) m.
Quick Tip: Remember the "inner dimensions must match" rule.
For AB': (n \(\times\) \textbf{m}) (\textbf{q} \(\times\) p) \(\implies\) \textbf{m = q}.
For B'A: (q \(\times\) \textbf{p}) (\textbf{n} \(\times\) m) \(\implies\) \textbf{p = n}.
So, B is p \(\times\) q, which is n \(\times\) m.

Step 1: Understanding the Question:


For the function f(x) to be continuous at x = 0, the limit of the function as x approaches 0 must be equal to the value of the function at x = 0.

We need to find \(k = \lim_{x \to 0} f(x)\).


Step 2: Key Formula or Approach:


We will use the standard limit formula: \(\lim_{u \to 0} \frac{\log(1+u)}{u} = 1\).

Alternatively, we can use L'Hôpital's Rule since the limit is of the indeterminate form 0/0.


Step 3: Detailed Explanation:


Using the standard limit approach:
\[ k = \lim_{x \to 0} \frac{\log(1+ax) + \log(1-bx)}{x} \]

We can split the limit into two parts:
\[ k = \lim_{x \to 0} \frac{\log(1+ax)}{x} + \lim_{x \to 0} \frac{\log(1-bx)}{x} \]

To match the standard form, we adjust each term:
\[ k = \lim_{x \to 0} \left(a \cdot \frac{\log(1+ax)}{ax}\right) + \lim_{x \to 0} \left(-b \cdot \frac{\log(1+(-bx))}{-bx}\right) \]

As x \(\to\) 0, both ax \(\to\) 0 and -bx \(\to\) 0. So, we can apply the standard limit:
\[ k = a(1) + (-b)(1) = a - b \]


Using L'Hôpital's Rule:
\[ k = \lim_{x \to 0} \frac{\frac{d}{dx}(\log(1+ax) + \log(1-bx))}{\frac{d}{dx}(x)} = \lim_{x \to 0} \frac{\frac{a}{1+ax} + \frac{-b}{1-bx}}{1} \]

Now, substitute x = 0:
\[ k = \frac{a}{1+0} - \frac{b}{1-0} = a - b \]


Step 4: Final Answer:


The value of k that makes the function continuous is a - b.
Quick Tip: For continuity problems involving piecewise functions at a point 'c', always set the function's value at 'c' equal to the limit of the function as x approaches 'c'.
L'Hôpital's Rule is often the quickest method for limits of the form 0/0 or \(\infty/\infty\).

Step 1: Understanding the Question:


We are given a function y and asked to find the value of an expression involving y and its first (y\(_1\)) and second (y\(_2\)) derivatives. This is a problem about forming a differential equation from a given solution.


Step 2: Detailed Explanation:


First, find the first derivative, y\(_1\):
\[ y_1 = \frac{dy}{dx} = a(-\sin(\log x)) \cdot \frac{1}{x} + b(\cos(\log x)) \cdot \frac{1}{x} \]

Multiply by x to simplify:
\[ xy_1 = -a \sin(\log x) + b \cos(\log x) \]


Now, differentiate this equation again with respect to x using the product rule on the left side:
\[ \frac{d}{dx}(xy_1) = \frac{d}{dx}(-a \sin(\log x) + b \cos(\log x)) \]
\[ (1)y_1 + x(y_2) = -a(\cos(\log x)) \cdot \frac{1}{x} + b(-\sin(\log x)) \cdot \frac{1}{x} \]
\[ xy_2 + y_1 = \frac{-1}{x} (a \cos(\log x) + b \sin(\log x)) \]


Multiply the entire equation by x to clear the denominator:
\[ x(xy_2 + y_1) = -(a \cos(\log x) + b \sin(\log x)) \]
\[ x^2y_2 + xy_1 = -(a \cos(\log x) + b \sin(\log x)) \]


Step 3: Final Answer:


We recognize that the term in the parenthesis on the right side is the original function y.

Therefore, \(x^2y_2 + xy_1 = -y\).
Quick Tip: This is a classic example of a Cauchy-Euler equation.
When you differentiate a function with log(x) terms, a factor of 1/x appears.
A good strategy is to multiply by x to clear the fraction before differentiating a second time. This often leads directly to the desired expression.

Step 1: Understanding the Question:


We need to evaluate an expression involving the principal values of inverse trigonometric functions.


Step 2: Key Formula or Approach:


We will use the properties and principal value ranges:

- Range of sec\(^{-1}\)(x) is [0, \(\pi\)] - {\(\pi\)/2.

- Property: sec\(^{-1}\)(-x) = \(\pi\) - sec\(^{-1}\)(x).

- Range of tan\(^{-1}\)(x) is (-\(\pi\)/2, \(\pi\)/2).


Step 3: Detailed Explanation:


First, evaluate sec\(^{-1}\)(-\(\sqrt{2}\)):
\[ sec^{-1}(-\sqrt{2}) = \pi - sec^{-1}(\sqrt{2}) \]

Since sec(\(\pi\)/4) = \(\sqrt{2}\), we have sec\(^{-1}\)(\(\sqrt{2}\)) = \(\pi\)/4.
\[ sec^{-1}(-\sqrt{2}) = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \]


Next, evaluate tan\(^{-1}\)(\(1/\sqrt{3}\)):

Since tan(\(\pi\)/6) = \(1/\sqrt{3}\), we have tan\(^{-1}\)(\(1/\sqrt{3}\)) = \(\pi\)/6.


Now, combine the results:
\[ \frac{3\pi}{4} - \frac{\pi}{6} \]

Find a common denominator (12):
\[ = \frac{9\pi}{12} - \frac{2\pi}{12} = \frac{7\pi}{12} \]


Step 4: Final Answer:


The value of the expression is \(\frac{7\pi}{12}\).
Quick Tip: Remember the rules for negative arguments in inverse trig functions.
For sin\(^{-1}\), csc\(^{-1}\), and tan\(^{-1}\), the negative sign comes out: e.g., sin\(^{-1}\)(-x) = -sin\(^{-1}\)(x).
For cos\(^{-1}\), sec\(^{-1}\), and cot\(^{-1}\), it's \(\pi\) minus the function: e.g., cos\(^{-1}\)(-x) = \(\pi\) - cos\(^{-1}\)(x).

Step 1: Understanding the Question:


We need to find the derivative \(\frac{dy}{dx}\) from an implicit equation involving an inverse trigonometric function.


Step 2: Key Formula or Approach:


The easiest way is to first simplify the equation to eliminate the inverse trig function, and then use implicit differentiation.


Step 3: Detailed Explanation:


Start with the given equation:
\[ \tan^{-1}(x^2 - y^2) = a \]

Apply the tangent function to both sides:
\[ \tan(\tan^{-1}(x^2 - y^2)) = \tan(a) \]
\[ x^2 - y^2 = \tan(a) \]

Since 'a' is a constant, tan(a) is also just a constant. Let's call it C.
\[ x^2 - y^2 = C \]

Now, differentiate both sides with respect to x:
\[ \frac{d}{dx}(x^2) - \frac{d}{dx}(y^2) = \frac{d}{dx}(C) \]
\[ 2x - 2y \frac{dy}{dx} = 0 \]

Solve for \(\frac{dy}{dx}\):
\[ 2x = 2y \frac{dy}{dx} \]
\[ \frac{dy}{dx} = \frac{2x}{2y} = \frac{x}{y} \]


Step 4: Final Answer:


The derivative \(\frac{dy}{dx}\) is \(\frac{x}{y}\).
Quick Tip: When an inverse trig function is equal to a constant, like \(f^{-1}(g(x,y)) = a\), it's almost always easier to rewrite it as \(g(x,y) = f(a)\) before differentiating.
This turns \(f(a)\) into a simple constant, making the differentiation much cleaner.

Step 1: Understanding the Question:


We need to analyze the properties of the function f(x) = x\(^2\) and identify the statement that is false.


Step 2: Detailed Explanation:


Let's evaluate each statement:

- (A) Minimum value of f does not exist.

The function f(x) = x\(^2\) is a parabola opening upwards. Its vertex is at (0, 0). The value of x\(^2\) is always greater than or equal to 0. The minimum value is 0, which occurs at x = 0. Therefore, the statement that the minimum value does not exist is incorrect.


- (B) There is no point of maximum value of f in R.

As x approaches \(\infty\) or \(-\infty\), f(x) = x\(^2\) approaches \(\infty\). The function is unbounded above, so it has no maximum value over the set of real numbers. This statement is correct.


- (C) f is continuous at x = 0.

The function f(x) = x\(^2\) is a polynomial. All polynomial functions are continuous everywhere, including at x = 0. This statement is correct.


- (D) f is differentiable at x = 0.

The derivative is f'(x) = 2x. At x = 0, the derivative f'(0) = 0 exists. Therefore, the function is differentiable at x = 0. This statement is correct.


Step 3: Final Answer:


The question asks for the incorrect statement. Statement (A) is incorrect.
Quick Tip: Remember the basic properties of the parabola \(y = ax^2\).
If a > 0, it opens upwards and has a global minimum at its vertex.
If a < 0, it opens downwards and has a global maximum at its vertex.
As a polynomial, it is continuous and differentiable everywhere.

Step 1: Understanding the Question:


We need to evaluate an integral that contains the product of e\(^x\) and a rational function. This structure suggests a special integration formula.


Step 2: Key Formula or Approach:


We will use the special integral form:
\[ \int e^x [f(x) + f'(x)] dx = e^x f(x) + C \]

Our goal is to manipulate the term \(\frac{x+5}{(x+6)^2}\) into the form \(f(x) + f'(x)\).


Step 3: Detailed Explanation:


Let's rewrite the rational function by adjusting the numerator:
\[ \frac{x+5}{(x+6)^2} = \frac{(x+6) - 1}{(x+6)^2} \]

Now, split the fraction:
\[ = \frac{x+6}{(x+6)^2} - \frac{1}{(x+6)^2} = \frac{1}{x+6} - \frac{1}{(x+6)^2} \]

Let's see if this fits the required form. Let \(f(x) = \frac{1}{x+6}\).

Now find the derivative, f'(x):
\[ f'(x) = \frac{d}{dx} (x+6)^{-1} = -1 \cdot (x+6)^{-2} = -\frac{1}{(x+6)^2} \]

The expression is exactly in the form \(f(x) + f'(x)\).

The integral becomes:
\[ \int e^x \left[ \frac{1}{x+6} + \left(-\frac{1}{(x+6)^2}\right) \right] dx \]

Applying the formula, the result is:
\[ e^x f(x) + C = e^x \cdot \frac{1}{x+6} + C = \frac{e^x}{x+6} + C \]


Step 4: Final Answer:


The value of the integral is \(\frac{e^x}{x+6} + C\).
Quick Tip: Whenever an integral contains \(e^x\) multiplied by a function, you should immediately suspect the \( \int e^x[f(x) + f'(x)]dx \) pattern.
The trick is almost always to manipulate the algebraic part to reveal the function and its derivative.

Step 1: Understanding the Question:


We are given the derivative of a function, f'(x), and an initial condition, f(1)=0. We need to find the original function f(x) by integration.


Step 2: Key Formula or Approach:


To find f(x) from f'(x), we must integrate f'(x) with respect to x.
\[ f(x) = \int f'(x) dx \]

We will use the power rule for integration: \(\int x^n dx = \frac{x^{n+1}}{n+1} + C\).


Step 3: Detailed Explanation:


First, rewrite f'(x) in a more integration-friendly form:
\[ f'(x) = 3x^2 + 6x - 4x^{-3} + 5 \]

Now, integrate term by term:
\[ f(x) = \int (3x^2 + 6x - 4x^{-3} + 5) dx \]
\[ f(x) = 3\frac{x^3}{3} + 6\frac{x^2}{2} - 4\frac{x^{-2}}{-2} + 5x + C \]
\[ f(x) = x^3 + 3x^2 + 2x^{-2} + 5x + C \]
\[ f(x) = x^3 + 3x^2 + \frac{2}{x^2} + 5x + C \]

Now, use the initial condition f(1) = 0 to find the constant C:
\[ f(1) = (1)^3 + 3(1)^2 + \frac{2}{(1)^2} + 5(1) + C = 0 \]
\[ 1 + 3 + 2 + 5 + C = 0 \implies 11 + C = 0 \implies C = -11 \]


Step 4: Final Answer:


Substitute C = -11 back into the expression for f(x):
\[ f(x) = x^3 + 3x^2 + \frac{2}{x^2} + 5x - 11 \]
Quick Tip: This is an initial value problem. The process is always:
1. Integrate the derivative to find the general form of the function with a constant C.
2. Use the given point (initial condition) to solve for C.
3. Write the final, particular solution.

Step 1: Understanding the Question:


We need to determine the order and degree of the given differential equation.


Step 2: Key Formula or Approach:


- Order: The order of the highest derivative present in the equation.

- Degree: The highest power of the highest order derivative, provided the equation is a polynomial in its derivatives. If derivatives appear inside functions like log, sin, exp, etc., the degree is not defined.


Step 3: Detailed Explanation:


The given differential equation is:
\[ \frac{d^2y}{dx^2} + 4\left(\frac{dy}{dx}\right)^2 = x \log\left(\frac{d^2y}{dx^2}\right) \]


Finding the Order:

The derivatives in the equation are \(\frac{d^2y}{dx^2}\) (second order) and \(\frac{dy}{dx}\) (first order).

The highest order is 2. Therefore, the order of the equation is 2.


Finding the Degree:

The equation contains the term \( \log\left(\frac{d^2y}{dx^2}\right) \). The highest order derivative, \(\frac{d^2y}{dx^2}\), is inside a logarithmic function.

Because of this, the differential equation cannot be expressed as a polynomial in its derivatives (y', y'', etc.).

Therefore, the degree is not defined.


Step 4: Final Answer:


The order is 2 and the degree is not defined.
Quick Tip: To determine the degree, always check if the equation is a polynomial in all its derivative terms.
If you see terms like \(\sin(y')\), \(e^{y''}\), or \(\log(y''')\), the degree is immediately "not defined".
The order is simply the highest 'tick mark' on any 'y'.

Step 1: Understanding the Question:


We need to find the optimal values of the objective function Z = x + 2y by evaluating it at the given corner points of the feasible region. Then we must check which statement is true.


Step 2: Key Formula or Approach:


The fundamental theorem of LPP states that the optimal solution (max or min) for a linear objective function over a bounded feasible region must occur at one of its corner points.


Step 3: Detailed Explanation:


We evaluate Z = x + 2y at each vertex:

- At P(\(\frac{3}{13}, \frac{24}{13}\)):


\( Z = \frac{3}{13} + 2\left(\frac{24}{13}\right) = \frac{3 + 48}{13} = \frac{51}{13} \approx 3.92 \)


- At Q(\(\frac{3}{2}, \frac{15}{4}\)):


\( Z = \frac{3}{2} + 2\left(\frac{15}{4}\right) = \frac{3}{2} + \frac{15}{2} = \frac{18}{2} = 9 \) (Maximum)


- At R(\(\frac{7}{2}, \frac{3}{4}\)):


\( Z = \frac{7}{2} + 2\left(\frac{3}{4}\right) = \frac{7}{2} + \frac{3}{2} = \frac{10}{2} = 5 \)


- At S(\(\frac{18}{7}, \frac{2}{7}\)):


\( Z = \frac{18}{7} + 2\left(\frac{2}{7}\right) = \frac{18 + 4}{7} = \frac{22}{7} \approx 3.14 \) (Minimum)


Now, we check the given statements:

(A) Z is minimum at S. This is true.

(B) Z is maximum at R. This is false (maximum is at Q).

(C) (Value at P) > (Value at Q) \(\implies\) 3.92 > 9. This is false.

(D) (Value at Q) < (Value at R) \(\implies\) 9 < 5. This is false.


Step 4: Final Answer:


The only correct statement is (A).
Quick Tip: For LPP problems where corner points are given, the task is purely computational.
Systematically substitute the coordinates of each point into the objective function.
Be careful with fraction arithmetic. The largest result is the maximum, and the smallest is the minimum.

Step 1: Understanding the Question:


We need to find the total area enclosed by the parabola y\(^2\) = x and the vertical lines x=0 and x=1.


Step 2: Key Formula or Approach:


The curve y\(^2\) = x is symmetric about the x-axis. It has an upper branch \(y = \sqrt{x}\) and a lower branch \(y = -\sqrt{x}\).

We can find the area of the upper region (in the first quadrant) and multiply it by 2 to get the total area.

Area = \(2 \times \int_{a}^{b} y \, dx\).


Step 3: Detailed Explanation:


The area is bounded from x=0 to x=1.
\[ Total Area = 2 \int_{0}^{1} \sqrt{x} \, dx = 2 \int_{0}^{1} x^{1/2} \, dx \]

Using the power rule for integration:
\[ = 2 \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{1} = 2 \left[ \frac{2}{3} x^{3/2} \right]_{0}^{1} \]
\[ = \frac{4}{3} \left[ x^{3/2} \right]_{0}^{1} \]

Apply the limits:
\[ = \frac{4}{3} (1^{3/2} - 0^{3/2}) = \frac{4}{3} (1 - 0) = \frac{4}{3} \]


Step 4: Final Answer:


The area of the region is \(\frac{4}{3}\) square units.
Quick Tip: When a curve is symmetric about the x-axis (e.g., contains only even powers of y), you can calculate the area in the top half and double it.
This avoids dealing with negative values from the lower half and simplifies the calculation.

Step 1: Understanding the Question:


We need to find the range of the magnitude of the vector \(\lambda\vec{a}\), given the magnitude of \(\vec{a}\) and the range of the scalar \(\lambda\).


Step 2: Key Formula or Approach:


We use the property of vector magnitudes:
\[ |\lambda \vec{a}| = |\lambda| |\vec{a}| \]

where \(|\lambda|\) is the absolute value of the scalar \(\lambda\).


Step 3: Detailed Explanation:


We are given \(|\vec{a}| = 5\) and \(-2 \leq \lambda \leq 1\).

First, we must find the range of \(|\lambda|\).

Since \(\lambda\) is in the interval [-2, 1], its absolute value, \(|\lambda|\), will range from its minimum possible value to its maximum possible value in that interval.

- The minimum value of \(|\lambda|\) occurs at \(\lambda = 0\), so min(\(|\lambda|\)) = 0.

- The maximum value of \(|\lambda|\) is max(|-2|, |1|) = max(2, 1) = 2.

So, the range of \(|\lambda|\) is [0, 2], or \(0 \leq |\lambda| \leq 2\).


Now, we find the range of \(|\lambda \vec{a}| = 5|\lambda|\):

Multiply the inequality for \(|\lambda|\) by 5:
\[ 5 \times 0 \leq 5|\lambda| \leq 5 \times 2 \]
\[ 0 \leq |\lambda \vec{a}| \leq 10 \]


Step 4: Final Answer:


The correct range for \(|\lambda \vec{a}|\) is [0, 10].

Since this is not an option, the question is likely flawed. If the question had asked for the range of \(\lambda|\vec{a}|\) (without the absolute value on \(\lambda\)), the answer would be the range of \(5\lambda\), which is [-10, 5], matching option (D).
Quick Tip: Pay very close attention to absolute value bars.
The magnitude of a vector \(|\vec{v}|\) is always non-negative.
The absolute value of a scalar \(|\lambda|\) is also always non-negative.
A common mistake is to confuse \(|\lambda \vec{a}|\) with \(\lambda |\vec{a}|\). They are not the same if \(\lambda\) can be negative.

Step 1: Understanding the Question:


We need to solve a first-order differential equation. The structure suggests using the method of separation of variables.


Step 2: Key Formula or Approach:


We will isolate \(\frac{dy}{dx}\), separate the x and y terms, and then integrate both sides.


Step 3: Detailed Explanation:


Start with the given equation:
\[ \log\left(\frac{dy}{dx}\right) = 3x + 4y \]

Exponentiate both sides to remove the log:
\[ \frac{dy}{dx} = e^{3x + 4y} = e^{3x} \cdot e^{4y} \]

Separate the variables by moving y terms to the left and x terms to the right:
\[ \frac{dy}{e^{4y}} = e^{3x} dx \]
\[ e^{-4y} dy = e^{3x} dx \]

Integrate both sides:
\[ \int e^{-4y} dy = \int e^{3x} dx \]
\[ \frac{e^{-4y}}{-4} = \frac{e^{3x}}{3} + C_1 \]

To match the options, clear the denominators by multiplying by -12:
\[ 3e^{-4y} = -4e^{3x} - 12C_1 \]

Rearrange all terms to one side:
\[ 3e^{-4y} + 4e^{3x} + 12C_1 = 0 \]

Letting \(12C_1\) be a new constant (represented as 12C in the option), this matches option (D).


Step 4: Final Answer:


The solution to the differential equation is \(3e^{-4y} + 4e^{3x} + 12C = 0\).
Quick Tip: If a differential equation can be written in the form \(f(y) dy = g(x) dx\), it is separable and this is usually the most direct method to solve it.
The form of the constant of integration (C, 12C, -C, etc.) can vary but doesn't change the solution. Match the variable terms first.

Step 1: Understanding the Question:


We need to analyze the given constraints to determine if a solution to the LPP exists. This requires checking for a non-empty feasible region.


Step 2: Detailed Explanation:


Let's analyze the constraints mathematically, ignoring the potentially misleading graph.

Constraint 1: \(x + y \leq 4\). This represents all points on or below the line x + y = 4.

Constraint 2: \(3x + 3y \geq 18\). Dividing by 3 simplifies this to \(x + y \geq 6\). This represents all points on or above the line x + y = 6.

Constraint 3: \(x \geq 0, y \geq 0\). This restricts points to the first quadrant.


We are looking for points (x, y) that satisfy both \(x+y \leq 4\) and \(x+y \geq 6\) simultaneously.

It is impossible for a number (the sum x+y) to be both less than or equal to 4 and greater than or equal to 6 at the same time.

These two constraints are contradictory.


Step 3: Final Answer:


Since there is no set of points (x, y) that can satisfy all constraints, the feasible region is empty.

Therefore, the LPP has no solution, and the correct option is that the solution does not exist.
Quick Tip: Always analyze the constraints mathematically before relying on a provided graph.
If the constraints are contradictory, there is no feasible region, which means there is no solution.
This is a common trick question in exams.

Step 1: Understanding the Question:


We need to find the probability of "at least one" of three independent events occurring. The easiest method is to use the complement rule.


Step 2: Key Formula or Approach:


P(at least one event) = 1 - P(no events).

The events of A, B, and C going to the market are independent.


Step 3: Detailed Explanation:


First, list the given probabilities and their complements (the probability of each person NOT going).

- P(A goes) = 30% = 0.3 \(\implies\) P(A does not go) = 1 - 0.3 = 0.7.

- P(B goes) = 60% = 0.6 \(\implies\) P(B does not go) = 1 - 0.6 = 0.4.

- P(C goes) = 50% = 0.5 \(\implies\) P(C does not go) = 1 - 0.5 = 0.5.


Now, calculate the probability that NONE of them go to the market. Since the events are independent, we multiply their probabilities:

P(none go) = P(A not) \(\times\) P(B not) \(\times\) P(C not)
\[ P(none go) = 0.7 \times 0.4 \times 0.5 = 0.14 \]


Finally, use the complement rule to find the probability that at least one goes:

P(at least one goes) = 1 - P(none go)
\[ P(at least one goes) = 1 - 0.14 = 0.86 \]


Step 4: Final Answer:


Convert the decimal to a fraction to match the options:
\(0.86 = \frac{86}{100} = \frac{43}{50}\).
Quick Tip: For probability questions with the phrase "at least one," the complement rule (1 - P(none)) is almost always the most efficient solution method.
It saves you from calculating and adding the probabilities of one person going, two people going, and all three people going.

Step 1: Understanding the Question:


We need to check the validity of an assertion about vector magnitudes using Lagrange's identity, and determine if the provided reason correctly explains it.


Step 2: Key Formula or Approach:


The statements in the Reason lead to Lagrange's identity: \(|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2|\vec{b}|^2\).


Step 3: Detailed Explanation:


First, analyze the Reason (R). All statements within it are fundamental and correct definitions from vector algebra and trigonometry. Thus, Reason (R) is true.

Now, let's use the reason to prove the identity:
\[ |\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = (|\vec{a}||\vec{b}|\sin\theta)^2 + (|\vec{a}||\vec{b}|\cos\theta)^2 \]
\[ = |\vec{a}|^2|\vec{b}|^2(\sin^2\theta + \cos^2\theta) = |\vec{a}|^2|\vec{b}|^2(1) = |\vec{a}|^2|\vec{b}|^2 \]

This shows that the Reason directly explains the identity needed.

Now, check the Assertion (A) using this identity:
\[ |\vec{a}|^2|\vec{b}|^2 = 256 \]

Substitute \(|\vec{b}| = 8 \implies |\vec{b}|^2 = 64\):
\[ |\vec{a}|^2 (64) = 256 \implies |\vec{a}|^2 = \frac{256}{64} = 4 \implies |\vec{a}| = 2 \]

The Assertion (A) is also true.


Step 4: Final Answer:


Both statements are true, and the Reason provides the exact definitions needed to derive the result in the Assertion. Therefore, (A) is the correct choice.
Quick Tip: Lagrange's identity is a powerful tool connecting the dot product, cross product, and magnitudes.
Recognizing it immediately confirms the relationship in the Assertion.
For A/R questions, always check if the Reason is the direct cause or formula behind the Assertion.

Step 1: Understanding the Question:


We need to check an assertion about the domain of the sum of two functions, and the reason providing the general rule for it.


Step 2: Detailed Explanation:


First, analyze the Reason (R). The rule that the domain of the sum of two functions is the intersection of their individual domains is correct. Thus, Reason (R) is true.


Now, let's use the rule from the Reason to check the Assertion (A).

- For f(x) = e\(^x\), the domain is all real numbers, Dom(f) = R.

- For g(x) = log x, the domain is all positive real numbers, Dom(g) = (0, \(\infty\)).


The domain of the sum (f + g) is the intersection:

Dom(f + g) = Dom(f) \(\cap\) Dom(g) = R \(\cap\) (0, \(\infty\)) = (0, \(\infty\)).


The Assertion (A) states that the domain of (f + g) is R. This is false.


Step 3: Final Answer:


Assertion (A) is false, but Reason (R) is true.
Quick Tip: The domain of a combined function is always restricted by the function with the more restrictive domain.
In this case, the `log x` term limits the domain of the sum to only positive numbers.
Always find the domain of each part separately before finding the intersection.

Step 1: Understanding the Question:


We are asked to find the derivative of one function with respect to another. This is a form of parametric differentiation.


Step 2: Key Formula or Approach:


Let \(u = e^{\sqrt{2x}}\) and \(v = \sqrt{e^{\sqrt{2x}}}\). We need to find \(\frac{du}{dv}\).

The simplest approach is to first express u directly in terms of v.


Step 3: Detailed Explanation:


By observing the expressions for u and v, we can see that:
\[ v = \sqrt{u} \]

Squaring both sides gives:
\[ v^2 = u \]

So, we have \(u = v^2\).

Now, we can differentiate u with respect to v:
\[ \frac{du}{dv} = \frac{d}{dv}(v^2) = 2v \]

Substitute the original expression for v back into the result:
\[ \frac{du}{dv} = 2\sqrt{e^{\sqrt{2x}}} \]


Step 4: Final Answer:


The derivative of \(e^{\sqrt{2x}}\) with respect to \(\sqrt{e^{\sqrt{2x}}}\) is \(2\sqrt{e^{\sqrt{2x}}}\).
Quick Tip: When asked to differentiate f(x) with respect to g(x), always check for a simple algebraic relationship between f and g first.
If you can write f as a function of g (or vice versa), the differentiation becomes much simpler than using the chain rule \(\frac{df/dx}{dg/dx}\).

Step 1: Understanding the Question:


We need to find \(\frac{dy}{dx}\) for an implicit equation where variables appear in both the base and the exponent. This requires logarithmic differentiation.


Step 2: Key Formula or Approach:


Take the natural logarithm (ln) of both sides to bring the exponents down, then differentiate implicitly with respect to x using the product rule.


Step 3: Detailed Explanation:


Start with the equation:
\[ x^y = y^x \]

Take the natural logarithm of both sides:
\[ \ln(x^y) = \ln(y^x) \]
\[ y \ln x = x \ln y \]

Now, differentiate both sides with respect to x, using the product rule:
\[ \frac{d}{dx}(y \ln x) = \frac{d}{dx}(x \ln y) \]
\[ \left(\frac{dy}{dx} \cdot \ln x + y \cdot \frac{1}{x}\right) = \left(1 \cdot \ln y + x \cdot \frac{1}{y} \cdot \frac{dy}{dx}\right) \]

Group all terms with \(\frac{dy}{dx}\) on one side:
\[ \frac{dy}{dx} \ln x - \frac{x}{y} \frac{dy}{dx} = \ln y - \frac{y}{x} \]

Factor out \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} \left(\ln x - \frac{x}{y}\right) = \ln y - \frac{y}{x} \]
\[ \frac{dy}{dx} \left(\frac{y \ln x - x}{y}\right) = \frac{x \ln y - y}{x} \]

Isolate \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{y(x \ln y - y)}{x(y \ln x - x)} \]


Step 4: Final Answer:


The derivative is \(\frac{dy}{dx} = \frac{y(x \ln y - y)}{x(y \ln x - x)}\).
Quick Tip: Logarithmic differentiation is essential for functions of the form \(f(x)^{g(x)}\).
The process is always:
1. Take ln of both sides.
2. Use log properties to simplify.
3. Differentiate implicitly, remembering the product rule and chain rule.
4. Isolate \(\frac{dy}{dx}\).

Step 1: Understanding the Question:


We are given the position vectors for points A (\(\vec{a}\)) and B (\(\vec{b}\)).

Point C lies on the line passing through B and A, extended beyond A.

The condition is given by the vector equation \(\vec{BC} = 3\vec{BA}\). We need to find the position vector of C, let's call it \(\vec{c}\).


Step 2: Key Formula or Approach:


We express the vectors in terms of the position vectors of their endpoints:
\(\vec{BC} = \vec{c} - \vec{b}\)
\(\vec{BA} = \vec{a} - \vec{b}\)

We will substitute these into the given equation and solve for \(\vec{c}\).


Step 3: Detailed Explanation:


The given vector relation is:
\[ \vec{BC} = 3\vec{BA} \]

Substitute the expressions from Step 2:
\[ \vec{c} - \vec{b} = 3(\vec{a} - \vec{b}) \]

Distribute the scalar 3 on the right side:
\[ \vec{c} - \vec{b} = 3\vec{a} - 3\vec{b} \]

To solve for \(\vec{c}\), add \(\vec{b}\) to both sides of the equation:
\[ \vec{c} = 3\vec{a} - 3\vec{b} + \vec{b} \]
\[ \vec{c} = 3\vec{a} - 2\vec{b} \]


Step 4: Final Answer:


The position vector of point C is \(3\vec{a} - 2\vec{b}\).
Quick Tip: Remember the formula for a vector between two points: the vector from point P to point Q is given by (position vector of Q) - (position vector of P).
Drawing a simple diagram can also help visualize the relative positions of A, B, and C to confirm your answer. C is an external point dividing the line segment.

Step 1: Understanding the Question:


A vector is inclined at equal angles to the coordinate axes, which gives information about its direction cosines.

We are also given its magnitude. We need to find the vector itself.


Step 2: Key Formula or Approach:


Let the equal angles be \(\alpha\). The direction cosines are l = cos(\(\alpha\)), m = cos(\(\alpha\)), n = cos(\(\alpha\)). So, l = m = n.

The fundamental identity for direction cosines is \(l^2 + m^2 + n^2 = 1\).

A vector \(\vec{r}\) can be expressed as \(\vec{r} = |\vec{r}| (l\hat{i} + m\hat{j} + n\hat{k})\).


Step 3: Detailed Explanation:


Since l = m = n, the identity becomes:
\[ l^2 + l^2 + l^2 = 1 \implies 3l^2 = 1 \implies l^2 = \frac{1}{3} \implies l = \pm \frac{1}{\sqrt{3}} \]

So, the direction cosines are \(l=m=n = \pm \frac{1}{\sqrt{3}}\).

We are given the magnitude \(|\vec{r}| = 5\sqrt{3}\).

Now we can write the vector \(\vec{r}\):
\[ \vec{r} = |\vec{r}| (l\hat{i} + m\hat{j} + n\hat{k}) \]
\[ \vec{r} = 5\sqrt{3} \left( \pm \frac{1}{\sqrt{3}}\hat{i} \pm \frac{1}{\sqrt{3}}\hat{j} \pm \frac{1}{\sqrt{3}}\hat{k} \right) \]

The sign must be the same for all components.
\[ \vec{r} = \pm \frac{5\sqrt{3}}{\sqrt{3}} (\hat{i} + \hat{j} + \hat{k}) = \pm 5 (\hat{i} + \hat{j} + \hat{k}) \]


Step 4: Final Answer:


The vector \(\vec{r}\) is \(5\hat{i} + 5\hat{j} + 5\hat{k}\) or \(-5\hat{i} - 5\hat{j} - 5\hat{k}\).
Quick Tip: A vector equally inclined to the coordinate axes is always parallel to the vector \(\hat{i} + \hat{j} + \hat{k}\) or its negative.
Its unit vector is always \(\pm \frac{1}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k})\). This can be used as a shortcut.

Step 1: Understanding the Question:


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

- f is increasing where f'(x) > 0.

- f is decreasing where f'(x) < 0.


Step 2: Key Formula or Approach:


We will differentiate f(x) with respect to x using the power rule.

Rewrite \(f(x) = 2x^{-1} - 5\).


Step 3: Detailed Explanation:


Find the derivative f'(x):
\[ f'(x) = \frac{d}{dx}(2x^{-1} - 5) = 2(-1)x^{-2} - 0 = -2x^{-2} = -\frac{2}{x^2} \]

Now, we analyze the sign of f'(x).

- The numerator is -2, which is always negative.

- The denominator is \(x^2\). For any non-zero real number x, \(x^2\) is always positive.

Therefore, f'(x) = \(\frac{negative}{positive}\), which is always negative for all x in the domain R - \{0\.


Step 4: Final Answer:


Since f'(x) < 0 for all x \(\neq\) 0, the function is decreasing throughout its domain.

It is decreasing on the intervals (-\(\infty\), 0) and (0, \(\infty\)).
Quick Tip: The sign of the first derivative tells you the behavior of the original function.
Always be careful with the domain of the function. For \(f(x) = 2/x\), the point x=0 is a discontinuity, so you should state the intervals of increase/decrease separately.

Step 1: Understanding the Question:


We need to find the set of all valid input values (x-values) for which the function f(x) is defined.


Step 2: Key Formula or Approach:


The domain of the standard inverse sine function, sin\(^{-1}\)(u), is the interval [-1, 1].

This means the argument 'u' must satisfy the inequality: \(-1 \leq u \leq 1\).

In this case, the argument is \(u = -x^2\).


Step 3: Detailed Explanation:


We apply the domain constraint to the argument of our function:
\[ -1 \leq -x^2 \leq 1 \]

This is a compound inequality, which we can split into two parts:

1) \(-1 \leq -x^2 \implies x^2 \leq 1 \implies |x| \leq 1 \implies -1 \leq x \leq 1\).

2) \(-x^2 \leq 1 \implies x^2 \geq -1\). This is true for all real numbers x, since a square is always non-negative.

The domain is the intersection of the solutions of both parts.

The intersection of [-1, 1] and (\(-\infty, \infty\)) is [-1, 1].


Step 4: Final Answer:


The domain of the function f(x) = sin\(^{-1}\)(– x\(^2\)) is [-1, 1].
Quick Tip: To find the domain of a composite function, start with the outer function.
Identify the domain requirement for the outer function (here, sin\(^{-1}\)) and apply it as an inequality to the inner function (here, -x\(^2\)).
Then, solve the resulting inequality for x.

Step 1: Understand the Condition for Perpendicular Lines


Two lines are perpendicular if and only if the dot product of their direction vectors (or direction ratios) is equal to zero.

Our first task is to find the correct direction ratios for each line by converting their equations to the standard form \(\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}\).


Step 2: Standardize the Line Equations


For line \(l_1\):
\[ \frac{1-x}{-3} = \frac{-(x-1)}{-3} = \frac{x-1}{3} \]
\[ \frac{3y-2}{2\lambda} = \frac{3(y-2/3)}{2\lambda} = \frac{y-2/3}{2\lambda/3} \]
\[ \frac{z-3}{3} is already in standard form. \]

So, the direction ratios for \(l_1\) are \(\langle a_1, b_1, c_1 \rangle = \langle 3, \frac{2\lambda}{3}, 3 \rangle\).


For line \(l_2\):
\[ \frac{x-1}{3\lambda} is already in standard form. \]
\[ \frac{1-y}{1} = \frac{-(y-1)}{1} = \frac{y-1}{-1} \]
\[ \frac{2z-5}{3} = \frac{2(z-5/2)}{3} = \frac{z-5/2}{3/2} \]

So, the direction ratios for \(l_2\) are \(\langle a_2, b_2, c_2 \rangle = \langle 3\lambda, -1, \frac{3}{2} \rangle\).


Step 3: Apply the Perpendicularity Condition


The condition for perpendicular lines is \(a_1a_2 + b_1b_2 + c_1c_2 = 0\).

Substituting the direction ratios we found:
\[ (3)(3\lambda) + \left(\frac{2\lambda}{3}\right)(-1) + (3)\left(\frac{3}{2}\right) = 0 \]


Step 4: Solve for \(\lambda\)

\[ 9\lambda - \frac{2\lambda}{3} + \frac{9}{2} = 0 \]

To simplify, let's move the constant term to the right side:
\[ 9\lambda - \frac{2\lambda}{3} = -\frac{9}{2} \]

Combine the terms with \(\lambda\) using a common denominator:
\[ \frac{27\lambda - 2\lambda}{3} = -\frac{9}{2} \]
\[ \frac{25\lambda}{3} = -\frac{9}{2} \]

Cross-multiply to solve for \(\lambda\):
\[ 2 \times 25\lambda = 3 \times (-9) \]
\[ 50\lambda = -27 \]
\[ \lambda = -\frac{27}{50} \]
Quick Tip: The most critical step in problems involving direction ratios is to ensure the line equations are in standard form.
The coefficients of x, y, and z in the numerators \textbf{must be +1}.
For example, \(\frac{1-x}{a}\) must be converted to \(\frac{x-1}{-a}\) before identifying the direction ratio.
Failing to do this is the most common source of error.

Step 1: Understand the Order of Operations and Dimensions


We need to calculate the product of three matrices: A, B, and C.

Matrix multiplication is associative, so we can compute this as (AB)C.

First, let's check the dimensions (orders) of the matrices:

- Order of A: 1 \(\times\) 3

- Order of B: 3 \(\times\) 3

- Order of C: 3 \(\times\) 1

The multiplications are possible, and the final result will be a (1 \(\times\) 3) \(\cdot\) (3 \(\times\) 1) \(\rightarrow\) 1 \(\times\) 1 matrix.


Step 2: Calculate the product AB


We multiply the row matrix A by the square matrix B. The result will be a 1 \(\times\) 3 matrix.
\[ AB = [1 \quad -1 \quad 0] \begin{bmatrix} 2 & 0 & 1
-1 & 3 & 4
0 & 5 & 1 \end{bmatrix} \]

Performing the row-by-column multiplication:
\[ = [(1)(2) + (-1)(-1) + (0)(0) \quad (1)(0) + (-1)(3) + (0)(5) \quad (1)(1) + (-1)(4) + (0)(1)] \]
\[ = [2+1+0 \quad 0-3+0 \quad 1-4+0] \]
\[ = [3 \quad -3 \quad -3] \]


Step 3: Calculate the final product (AB)C


Now we multiply the resulting 1 \(\times\) 3 matrix (AB) by the column matrix C. The result will be a 1 \(\times\) 1 matrix.
\[ (AB)C = [3 \quad -3 \quad -3] \begin{bmatrix} 2
3
4 \end{bmatrix} \]

Performing the multiplication:
\[ = [(3)(2) + (-3)(3) + (-3)(4)] \]
\[ = [6 - 9 - 12] \]
\[ = [-3 - 12] \]
\[ = [-15] \]


Step 4: Final Answer:


The product of the three matrices, ABC, is the 1 \(\times\) 1 matrix \([-15]\).
Quick Tip: Matrix multiplication is associative, meaning \( (AB)C = A(BC) \). You can choose the order that seems easier.
Always verify that the "inner dimensions" match before you multiply. For AB, the inner dimensions are 3 and 3. For (AB)C, they are 3 and 3.
The final result's dimension is determined by the "outer dimensions". For (AB)C, this is (1 \(\times\) 3) \(\cdot\) (3 \(\times\) 1) \(\rightarrow\) 1 \(\times\) 1.

Step 1: Identify the Problem and Boundary Lines


The problem is to minimize Z = x + 4y subject to the given constraints.

First, we plot the boundary lines:

- L1: \(2x + y = 1000\). Intercepts are (500, 0) and (0, 1000).

- L2: \(x + 2y = 800\). Intercepts are (800, 0) and (0, 400).

The feasible region is the unbounded area on or above both lines in the first quadrant.


Step 2: Find the Corner Points of the Feasible Region


The vertices of the feasible region are:

- Point A (y-intercept): The highest y-intercept on the boundary is from L1. A = (0, 1000).

- Point B (Intersection): We solve the system \(2x + y = 1000\) and \(x + 2y = 800\).

This yields the point B = (400, 200).

- Point C (x-intercept): The rightmost x-intercept on the boundary is from L2. C = (800, 0).


Step 3: Evaluate the Objective Function at Corner Points


We evaluate Z = x + 4y at each vertex:

- At A(0, 1000): \(Z = 0 + 4(1000) = 4000\)

- At B(400, 200): \(Z = 400 + 4(200) = 1200\)

- At C(800, 0): \(Z = 800 + 4(0) = 800\)

The minimum value among the vertices is M = 800.


Step 4: Check for Unbounded Region


Since the region is unbounded, we must test if a value smaller than 800 is possible by checking the open half-plane \(x + 4y < 800\).

For any point (x, y) in the feasible region, we know \(x + 2y \geq 800\). Since \(y \geq 0\), it follows that \(2y \geq 0\).

Therefore, \(Z = x + 4y = (x + 2y) + 2y \geq 800 + 2y \geq 800\).

This confirms that no value of Z can be less than 800. The minimum is valid.


Step 5: Final Answer:


The minimum value of Z is 800.




\begin{figure[h!]
\centering
\begin{tikzpicture[
scale=0.9,
x=0.007cm, y=0.007cm, % Scale coordinates to fit page
every node/.style={font=\small
]

% Draw axes
\draw[->, thick] (0,0) -- (1150,0) node[right] {\(x\);
\draw[->, thick] (0,0) -- (0,1150) node[above] {\(y\);

% Draw ticks and labels for axes
\foreach \x in {200, 400, 600, 800, 1000 {
\draw (\x, 15) -- (\x, -15) node[below] {\x;

\foreach \y in {200, 400, 600, 800, 1000 {
\draw (15, \y) -- (-15, \y) node[left] {\y;

\node[below left] at (0,0) {0;

% Define coordinates for corner points
\coordinate (A) at (0, 1000);
\coordinate (B) at (400, 200);
\coordinate (C) at (800, 0);

% Shade the feasible region (unbounded)
\fill[cyan!20, opacity=0.6] (A) -- (B) -- (C) -- (1150,0) -- (1150,1150) -- (0,1150) -- cycle;

% Draw the boundary lines of the constraints
% Line 1: 2x + y = 1000 (intercepts at (500,0) and (0,1000))
\draw[blue, thick] (0, 1000) -- (500, 0)
node[pos=0.7, above, sloped, rotate=0, black, xshift=-15pt] {\(2x+y \geq 1000\);

% Line 2: x + 2y = 800 (intercepts at (800,0) and (0,400))
\draw[red, thick] (0, 400) -- (800, 0)
node[pos=0.3, below, sloped, rotate=0, black, xshift=0pt] {\(x+2y \geq 800\);

% Highlight the boundary of the feasible region
\draw[black, ultra thick] (A) -- (B) -- (C);

% Mark and label the corner points
\filldraw[black] (A) circle (3pt) node[above right=2pt] {A(0, 1000);
\filldraw[black] (B) circle (3pt) node[below right=5pt] {B(400, 200);
\filldraw[black] (C) circle (3pt) node[above right=2pt] {C(800, 0);

% Label the feasible region
\node[text=black, align=center] at (750, 600) {Feasible Region;

\end{tikzpicture
\end{figure Quick Tip: For minimization problems with an unbounded feasible region, the minimum value M found at a vertex is only valid if the open half-plane \(Z < M\) has no points in common with the feasible region.
Always perform this check.

Step 1: Understanding the Problem


We need to find the shortest distance from a given point to a given line in 3D space.


Step 2: Key Formula or Approach


The distance (d) of a point P with position vector \(\vec{p}\) from a line \(\vec{r} = \vec{a} + \lambda\vec{b}\) is given by:
\[ d = \frac{|(\vec{p} - \vec{a}) \times \vec{b}|}{|\vec{b}|} \]

From the line equation, a point on the line is A(-5, -3, 6) and the direction vector is \(\vec{b} = \hat{i} + 4\hat{j} - 9\hat{k}\).


Step 3: Detailed Explanation


The position vector of P is \(\vec{p} = 2\hat{i} + 4\hat{j} - \hat{k}\).

The position vector of A is \(\vec{a} = -5\hat{i} - 3\hat{j} + 6\hat{k}\).

First, calculate the vector \(\vec{AP} = \vec{p} - \vec{a}\):
\[ \vec{p} - \vec{a} = (2 - (-5))\hat{i} + (4 - (-3))\hat{j} + (-1 - 6)\hat{k} = 7\hat{i} + 7\hat{j} - 7\hat{k} \]

Next, calculate the cross product \((\vec{p} - \vec{a}) \times \vec{b}\):
\[ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
7 & 7 & -7
1 & 4 & -9 \end{vmatrix} = \hat{i}(-63 + 28) - \hat{j}(-63 + 7) + \hat{k}(28 - 7) = -35\hat{i} + 56\hat{j} + 21\hat{k} \]

Find the magnitude of this cross product:
\[ |(\vec{p} - \vec{a}) \times \vec{b}| = \sqrt{(-35)^2 + 56^2 + 21^2} = \sqrt{1225 + 3136 + 441} = \sqrt{4802} \]

Find the magnitude of the direction vector \(\vec{b}\):
\[ |\vec{b}| = \sqrt{1^2 + 4^2 + (-9)^2} = \sqrt{1 + 16 + 81} = \sqrt{98} \]

Finally, calculate the distance:
\[ d = \frac{\sqrt{4802}}{\sqrt{98}} = \sqrt{\frac{4802}{98}} = \sqrt{49} = 7 \]


Step 4: Final Answer:


The distance of the point P from the line is 7 units.
Quick Tip: The vector formula \(d = \frac{|\vec{AP} \times \vec{b}|}{|\vec{b}|}\) is generally the fastest method for this type of problem.
The numerator represents the area of the parallelogram formed by vectors \(\vec{AP}\) and \(\vec{b}\). Dividing by the base \(|\vec{b}|\) gives the height, which is the perpendicular distance.

Step 1: Understanding the Problem


We need to find the equation of a line that passes through a given point A and has a direction parallel to the vector \(\vec{BC}\).


Step 2: Key Formula or Approach


The vector equation of a line is \(\vec{r} = \vec{a} + \lambda\vec{d}\), where \(\vec{a}\) is the position vector of a point on the line and \(\vec{d}\) is its direction vector.


Step 3: Detailed Explanation


The line passes through point A, so its position vector is given:
\[ \vec{a} = 3\hat{i} - \hat{j} - 2\hat{k} \]

The line is parallel to BC, so its direction vector \(\vec{d}\) is \(\vec{BC}\):
\[ \vec{d} = \vec{BC} = (position vector of C) - (position vector of B) \]
\[ \vec{d} = (\hat{i} + 5\hat{j} + 3\hat{k}) - (\hat{i} + 2\hat{j} - \hat{k}) = 0\hat{i} + 3\hat{j} + 4\hat{k} \]

Vector Equation:
\[ \vec{r} = (3\hat{i} - \hat{j} - 2\hat{k}) + \lambda(3\hat{j} + 4\hat{k}) \]

Cartesian Equation:

The point is A(3, -1, -2) and the direction ratios are \(<0, 3, 4>\).

The equation is \(\frac{x-3}{0} = \frac{y-(-1)}{3} = \frac{z-(-2)}{4}\).

When a direction ratio is 0, we write it separately:
\[ x-3 = 0 \implies x = 3 \]

So the equations are \(x = 3, \frac{y+1}{3} = \frac{z+2}{4}\).


Step 4: Final Answer:


Vector Equation: \(\vec{r} = (3\hat{i} - \hat{j} - 2\hat{k}) + \lambda(3\hat{j} + 4\hat{k})\).

Cartesian Equations: \(x = 3, \frac{y+1}{3} = \frac{z+2}{4}\).
Quick Tip: When a direction ratio in a Cartesian equation is zero (e.g., for the x-component), it signifies that the line is parallel to the yz-plane.
This means all points on the line share the same x-coordinate, which is expressed as \(x = x_1\).

Step 1: Understanding the Question:


The expression inside the sin\(^{-1}\) function matches the triple angle formula for sine, suggesting a trigonometric substitution.


Step 2: Key Formula or Approach:


We use the identity: \(\sin(3\theta) = 3\sin\theta - 4\sin^3\theta\).

Let \(x = \sin\theta\), which means \(\theta = \sin^{-1}x\).

We must check that the given domain for x keeps \(3\theta\) within the principal value branch of sin\(^{-1}\), which is [-\(\pi/2\), \(\pi/2\)].


Step 3: Detailed Explanation:


Given \(-\frac{1}{2} < x < \frac{1}{2}\), we substitute \(x = \sin\theta\):
\[ -\frac{1}{2} < \sin\theta < \frac{1}{2} \implies -\frac{\pi}{6} < \theta < \frac{\pi}{6} \]

Multiplying by 3 gives:
\[ -\frac{\pi}{2} < 3\theta < \frac{\pi}{2} \]

This range is valid for the simplification.

Substitute into the function:
\[ y = \sin^{-1}(3\sin\theta - 4\sin^3\theta) = \sin^{-1}(\sin(3\theta)) \]

Since \(3\theta\) is in the principal branch, we have \(y = 3\theta\).

Substitute back \(\theta = \sin^{-1}x\):
\[ y = 3\sin^{-1}x \]

Now, differentiate:
\[ \frac{dy}{dx} = 3 \cdot \frac{d}{dx}(\sin^{-1}x) = \frac{3}{\sqrt{1-x^2}} \]


Step 4: Final Answer:


The derivative is \(\frac{dy}{dx} = \frac{3}{\sqrt{1-x^2}}\).
Quick Tip: Recognizing trigonometric identities is essential for simplifying the differentiation of inverse trig functions.
The expression \(3x - 4x^3\) is a strong indicator for the substitution \(x = \sin\theta\).
Always verify the domain to ensure the simplification \( \sin^{-1}(\sin u) = u \) is valid.

Step 1: Understanding the Question:


The expression inside the cos\(^{-1}\) function matches the double angle formula for cosine, suggesting a trigonometric substitution.


Step 2: Key Formula or Approach:


We use the identity: \(\cos(2\theta) = \frac{1-\tan^2\theta}{1+\tan^2\theta}\).

Let \(x = \tan\theta\), which means \(\theta = \tan^{-1}x\).

We check the domain of x to ensure \(2\theta\) is in the principal value branch of cos\(^{-1}\), which is [0, \(\pi\)].


Step 3: Detailed Explanation:


Given \(0 < x < 1\), we substitute \(x = \tan\theta\):
\[ 0 < \tan\theta < 1 \implies 0 < \theta < \frac{\pi}{4} \]

Multiplying by 2 gives:
\[ 0 < 2\theta < \frac{\pi}{2} \]

This range is valid for the simplification.

Substitute into the function:
\[ y = \cos^{-1}\left(\frac{1-\tan^2\theta}{1+\tan^2\theta}\right) = \cos^{-1}(\cos(2\theta)) \]

Since \(2\theta\) is in the principal branch, we have \(y = 2\theta\).

Substitute back \(\theta = \tan^{-1}x\):
\[ y = 2\tan^{-1}x \]

Now, differentiate:
\[ \frac{dy}{dx} = 2 \cdot \frac{d}{dx}(\tan^{-1}x) = \frac{2}{1+x^2} \]


Step 4: Final Answer:


The derivative is \(\frac{dy}{dx} = \frac{2}{1+x^2}\).
Quick Tip: The form \(\frac{1-x^2}{1+x^2}\) is a classic indicator for the substitution \(x = \tan\theta\).
Knowing your double and triple angle formulas for sin, cos, and tan is key to solving these problems quickly.

Step 1: Domain and Range


The relation R consists of pairs (x, y) such that y = 41 - 2x, with x, y \(\in\) N (natural numbers \(\geq 1\)).

For y to be a natural number, \(41 - 2x \geq 1 \implies 40 \geq 2x \implies 20 \geq x\).

Since x must also be a natural number, the domain is Domain = {1, 2, 3, ..., 20}.

The corresponding y values range from y(1)=39 down to y(20)=1.

The range is the set of odd numbers from 1 to 39. Range = {1, 3, 5, ..., 39}.


Step 2: Checking Properties


- Reflexive: For (a, a) \(\in\) R, we need 2a + a = 41, or 3a = 41. This gives a = 41/3, which is not a natural number. Thus, the relation is not reflexive.


- Symmetric: If (a, b) \(\in\) R, then (b, a) \(\in\) R.

Let's take a counterexample. (1, 39) is in R because 2(1) + 39 = 41.

But for (39, 1) to be in R, we need 2(39) + 1 = 41, which is false (79 \(\neq\) 41). Thus, the relation is not symmetric.


- Transitive: If (a, b) \(\in\) R and (b, c) \(\in\) R, then (a, c) \(\in\) R.

Let (a, b) \(\in\) R \(\implies\) b = 41 - 2a. Let (b, c) \(\in\) R \(\implies\) c = 41 - 2b.

Substitute b: \(c = 41 - 2(41 - 2a) = 4a - 41\).

For (a, c) to be in R, we need 2a + c = 41. Let's check: \(2a + (4a - 41) = 6a - 41\). This is not equal to 41 in general. Thus, the relation is not transitive.


Step 3: Final Answer:


An equivalence relation must be reflexive, symmetric, AND transitive. Since this relation is none of these, it is not an equivalence relation.
Quick Tip: To disprove a property (reflexive, symmetric, or transitive), you only need to find one single counterexample.
To prove it, you must show it holds true for all general cases.

Step 1: Understanding the Question


To prove a function is a bijection, we must prove it is both one-one (injective) and onto (surjective).


Step 2: Proving One-one (Injective)


We consider three cases for inputs n\(_1\) and n\(_2\).

- Case 1 (Both even): If f(n\(_1\)) = f(n\(_2\)), then n\(_1\) - 1 = n\(_2\) - 1 \(\implies\) n\(_1\) = n\(_2\).

- Case 2 (Both odd): If f(n\(_1\)) = f(n\(_2\)), then n\(_1\) + 1 = n\(_2\) + 1 \(\implies\) n\(_1\) = n\(_2\).

- Case 3 (One even, one odd): Let n\(_1\) be even and n\(_2\) be odd. Then f(n\(_1\)) = n\(_1\) - 1 (which is odd) and f(n\(_2\)) = n\(_2\) + 1 (which is even). An odd number can never equal an even number, so f(n\(_1\)) \(\neq\) f(n\(_2\)).

In all cases, different inputs give different outputs. Thus, f is one-one.


Step 3: Proving Onto (Surjective)


We must show that for any natural number y in the codomain, there exists a pre-image n in the domain.

- Case 1 (y is odd): We are looking for an input n such that f(n) = y. Since y is odd, the input must have been even (as f(even) = even-1 = odd). So, we solve \(n-1 = y \implies n = y+1\). If y is an odd natural number, n = y+1 is an even natural number. So a pre-image exists.

- Case 2 (y is even): We are looking for an input n such that f(n) = y. Since y is even, the input must have been odd (as f(odd) = odd+1 = even). So, we solve \(n+1 = y \implies n = y-1\). If y is an even natural number (\(\geq 2\)), n = y-1 is an odd natural number. So a pre-image exists.

Every natural number in the codomain has a pre-image. Thus, f is onto.


Step 4: Final Answer:


Since the function is both one-one and onto, it is a bijection.
Quick Tip: For piecewise functions defined by properties like even/odd, always structure your proofs for injectivity and surjectivity into cases.
For surjectivity (onto), work backwards: pick an arbitrary 'y' from the codomain and find the 'x' from the domain that maps to it.

Step 1: Find Probabilities for a Single Toss


Let P(H) be the probability of getting a Head and P(T) be the probability of getting a Tail.

We are given P(H) = 3 * P(T).

Since P(H) + P(T) = 1, we have 3 * P(T) + P(T) = 1 \(\implies\) 4 * P(T) = 1.

So, P(T) = 1/4 and P(H) = 3/4.


Step 2: Define the Random Variable and Find its Probability Distribution


Let X be the random variable representing the number of tails in 3 tosses. X can take values {0, 1, 2, 3. This is a binomial distribution with n=3 and p=P(T)=1/4.


The probability mass function is \(P(X=k) = \binom{3}{k} (\frac{1}{4})^k (\frac{3}{4})^{3-k}\).


- P(X=0): (No tails, HHH) \( = \binom{3}{0} (\frac{1}{4})^0 (\frac{3}{4})^3 = 1 \cdot 1 \cdot \frac{27}{64} = \frac{27}{64} \)


- P(X=1): (One tail, T HH, HT H, HHT) \( = \binom{3}{1} (\frac{1}{4})^1 (\frac{3}{4})^2 = 3 \cdot \frac{1}{4} \cdot \frac{9}{16} = \frac{27}{64} \)


- P(X=2): (Two tails, TTH, THT, HTT) \( = \binom{3}{2} (\frac{1}{4})^2 (\frac{3}{4})^1 = 3 \cdot \frac{1}{16} \cdot \frac{3}{4} = \frac{9}{64} \)


- P(X=3): (Three tails, TTT) \( = \binom{3}{3} (\frac{1}{4})^3 (\frac{3}{4})^0 = 1 \cdot \frac{1}{64} \cdot 1 = \frac{1}{64} \)


The probability distribution is:


\begin{tabular{|c|c|c|c|c|
\hline
X & 0 & 1 & 2 & 3

\hline
P(X) & 27/64 & 27/64 & 9/64 & 1/64

\hline
\end{tabular


Step 3: Calculate the Mean of the Distribution


The mean (\(\mu\) or E[X]) is given by the formula \(\mu = \sum x_i P(x_i)\).
\[ \mu = (0)\left(\frac{27}{64}\right) + (1)\left(\frac{27}{64}\right) + (2)\left(\frac{9}{64}\right) + (3)\left(\frac{1}{64}\right) \]
\[ \mu = 0 + \frac{27}{64} + \frac{18}{64} + \frac{3}{64} = \frac{27 + 18 + 3}{64} = \frac{48}{64} \]

Simplify the fraction:
\[ \mu = \frac{3 \times 16}{4 \times 16} = \frac{3}{4} \]

Alternatively, for a binomial distribution, the mean is simply \(n \cdot p = 3 \cdot (1/4) = 3/4\).


Step 4: Final Answer:


The probability distribution is P(X=0)=27/64, P(X=1)=27/64, P(X=2)=9/64, P(X=3)=1/64.

The mean of the distribution is 3/4.
Quick Tip: For binomial distributions (fixed number of independent trials with two outcomes), you can use shortcuts for the mean and variance.
Mean (\(\mu\)) = np.
Variance (\(\sigma^2\)) = npq.
Recognizing a problem as a binomial distribution can save a lot of calculation time.

Step 1: Identifying the Type of Equation


Rearrange the equation to find \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{x^2y}{x^3 + y^3} \]

This is a homogeneous differential equation because each term has the same degree (3).


Step 2: Substitution


Use the substitution \(y = vx\). This implies \(\frac{dy}{dx} = v + x\frac{dv}{dx}\).

Substitute into the equation:
\[ v + x\frac{dv}{dx} = \frac{x^2(vx)}{x^3 + (vx)^3} = \frac{vx^3}{x^3(1+v^3)} = \frac{v}{1+v^3} \]

Isolate the terms for separation:
\[ x\frac{dv}{dx} = \frac{v}{1+v^3} - v = \frac{v - v(1+v^3)}{1+v^3} = \frac{-v^4}{1+v^3} \]


Step 3: Separation and Integration


Separate the variables:
\[ \frac{1+v^3}{v^4} dv = -\frac{1}{x} dx \implies \left( v^{-4} + \frac{1}{v} \right) dv = -\frac{1}{x} dx \]

Integrate both sides:
\[ \int (v^{-4} + \frac{1}{v}) dv = \int -\frac{1}{x} dx \]
\[ \frac{v^{-3}}{-3} + \ln|v| = -\ln|x| + C \]


Step 4: Back-Substitution


Substitute back \(v = \frac{y}{x}\):
\[ -\frac{1}{3(y/x)^3} + \ln\left|\frac{y}{x}\right| = -\ln|x| + C \]
\[ -\frac{x^3}{3y^3} + \ln|y| - \ln|x| = -\ln|x| + C \]
\[ -\frac{x^3}{3y^3} + \ln|y| = C \]


Step 5: Final Answer:


The general solution is \(-\frac{x^3}{3y^3} + \ln|y| = C\).
Quick Tip: To check if a DE is homogeneous, replace x with \(\lambda x\) and y with \(\lambda y\). If the \(\lambda\)s cancel out, it is homogeneous.
The substitution \(y=vx\) will always transform it into a separable equation.

Step 1: Identifying the Type of Equation


Rearrange the equation into the standard linear form, \(\frac{dy}{dx} + P(x)y = Q(x)\).
\[ \frac{dy}{dx} + \frac{2x}{1+x^2}y = \frac{4x^2}{1+x^2} \]

This is a linear DE with \(P(x) = \frac{2x}{1+x^2}\) and \(Q(x) = \frac{4x^2}{1+x^2}\).


Step 2: Finding the Integrating Factor (I.F.)


I.F. = \(e^{\int P(x) dx} = e^{\int \frac{2x}{1+x^2} dx}\).

The integral is \(\ln(1+x^2)\). So, I.F. = \(e^{\ln(1+x^2)} = 1+x^2\).


Step 3: Finding the General Solution


The solution is \(y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx\).
\[ y(1+x^2) = \int \frac{4x^2}{1+x^2} \cdot (1+x^2) dx = \int 4x^2 dx \]
\[ y(1+x^2) = \frac{4x^3}{3} + C \]


Step 4: Applying the Initial Condition


Given y(0) = 0, substitute x = 0, y = 0:
\[ 0(1+0) = \frac{4(0)}{3} + C \implies C = 0 \]

The particular solution is \(y(1+x^2) = \frac{4x^3}{3}\).


Step 5: Final Answer:


The particular solution is \(y = \frac{4x^3}{3(1+x^2)}\).
Quick Tip: The process for solving first-order linear DEs is algorithmic:
1. Put in standard form \(\frac{dy}{dx} + P(x)y = Q(x)\).
2. Find Integrating Factor \(I.F. = e^{\int P(x)dx}\).
3. Solution is \(y \cdot I.F. = \int Q(x) \cdot I.F. dx\).
4. Use the initial condition to find C.

Step 1: Understanding the Region


The curve is \(y = -x^2\), which is a downward-opening parabola with its vertex at the origin (0,0).

The region is bounded by this parabola, the x-axis (y=0), and the vertical lines x=-3 and x=2.

The entire region lies below the x-axis.


Step 2: Setting up the Integral for Area


Since the region is below the x-axis (\(y \leq 0\)), the area is given by the absolute value of the definite integral, or by integrating the top curve minus the bottom curve.


Area \(A = \int_{-3}^{2} (0 - (-x^2)) dx = \int_{-3}^{2} x^2 dx \).


Alternatively, \(A = \left| \int_{-3}^{2} (-x^2) dx \right|\). Let's use the first method.


Step 3: Evaluating the Integral

\[ A = \int_{-3}^{2} x^2 dx = \left[ \frac{x^3}{3} \right]_{-3}^{2} \]

Apply the limits of integration:
\[ A = \left( \frac{2^3}{3} \right) - \left( \frac{(-3)^3}{3} \right) \]
\[ A = \frac{8}{3} - \frac{-27}{3} = \frac{8}{3} + \frac{27}{3} = \frac{35}{3} \]


Step 4: Final Answer:

\begin{tikzpicture[
scale=0.6,
every node/.style={font=\small
]
\draw[->, thick] (-4,0) -- (4,0) node[right] {\(x\);
\draw[->, thick] (0,2) -- (0,-10) node[below] {\(y\);
\foreach \x in {-3, -2, -1, 1, 2, 3 {
\draw (\x, 2pt) -- (\x, -2pt) node[below] {\x;

\foreach \y in {-2, -4, -6, -8 {
\draw (2pt, \y) -- (-2pt, \y) node[left] {\y;

\node[below left] at (0,0) {0;

% Draw the parabola y = -x^2
\draw[blue, ultra thick, domain=-3.1:3.1, samples=100] plot (\x, {-\x*\x);
\node[blue, below] at (2.5, -6.25) {\(y=-x^2\);

% Define coordinates for boundaries
\coordinate (L1) at (-3, 0);
\coordinate (L2) at (-3, -9);
\coordinate (R1) at (2, 0);
\coordinate (R2) at (2, -4);

% Shade the region
\fill[orange!20, domain=-3:2, variable=\x]
(L1) -- plot (\x, {-\x*\x) -- (R1) -- cycle;

% Draw vertical boundary lines
\draw[black, dashed] (L1) -- (L2);
\draw[black, dashed] (R1) -- (R2);

% Label the region
\node[align=center] at (-0.5, -4) {Bounded Region
Area;

\end{tikzpicture


The area of the region is \(\frac{35}{3}\) square units.
Quick Tip: When calculating area, always sketch the graph first to visualize the region.
If the region is below the x-axis, the definite integral \(\int y \, dx\) will be negative. The area must be positive, so you should calculate \(\int (upper curve - lower curve) \, dx\).
In this case, the upper curve is y=0 and the lower curve is y=-x².

Step 1: Understanding the Problem


This is an integral of a rational function where the degree of the numerator (2) is less than the degree of the denominator (3). We use partial fraction decomposition.


Step 2: Partial Fraction Decomposition


The denominator has a repeated linear factor \((x-1)^2\) and a distinct linear factor \((x+3)\). The decomposition is:
\[ \frac{x^2 + 1}{(x-1)^2(x+3)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+3} \]

Multiplying by the denominator gives:
\[ x^2 + 1 = A(x-1)(x+3) + B(x+3) + C(x-1)^2 \]

- Set x = 1: \( 2 = 4B \implies B = \frac{1}{2} \)

- Set x = -3: \( 10 = 16C \implies C = \frac{5}{8} \)

- Equate coefficients of \(x^2\): \( 1 = A + C \implies A = 1 - \frac{5}{8} = \frac{3}{8} \)


Step 3: Integration


Substitute the constants and integrate:
\[ \int \left( \frac{3/8}{x-1} + \frac{1/2}{(x-1)^2} + \frac{5/8}{x+3} \right) dx \]
\[ = \frac{3}{8} \int \frac{1}{x-1} dx + \frac{1}{2} \int (x-1)^{-2} dx + \frac{5}{8} \int \frac{1}{x+3} dx \]
\[ = \frac{3}{8} \ln|x-1| + \frac{1}{2} \frac{(x-1)^{-1}}{-1} + \frac{5}{8} \ln|x+3| + K \]
\[ = \frac{3}{8} \ln|x-1| - \frac{1}{2(x-1)} + \frac{5}{8} \ln|x+3| + K \]


Step 4: Final Answer:


The integral is \(\frac{3}{8} \ln|x-1| + \frac{5}{8} \ln|x+3| - \frac{1}{2(x-1)} + K\).
Quick Tip: For partial fractions with repeated linear factors like \((x-a)^n\), the decomposition must include terms for all powers from 1 to n.
e.g., \(\frac{A}{x-a} + \frac{B}{(x-a)^2} + \dots + \frac{Z}{(x-a)^n}\).

Step 1: Understanding the Problem


The 'x' in the numerator of a definite integral from 0 to 'a' is a strong hint to use the "King's property".


Step 2: Applying Properties of Definite Integrals


Let \( I = \int_{0}^{\pi/2} \frac{x}{\sin x + \cos x} dx \quad (1) \).

Using the property \(\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx\):
\[ I = \int_{0}^{\pi/2} \frac{\frac{\pi}{2} - x}{\sin(\frac{\pi}{2}-x) + \cos(\frac{\pi}{2}-x)} dx = \int_{0}^{\pi/2} \frac{\frac{\pi}{2} - x}{\cos x + \sin x} dx \quad (2) \]

Adding equations (1) and (2):
\[ 2I = \int_{0}^{\pi/2} \frac{x + (\frac{\pi}{2} - x)}{\sin x + \cos x} dx = \frac{\pi}{2} \int_{0}^{\pi/2} \frac{1}{\sin x + \cos x} dx \]


Step 3: Evaluating the New Integral


To evaluate \(\int \frac{1}{\sin x + \cos x} dx\), we write \(\sin x + \cos x = \sqrt{2}\sin(x + \frac{\pi}{4})\).
\[ 2I = \frac{\pi}{2\sqrt{2}} \int_{0}^{\pi/2} \csc(x + \frac{\pi}{4}) dx \]
\[ 2I = \frac{\pi}{2\sqrt{2}} [\ln|\csc(x + \frac{\pi}{4}) - \cot(x + \frac{\pi}{4})|]_{0}^{\pi/2} \]
\[ 2I = \frac{\pi}{2\sqrt{2}} \left[ \ln|\csc(\frac{3\pi}{4}) - \cot(\frac{3\pi}{4})| - \ln|\csc(\frac{\pi}{4}) - \cot(\frac{\pi}{4})| \right] \]
\[ 2I = \frac{\pi}{2\sqrt{2}} \left[ \ln|\sqrt{2} - (-1)| - \ln|\sqrt{2} - 1| \right] = \frac{\pi}{2\sqrt{2}} \ln\left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right) \]

Rationalizing the argument of the log: \( \frac{\sqrt{2}+1}{\sqrt{2}-1} \times \frac{\sqrt{2}+1}{\sqrt{2}+1} = (\sqrt{2}+1)^2 \).
\[ 2I = \frac{\pi}{2\sqrt{2}} \ln((\sqrt{2}+1)^2) = \frac{\pi}{2\sqrt{2}} \cdot 2\ln(\sqrt{2}+1) = \frac{\pi}{\sqrt{2}} \ln(\sqrt{2}+1) \]
\[ I = \frac{\pi}{2\sqrt{2}} \ln(\sqrt{2}+1) \]


Step 4: Final Answer:


The value of the integral is \(\frac{\pi}{2\sqrt{2}} \ln(\sqrt{2}+1)\).
Quick Tip: The King's property is one of the most useful tools for definite integrals.
When adding the original integral (I) and the transformed integral (I), the 'x' term often cancels, leaving a much simpler integral to solve.

Step 1: Identify the Given Information and Goal


We are given a point P(2, -1, 5) and a line in Cartesian form.

Our goal is to find the coordinates of the point M on the line such that the line segment PM is perpendicular to the given line. This point M is called the foot of the perpendicular. We also need to find the length of PM.


Step 2: Parameterize the Line


The equation of the line is \(\frac{x-11}{10} = \frac{y+2}{-4} = \frac{z+8}{-11}\).

Let's set these ratios equal to a parameter, \(\lambda\).
\[ \frac{x-11}{10} = \frac{y+2}{-4} = \frac{z+8}{-11} = \lambda \]

Any point M on this line can be written in terms of \(\lambda\):
\(x = 11 + 10\lambda\)
\(y = -2 - 4\lambda\)
\(z = -8 - 11\lambda\)

So, the coordinates of M are \((11 + 10\lambda, -2 - 4\lambda, -8 - 11\lambda)\).


Step 3: Use the Perpendicularity Condition


The direction vector of the given line is \(\vec{b} = \langle 10, -4, -11 \rangle\).

The vector from P to M, \(\vec{PM}\), is found by subtracting the coordinates of P from M:
\[ \vec{PM} = \langle (11 + 10\lambda - 2), (-2 - 4\lambda - (-1)), (-8 - 11\lambda - 5) \rangle \]
\[ \vec{PM} = \langle 9 + 10\lambda, -1 - 4\lambda, -13 - 11\lambda \rangle \]

Since PM is perpendicular to the line, the vector \(\vec{PM}\) must be perpendicular to the direction vector \(\vec{b}\). This means their dot product is zero:
\[ \vec{PM} \cdot \vec{b} = 0 \]
\[ (10)(9 + 10\lambda) + (-4)(-1 - 4\lambda) + (-11)(-13 - 11\lambda) = 0 \]


Step 4: Solve for \(\lambda\)

\[ 90 + 100\lambda + 4 + 16\lambda + 143 + 121\lambda = 0 \]

Combine like terms:
\[ (100 + 16 + 121)\lambda + (90 + 4 + 143) = 0 \]
\[ 237\lambda + 237 = 0 \]
\[ \lambda = -1 \]


Step 5: Find the Coordinates of the Foot (M)


Substitute \(\lambda = -1\) back into the parametric coordinates of M:
\(x = 11 + 10(-1) = 11 - 10 = 1\)
\(y = -2 - 4(-1) = -2 + 4 = 2\)
\(z = -8 - 11(-1) = -8 + 11 = 3\)

The foot of the perpendicular is M(1, 2, 3).


Step 6: Find the Length of the Perpendicular (PM)


Now, we find the distance between P(2, -1, 5) and M(1, 2, 3) using the distance formula:
\[ Length = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]
\[ |\vec{PM}| = \sqrt{(1 - 2)^2 + (2 - (-1))^2 + (3 - 5)^2} \]
\[ |\vec{PM}| = \sqrt{(-1)^2 + (3)^2 + (-2)^2} \]
\[ |\vec{PM}| = \sqrt{1 + 9 + 4} = \sqrt{14} \]


Step 7: Final Answer:


The foot of the perpendicular is the point (1, 2, 3).

The length of the perpendicular is \(\sqrt{14}\) units.
Quick Tip: This is a standard and very important problem in 3D geometry. The core idea is to use the dot product property of perpendicular vectors.
The steps are always:
1. Parameterize a general point M on the line.
2. Form the vector \(\vec{PM}\).
3. Set the dot product of \(\vec{PM}\) and the line's direction vector to zero.
4. Solve for the parameter and find the specific point M and the required length.

Step 1: Define events and list probabilities.

Let A, B, and C be the events that the chosen smartphone is from company A, B, and C, respectively.

Let D be the event that the chosen smartphone is defective.

We are given the following probabilities from the inventory:
- P(A) = 25% = 0.25
- P(B) = 35% = 0.35
- P(C) = 40% = 0.40
We are also given the conditional probabilities of a phone being defective, given the company:
- P(D|A) = 5% = 0.05
- P(D|B) = 4% = 0.04
- P(D|C) = 2% = 0.02

Step 2: Apply the Law of Total Probability.

The probability of the phone being defective, P(D), is the sum of the probabilities of it being a defective phone from each company. \[ P(D) = P(A)P(D|A) + P(B)P(D|B) + P(C)P(D|C) \] \[ P(D) = (0.25)(0.05) + (0.35)(0.04) + (0.40)(0.02) \] \[ P(D) = 0.0125 + 0.0140 + 0.0080 \] \[ P(D) = 0.0345 \]

Final Answer:

The probability that the smartphone was defective is 0.0345 or 3.45%.
Quick Tip: The Law of Total Probability is perfect for finding the overall probability of an event that can occur via several distinct paths. Structure the problem by listing the probability of each path and the conditional probability of the event along that path.

Step 1: Identify the required conditional probability.

We need to find the probability that the phone was made by company B, given that it is defective. This is the conditional probability P(B|D).


Step 2: Apply Bayes' Theorem.

Bayes' theorem states: \[ P(B|D) = \frac{P(B) P(D|B)}{P(D)} \]
We have all the necessary values from the problem statement and part (i).
- P(B) = 0.35
- P(D|B) = 0.04
- P(D) = 0.0345 (calculated in part i)

Step 3: Calculate the probability.
\[ P(B|D) = \frac{0.35 \times 0.04}{0.0345} \] \[ P(B|D) = \frac{0.0140}{0.0345} \]
To simplify the fraction, multiply the numerator and denominator by 10000: \[ P(B|D) = \frac{140}{345} \]
Divide both by 5: \[ P(B|D) = \frac{28}{69} \]

Final Answer:

The probability that the defective smartphone was manufactured by company B is \(\frac{28}{69}\).
Quick Tip: Bayes' Theorem helps "reverse" conditional probability. Use it when you know the outcome (the phone is defective) and want to find the probability of a specific cause (it came from company B). The denominator is almost always the total probability calculated in the first part of the question.

Step 1: Define the variables.

Let the price of one pen be Rs x.

Let the price of one notepad be Rs y.

Let the price of one eraser be Rs z.


Step 2: Form the linear equations based on the given information.

From Neha's purchase: \(4x + 3y + 2z = 60\)

From Rani's purchase: \(2x + 4y + 6z = 90\)

From Sam's purchase: \(6x + 2y + 3z = 70\)


Step 3: Express the system of equations in matrix form AX = B.

The system can be written as: \[ \begin{bmatrix} 4 & 3 & 2
2 & 4 & 6
6 & 2 & 3 \end{bmatrix} \begin{bmatrix} x
y
z \end{bmatrix} = \begin{bmatrix} 60
90
70 \end{bmatrix} \]
Where: \[ A = \begin{bmatrix} 4 & 3 & 2
2 & 4 & 6
6 & 2 & 3 \end{bmatrix}, \quad X = \begin{bmatrix} x
y
z \end{bmatrix}, \quad B = \begin{bmatrix} 60
90
70 \end{bmatrix} \] Quick Tip: When converting a word problem to a matrix equation, ensure the variables are consistent in each equation. The coefficients of these variables form the rows of matrix A, the variables form matrix X, and the constants on the right side form matrix B.

Step 1: Write the matrix A. \[ A = \begin{bmatrix} 4 & 3 & 2
2 & 4 & 6
6 & 2 & 3 \end{bmatrix} \]

Step 2: Calculate the determinant \(|A|\).

We expand along the first row: \[ |A| = 4(4 \cdot 3 - 6 \cdot 2) - 3(2 \cdot 3 - 6 \cdot 6) + 2(2 \cdot 2 - 4 \cdot 6) \] \[ |A| = 4(12 - 12) - 3(6 - 36) + 2(4 - 24) \] \[ |A| = 4(0) - 3(-30) + 2(-20) \] \[ |A| = 0 + 90 - 40 = 50 \]

Step 3: Confirm if A\(^{-1}\) exists.

The inverse of a matrix A exists if and only if its determinant is non-zero (\(|A| \neq 0\)).

Since \(|A| = 50 \neq 0\), the matrix A is non-singular.

Therefore, it is possible to find A\(^{-1}\).
Quick Tip: A matrix is called "singular" if its determinant is zero, and "non-singular" otherwise. Only non-singular matrices are invertible. This is a fundamental concept in matrix algebra.

Step 1: Find the cofactor matrix of A.

The cofactors are:

\( C_{11} = (12-12) = 0 \)
\( C_{12} = -(6-36) = 30 \)
\( C_{13} = (4-24) = -20 \)
\( C_{21} = -(9-4) = -5 \)
\( C_{22} = (12-12) = 0 \)
\( C_{23} = -(8-18) = 10 \)
\( C_{31} = (18-8) = 10 \)
\( C_{32} = -(24-4) = -20 \)
\( C_{33} = (16-6) = 10 \)

The cofactor matrix is \(\begin{bmatrix} 0 & 30 & -20
-5 & 0 & 10
10 & -20 & 10 \end{bmatrix}\).


Step 2: Find the adjugate (adjoint) of A.

The adjugate of A is the transpose of the cofactor matrix. \[ adj(A) = \begin{bmatrix} 0 & -5 & 10
30 & 0 & -20
-20 & 10 & 10 \end{bmatrix} \]

Step 3: Find the inverse of A.

The formula for the inverse is \(A^{-1} = \frac{1}{|A|} adj(A)\). We know \(|A| = 50\). \[ A^{-1} = \frac{1}{50} \begin{bmatrix} 0 & -5 & 10
30 & 0 & -20
-20 & 10 & 10 \end{bmatrix} \]

Step 4: Write the formula to find X.

The solution to the system of equations AX = B is given by: \[ X = A^{-1}B \] Quick Tip: The most common error in finding a matrix inverse is confusing the cofactor matrix with the adjugate matrix. Remember: adjugate is the \textbf{transpose} of the cofactors. The formula \(X = A^{-1}B\) is crucial for solving linear systems.

Step 1: Calculate A\(^2\). \[ A^2 = A \cdot A = \begin{bmatrix} 4 & 3 & 2
2 & 4 & 6
6 & 2 & 3 \end{bmatrix} \begin{bmatrix} 4 & 3 & 2
2 & 4 & 6
6 & 2 & 3 \end{bmatrix} \] \[ A^2 = \begin{bmatrix} (16+6+12) & (12+12+4) & (8+18+6)
(8+8+36) & (6+16+12) & (4+24+18)
(24+4+18) & (18+8+6) & (12+12+9) \end{bmatrix} \] \[ A^2 = \begin{bmatrix} 34 & 28 & 32
52 & 34 & 46
46 & 32 & 33 \end{bmatrix} \]

Step 2: Calculate 8I.

I is the 3x3 identity matrix. \[ 8I = 8 \begin{bmatrix} 1 & 0 & 0
0 & 1 & 0
0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 8 & 0 & 0
0 & 8 & 0
0 & 0 & 8 \end{bmatrix} \]

Step 3: Calculate A\(^2\) - 8I. \[ A^2 - 8I = \begin{bmatrix} 34 & 28 & 32
52 & 34 & 46
46 & 32 & 33 \end{bmatrix} - \begin{bmatrix} 8 & 0 & 0
0 & 8 & 0
0 & 0 & 8 \end{bmatrix} \] \[ A^2 - 8I = \begin{bmatrix} (34-8) & (28-0) & (32-0)
(52-0) & (34-8) & (46-0)
(46-0) & (32-0) & (33-8) \end{bmatrix} \] \[ A^2 - 8I = \begin{bmatrix} 26 & 28 & 32
52 & 26 & 46
46 & 32 & 25 \end{bmatrix} \] Quick Tip: Matrix multiplication (\(A^2\)) is done row-by-column. Scalar multiplication (8I) multiplies every element by the scalar. Matrix subtraction is performed element-wise. Keep the operations distinct to avoid errors.

Step 1: Express y in terms of h and x.

The ladder, the wall, and the ground form a right-angled triangle.

- The length of the ladder is the hypotenuse, 'h'.

- The height on the wall is one leg, 'x'.

- The distance between the wall and the foot of the ladder is the other leg, 'y'.

By the Pythagorean theorem: \[ x^2 + y^2 = h^2 \]
Solving for y: \[ y^2 = h^2 - x^2 \] \[ y = \sqrt{h^2 - x^2} \quad (since distance y must be positive) \]

Step 2: Write an expression for the area (A).

The area of the right triangle is given by A = \(\frac{1}{2} \times base \times height\).

Here, the base is y and the height is x. \[ A = \frac{1}{2} yx \]
Substitute the expression for y from Step 1: \[ A = \frac{1}{2} x \sqrt{h^2 - x^2} \] Quick Tip: Drawing a simple diagram is the best way to start any geometry-based word problem. Clearly labeling the sides helps in correctly applying fundamental theorems like the Pythagorean theorem.

Step 1: Find the derivative of A with respect to x.

The area function is \(A(x) = \frac{1}{2} x \sqrt{h^2 - x^2}\). We use the product rule \((uv)' = u'v + uv'\) to differentiate.

Let \(u = x\) and \(v = \sqrt{h^2 - x^2}\). Then \(u' = 1\) and \(v' = \frac{1}{2\sqrt{h^2 - x^2}}(-2x) = \frac{-x}{\sqrt{h^2 - x^2}}\). \[ \frac{dA}{dx} = \frac{1}{2} \left[ (1) \sqrt{h^2 - x^2} + x \left( \frac{-x}{\sqrt{h^2 - x^2}} \right) \right] \] \[ \frac{dA}{dx} = \frac{1}{2} \left[ \sqrt{h^2 - x^2} - \frac{x^2}{\sqrt{h^2 - x^2}} \right] \]
Combine the terms by finding a common denominator: \[ \frac{dA}{dx} = \frac{1}{2} \left[ \frac{(h^2 - x^2) - x^2}{\sqrt{h^2 - x^2}} \right] = \frac{h^2 - 2x^2}{2\sqrt{h^2 - x^2}} \]

Step 2: Find the critical point.

Critical points occur where \(\frac{dA}{dx} = 0\) or is undefined. We set the numerator to zero: \[ h^2 - 2x^2 = 0 \] \[ 2x^2 = h^2 \] \[ x^2 = \frac{h^2}{2} \] \[ x = \frac{h}{\sqrt{2}} \quad (since height x must be positive) \]
This is the critical point.
Quick Tip: For optimization problems, critical points are the candidates for maxima or minima. They are found by setting the first derivative equal to zero. Remember to use the product and chain rules correctly.

Step 1: Use the Second Derivative Test.

We need to find the second derivative, \(\frac{d^2A}{dx^2}\), and evaluate its sign at the critical point \(x = \frac{h}{\sqrt{2}}\).

Our first derivative is \(\frac{dA}{dx} = \frac{h^2 - 2x^2}{2\sqrt{h^2 - x^2}}\).

Using the quotient rule \(\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}\), where \(u = h^2 - 2x^2\) and \(v = 2\sqrt{h^2 - x^2}\).
\(u' = -4x\)
\(v' = 2 \cdot \frac{-x}{\sqrt{h^2 - x^2}} = \frac{-2x}{\sqrt{h^2 - x^2}}\)
\[ \frac{d^2A}{dx^2} = \frac{(-4x)(2\sqrt{h^2 - x^2}) - (h^2 - 2x^2)\left(\frac{-2x}{\sqrt{h^2 - x^2}}\right)}{(2\sqrt{h^2 - x^2})^2} \] \[ = \frac{-8x\sqrt{h^2 - x^2} + \frac{2x(h^2 - 2x^2)}{\sqrt{h^2 - x^2}}}{4(h^2 - x^2)} \]
Multiply numerator and denominator by \(\sqrt{h^2 - x^2}\): \[ = \frac{-8x(h^2 - x^2) + 2x(h^2 - 2x^2)}{4(h^2 - x^2)^{3/2}} = \frac{-8xh^2 + 8x^3 + 2xh^2 - 4x^3}{4(h^2 - x^2)^{3/2}} = \frac{4x^3 - 6xh^2}{4(h^2 - x^2)^{3/2}} = \frac{2x^3 - 3xh^2}{2(h^2 - x^2)^{3/2}} \]

Step 2: Evaluate the second derivative at the critical point.

At the critical point, \(x = \frac{h}{\sqrt{2}}\), we have \(x^2 = \frac{h^2}{2}\). The term \(h^2 - 2x^2\) in the numerator of \(\frac{d^2A}{dx^2}\) becomes \(h^2 - 2(\frac{h^2}{2}) = 0\).
Let's re-examine the expression for \(\frac{d^2A}{dx^2}\) before simplifying the numerator: \[ \frac{d^2A}{dx^2} = \frac{(-4x)(2\sqrt{h^2 - x^2}) - (h^2 - 2x^2)\left(\frac{-2x}{\sqrt{h^2 - x^2}}\right)}{4(h^2 - x^2)} \]
When \(x^2 = h^2/2\), the term \(h^2 - 2x^2 = 0\). So the second part of the numerator vanishes. \[ \frac{d^2A}{dx^2}\bigg|_{x=h/\sqrt{2}} = \frac{(-4(h/\sqrt{2}))(2\sqrt{h^2 - h^2/2}) - 0}{4(h^2 - h^2/2)} = \frac{(-4h/\sqrt{2})(2\sqrt{h^2/2})}{4(h^2/2)} = \frac{(-4h/\sqrt{2})(2h/\sqrt{2})}{2h^2}\]

\[= \frac{-8h^2/2}{2h^2} = \frac{-4h^2}{2h^2} = -2 \]


Since \(\frac{d^2A}{dx^2} = -2 < 0\) at the critical point, the area A is maximum at \(x = \frac{h}{\sqrt{2}}\).
Quick Tip: The First Derivative Test is often algebraically simpler than the Second Derivative Test. If the sign of the derivative changes from positive to negative at a critical point, it's a maximum. If it changes from negative to positive, it's a minimum.

Step 1: Set up the related rates problem.

We are given:
- Length of ladder, h = 5 m.
- The distance y is decreasing at 2 m/s, so \(\frac{dy}{dt} = -2\) m/s.
- We need to find \(\frac{dx}{dt}\) when y = 3 m.
The relationship between x and y is given by the Pythagorean theorem: \[ x^2 + y^2 = h^2 \implies x^2 + y^2 = 5^2 = 25 \]

Step 2: Differentiate with respect to time (t).

Differentiate the equation \(x^2 + y^2 = 25\) implicitly with respect to time t: \[ \frac{d}{dt}(x^2) + \frac{d}{dt}(y^2) = \frac{d}{dt}(25) \] \[ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \]

Step 3: Find the value of x at the given instant.

We need to find x when y = 3 m. \[ x^2 + 3^2 = 25 \] \[ x^2 + 9 = 25 \] \[ x^2 = 16 \implies x = 4 m \quad (since height x must be positive) \]

Step 4: Solve for \(\frac{dx}{dt}\).

Substitute the known values (x=4, y=3, \(\frac{dy}{dt}=-2\)) into the differentiated equation: \[ 2(4) \frac{dx}{dt} + 2(3)(-2) = 0 \] \[ 8 \frac{dx}{dt} - 12 = 0 \] \[ 8 \frac{dx}{dt} = 12 \] \[ \frac{dx}{dt} = \frac{12}{8} = \frac{3}{2} = 1.5 m/s \]
Since the result is positive, the height x is increasing.


Final Answer:

The height on the wall (x) is increasing at a rate of 1.5 m/s.
Quick Tip: In related rates problems, the key is to find an equation connecting the variables and then differentiate it implicitly with respect to time, \(t\). Pay close attention to the signs of the rates (positive for increasing quantities, negative for decreasing).