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 -\(\pi/2\) and \(\pi/2\). The graph has horizontal asymptotes at y = -\(\pi/2\) and y = \(\pi/2\). The range is (-\(\pi/2, \pi/2\)).

- Behavior: The function is strictly increasing over its entire domain.

- Key Point: The graph passes through the origin (0, 0).


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

- (A) y = tan\(^{-1}\) x: The principal value range of the inverse tangent function is (-\(\pi/2, \pi/2\)). Its domain is (\(-\infty, \infty\)), it is an increasing function, and it passes through (0,0). This perfectly matches all features of the graph.

- (B) y = cosec\(^{-1}\) x: Has a domain of \(|x| \geq 1\) and a range of \([-\pi/2, \pi/2] - \{0\}\). This does not match.

- (C) y = cot\(^{-1}\) x: Has a range of (0, \(\pi\)). This does not match.

- (D) y = sec\(^{-1}\) x: Has a domain of \(|x| \geq 1\) and a range of \([0, \pi] - \{\pi/2\}\). This does not match.


Step 3: Final Answer:


The graph shown depicts the function y = tan\(^{-1}\) 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.
The graph of tan\(^{-1}\)(x) is particularly distinctive with its "S" shape and horizontal asymptotes at y = \(\pm\pi/2\).

Step 1: Understanding the Question:


We need to find the domain of a function that is the sum of two functions, \(g(x) = \cos^{-1}x\) and \(h(x) = \sin x\).


Step 2: Key Formula or Approach:


The domain of the sum of two functions, \(f(x) = g(x) + h(x)\), is the intersection of their individual domains.

Domain(f) = Domain(g) \(\cap\) Domain(h).


Step 3: Detailed Explanation:


First, find the domain of \(g(x) = \cos^{-1}x\).

The inverse cosine function is defined only for inputs between -1 and 1, inclusive.

So, Domain(g) = [-1, 1].


Next, find the domain of \(h(x) = \sin x\).

The sine function is defined for all real numbers.

So, Domain(h) = R (the set of all real numbers, or (\(-\infty, \infty\))).


Now, find the intersection of these two domains:

Domain(f) = [-1, 1] \(\cap\) R = [-1, 1].


Step 4: Final Answer:


The domain of f(x) = cos\(^{-1}\) x + sin x is [-1, 1].
Quick Tip: When combining functions (adding, subtracting, multiplying, etc.), the domain of the resulting function is always governed by the function with the \textbf{most restrictive} domain.
In this case, \(\cos^{-1}x\) restricts the domain, while \(\sin x\) does not add any further restrictions.

Step 1: Understanding the Question:


We need to find the total number of different 3 \(\times\) 3 matrices that can be created where every element must be either \(\sqrt{2}\) or \(\sqrt{3}\).


Step 2: Key Formula or Approach:


This is a counting problem based on the multiplication principle.

First, determine the total number of entries in the matrix.

Then, for each entry, determine the number of choices available.

The total number of possible matrices is the product of the number of choices for each entry.


Step 3: Detailed Explanation:


A 3 \(\times\) 3 matrix has \(3 \times 3 = 9\) entries (elements).

For each of these 9 entries, there are 2 possible choices: it can be either \(\sqrt{2}\) or \(\sqrt{3}\).

Using the multiplication principle, the total number of possible matrices is the number of choices for each entry multiplied together:

Total matrices = (Choices for entry 1) \(\times\) (Choices for entry 2) \(\times\) ... \(\times\) (Choices for entry 9)
\[ Total matrices = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \]
\[ = 2^9 \]

Now, calculate \(2^9\):
\[ 2^9 = 2 \times 2^8 = 2 \times (2^4)^2 = 2 \times 16^2 = 2 \times 256 = 512 \]


Step 4: Final Answer:


There are 512 possible matrices of the specified order and entries.
Quick Tip: This is a classic combinatorial problem. The general formula is:
(Number of choices per entry)\(^{(Total number of entries)}\).
In this case, it's \(2^9\). If the matrix were 2 \(\times\) 2, the answer would be \(2^4 = 16\).

Step 1: Understanding the Question:


We need to classify the given matrix A by checking its properties against the definitions of the given options.


Step 2: Detailed Explanation:


Let's analyze the matrix A = \(\begin{bmatrix} \sqrt{3} & 0 & 0
0 & \sqrt{2} & 0
0 & 0 & \sqrt{5} \end{bmatrix}\).

This is a diagonal matrix because all of its off-diagonal elements are zero.


Now let's check the options:

- (A) Scalar matrix: This is a diagonal matrix where all diagonal elements are equal. Here, the diagonal elements are \(\sqrt{3}, \sqrt{2}, \sqrt{5}\), which are not equal. So, it is not a scalar matrix.

- (B) Identity matrix: This is a scalar matrix where all diagonal elements are 1. This is not the case here.

- (C) Null matrix: This is a matrix where all elements are zero. This is not the case here.

- (D) Symmetric matrix: A matrix is symmetric if its transpose is equal to itself (\(A' = A\)). Let's find the transpose of A:

\[ A' = \begin{bmatrix} \sqrt{3} & 0 & 0
0 & \sqrt{2} & 0
0 & 0 & \sqrt{5} \end{bmatrix}' = \begin{bmatrix} \sqrt{3} & 0 & 0
0 & \sqrt{2} & 0
0 & 0 & \sqrt{5} \end{bmatrix} = A \]

Since \(A' = A\), the matrix is symmetric.


Step 3: Final Answer:


The matrix A is a diagonal matrix, and every diagonal matrix is also a symmetric matrix. Therefore, from the given options, "symmetric matrix" is the correct classification.
Quick Tip: Remember the hierarchy of matrix types:
Identity Matrix \(\subset\) Scalar Matrix \(\subset\) Diagonal Matrix \(\subset\) Symmetric Matrix.
Any diagonal matrix is automatically symmetric because transposing it does not change any elements.
If "diagonal matrix" were an option, it would be a more specific and better description, but "symmetric matrix" is also correct.

Step 1: Understanding the Question:


We need to find the determinant of the matrix product 2AB, using the properties of determinants.


Step 2: Key Formula or Approach:


We will use two main properties of determinants:

1. \(|kC| = k^n |C|\), where C is a square matrix of order n and k is a scalar.

2. \(|CD| = |C| |D|\), for square matrices C and D of the same order.


Step 3: Detailed Explanation:


We need to find \(|2AB|\). Let's apply the properties.

First, apply the scalar multiplication property. Here, the scalar is k=2 and the matrix is AB. The order of the matrices is n=3.
\[ |2(AB)| = 2^3 |AB| \]

Now, apply the product property to \(|AB|\):
\[ 2^3 |AB| = 8 \cdot (|A| |B|) \]

We are given \(|A|\) = 3 and \(|B|\) = 5. Substitute these values:
\[ = 8 \cdot (3 \times 5) = 8 \cdot 15 \]
\[ = 120 \]


Step 4: Final Answer:


The value of \(|2AB|\) is 120.
Quick Tip: The most common mistake in this type of problem is forgetting to raise the scalar to the power of the matrix order.
Remember, \(|kA| = k^n|A|\), not \(k|A|\).
This property arises because the scalar 'k' multiplies every row of the matrix, and each time a row is multiplied by k, the determinant is multiplied by k.

Step 1: Understanding the Question:


We need to find the determinant of the adjugate (or adjoint) of a 3x3 matrix, given the determinant of the matrix itself.


Step 2: Key Formula or Approach:


There is a standard property relating the determinant of a matrix and the determinant of its adjugate.

For any square matrix A of order n, the formula is:
\[ |adj(A)| = |A|^{n-1} \]


Step 3: Detailed Explanation:


We are given:

- The order of the matrix, n = 3.

- The determinant of the matrix, \(|A|\) = 5.

Substitute these values into the formula:
\[ |adj(A)| = (5)^{3-1} = 5^2 \]
\[ = 25 \]


Step 4: Final Answer:


The value of \(|adj A|\) is 25.
Quick Tip: This is a direct application of a very important determinant property. Memorize it to solve these questions in seconds.
Formula: \(|adj(A)| = |A|^{n-1}\).
This property is derived from the identity \(A \cdot adj(A) = |A| \cdot I\).

Step 1: Understanding the Question:


We are given two equal matrices. By definition, two matrices are equal if and only if their corresponding elements are equal. We will use this property to form equations and solve for the variables x and y.


Step 2: Form Equations from Corresponding Elements


Comparing the elements at the same positions in both matrices:

1. From element \(a_{11}\): \(2x - 1 = x + 3\)

2. From element \(a_{12}\): \(3x = 12\)

3. From element \(a_{22}\): \(y^2 - 1 = 35\)


Step 3: Solve for x


Using the second equation:
\[ 3x = 12 \implies x = 4 \]

Let's check this value with the first equation to ensure consistency:

LHS: \(2(4) - 1 = 8 - 1 = 7\)

RHS: \(4 + 3 = 7\)

Since LHS = RHS, the value \(x = 4\) is correct.


Step 4: Solve for y


Using the third equation:
\[ y^2 - 1 = 35 \]
\[ y^2 = 36 \]

Taking the square root of both sides:
\[ y = \pm\sqrt{36} \]

So, \(y = 6\) or \(y = -6\).


Step 5: Calculate the Value of (x - y)


We have \(x = 4\) and two possible values for y.

Case 1: If \(y = 6\):
\[ x - y = 4 - 6 = -2 \]

Case 2: If \(y = -6\):
\[ x - y = 4 - (-6) = 4 + 6 = 10 \]


The possible values for \((x - y)\) are \(-2\) and \(10\).


Step 6: Final Answer:


Comparing our results with the given options, \(-2\) or \(10\) matches option (B).
Quick Tip: When solving for variables from matrix equality, always use the simplest element to find the value first (like \(3x=12\) here).
Crucially, when solving \(y^2 = a\), remember to consider both positive and negative roots (\(y = \pm\sqrt{a}\)). Missing the negative root is a common error that would lead to missing half of the answer.

Step 1: Understanding the Question:


For the piecewise function f(x) to be continuous everywhere, it must be continuous at the points where the definition of the function changes. These points are x = 3 and x = 5.


Step 2: Key Formula or Approach:


For continuity at a point 'c', the left-hand limit (LHL), right-hand limit (RHL), and the function's value must all be equal.

LHL = \(\lim_{x \to c^-} f(x)\)

RHL = \(\lim_{x \to c^+} f(x)\)

Value = \(f(c)\)

We need LHL = RHL = f(c) at both x=3 and x=5.


Step 3: Detailed Explanation:


Continuity at x = 3:

- LHL = \(\lim_{x \to 3^-} f(x)\). For x < 3, f(x) = 1. So, LHL = 1.

- RHL = \(\lim_{x \to 3^+} f(x)\). For x > 3, f(x) = ax + b. So, RHL = a(3) + b = 3a + b.

- f(3) = 1.

For continuity, LHL = RHL \(\implies\) 1 = 3a + b. (Equation 1)


Continuity at x = 5:

- LHL = \(\lim_{x \to 5^-} f(x)\). For x < 5, f(x) = ax + b. So, LHL = a(5) + b = 5a + b.

- RHL = \(\lim_{x \to 5^+} f(x)\). For x > 5, f(x) = 7. So, RHL = 7.

- f(5) = 7.

For continuity, LHL = RHL \(\implies\) 5a + b = 7. (Equation 2)


Now we have a system of two linear equations:

1) 3a + b = 1

2) 5a + b = 7

Subtract Equation 1 from Equation 2:

(5a + b) - (3a + b) = 7 - 1
\(2a = 6 \implies a = 3\).

Substitute a = 3 into Equation 1:
\(3(3) + b = 1 \implies 9 + b = 1 \implies b = -8\).


Step 4: Final Answer:


The values are a = 3 and b = -8.
Quick Tip: For piecewise functions, continuity must be checked at every "breakpoint" where the function definition changes.
This will give you a system of linear equations, one for each breakpoint, which you can then solve for the unknown constants.

Step 1: Understand the Question and the Properties of the Function


We are given the function \(f(x) = -2x^8\) and need to determine which of the given statements about its derivative is correct. We can approach this by either direct calculation or by analyzing the properties of even and odd functions.


Method 1: Using Properties of Even and Odd Functions


Step 2: Analyze the Parity of f(x) and f'(x)


- The function \(f(x) = -2x^8\) is an even function because the exponent of x is even. This means \(f(-x) = -2(-x)^8 = -2x^8 = f(x)\).

- A key property of differentiation is that the derivative of an even function is an odd function. Let's find the derivative:

\[ f'(x) = \frac{d}{dx}(-2x^8) = -2 \cdot (8x^7) = -16x^7 \]

The derivative \(f'(x) = -16x^7\) is an odd function because the exponent of x is odd.

- The definition of an odd function is \(g(-x) = -g(x)\). Applying this to our derivative, we have:

\[ f'(-x) = -f'(x) \]

If we let \(x = 1/2\), this general property becomes:

\[ f'(-1/2) = -f'(1/2) \]

- Now, let's check which option matches this property.
- Option (A) is the definition of an even function, which is false for f'(x).
- Option (B) is \(f'(1/2) = -f'(-1/2)\). If we multiply our property \(f'(-1/2) = -f'(1/2)\) by -1, we get \(-f'(-1/2) = f'(1/2)\), which is exactly the statement in option (B). So, (B) is correct.
- Option (C) is \(-f'(1/2) = f'(-1/2)\). This is exactly our property. So, (C) is also correct.
- Option (D) compares the derivative to the original function, which is generally not a simple identity.


Method 2: Direct Calculation


Step 3: Calculate the Derivative at the Specific Points


We already found the derivative: \(f'(x) = -16x^7\).

- At x = 1/2:

\[ f'(1/2) = -16\left(\frac{1}{2}\right)^7 = -16 \cdot \frac{1}{128} = -\frac{16}{128} = -\frac{1}{8} \]

- At x = -1/2:

\[ f'(-1/2) = -16\left(-\frac{1}{2}\right)^7 = -16 \cdot \left(-\frac{1}{128}\right) = \frac{16}{128} = \frac{1}{8} \]


Step 4: Verify the Options with Calculated Values


- (A) \(f'(1/2) = f'(-1/2)\) \(\implies -1/8 = 1/8\). This is False.
- (B) \(f'(1/2) = -f'(-1/2)\) \(\implies -1/8 = -(1/8)\). This is True.
- (C) \(-f'(1/2) = f'(-1/2)\) \(\implies -(-1/8) = 1/8 \implies 1/8 = 1/8\). This is True.
- (D) \(f'(1/2) = -f(-1/2)\). \(f(-1/2) = -2(-1/2)^8 = -2(1/256) = -1/128\). The statement becomes \(-1/8 = -(-1/128)\), which is false.


Step 5: Final Answer and Conclusion


Both direct calculation and property analysis show that options (B) and (C) are correct statements. They are mathematically equivalent rearrangements of the same identity for an odd function. In an exam context where only one answer can be chosen, this points to a flawed question design. However, both statements are valid.
Quick Tip: A key property of polynomials:
- The derivative of an even function (all powers of x are even) is an odd function (all powers of x are odd).
- The derivative of an odd function is an even function.
Recognizing this can allow you to answer the question without any calculation. Since f(x) is an even function, its derivative f'(x) must be an odd function, meaning \(f'(-x) = -f'(x)\), which is equivalent to the statements in options B and C.

Step 1: Understanding the Question:


We need to find the rate of change of the volume of a sphere with respect to x, which is \(\frac{dV}{dx}\).

We are given the diameter in terms of x, so we first need to find the radius and then the volume formula in terms of x.


Step 2: Express Volume in Terms of x


Diameter \(D = \frac{5}{2}(3x+1)\).

Radius \(r = \frac{D}{2} = \frac{1}{2} \cdot \frac{5}{2}(3x+1) = \frac{5}{4}(3x+1)\).

The volume of a sphere is \(V = \frac{4}{3}\pi r^3\).

Substitute the expression for r:
\[ V = \frac{4}{3}\pi \left( \frac{5}{4}(3x+1) \right)^3 = \frac{4}{3}\pi \cdot \frac{5^3}{4^3}(3x+1)^3 \]
\[ V = \frac{4}{3}\pi \cdot \frac{125}{64}(3x+1)^3 = \frac{125\pi}{3 \cdot 16}(3x+1)^3 = \frac{125\pi}{48}(3x+1)^3 \]


Step 3: Differentiate Volume with Respect to x


Now, find \(\frac{dV}{dx}\) using the chain rule:
\[ \frac{dV}{dx} = \frac{125\pi}{48} \cdot \frac{d}{dx}(3x+1)^3 \]
\[ \frac{dV}{dx} = \frac{125\pi}{48} \cdot [3(3x+1)^2 \cdot (3)] = \frac{125\pi}{48} \cdot 9(3x+1)^2 \]

Simplify the constant: \(\frac{9}{48} = \frac{3}{16}\).
\[ \frac{dV}{dx} = \frac{125\pi \cdot 3}{16}(3x+1)^2 = \frac{375\pi}{16}(3x+1)^2 \]


Step 4: Evaluate the Derivative at x = 1


Substitute x = 1 into the expression for \(\frac{dV}{dx}\):
\[ \frac{dV}{dx}\bigg|_{x=1} = \frac{375\pi}{16}(3(1)+1)^2 = \frac{375\pi}{16}(4)^2 = \frac{375\pi}{16}(16) \]
\[ = 375\pi \]


Step 5: Final Answer:


The rate of change of the volume is 375\(\pi\).
Quick Tip: This is a related rates problem where the variable is x, not time t.
1. Find the formula for the quantity you need (Volume) in terms of the given variable (x).
2. Differentiate this formula with respect to the variable.
3. Substitute the specific value of the variable to find the rate at that instant.
Be careful to distinguish between diameter and radius.

Step 1: Understanding the Question:


To determine if the function is increasing or decreasing, we need to analyze the sign of its first derivative, f'(x).

If f'(x) > 0 for all x, the function is strictly increasing.

If f'(x) < 0 for all x, the function is strictly decreasing.


Step 2: Find the Derivative


The function is \(f(x) = 2x - \sin x\).

The derivative is:
\[ f'(x) = \frac{d}{dx}(2x - \sin x) = 2 - \cos x \]


Step 3: Analyze the Sign of the Derivative


We know that the range of the cosine function is [-1, 1]. That is:
\[ -1 \leq \cos x \leq 1 \]

Now, let's analyze the expression for f'(x) = 2 - cos x.

- The maximum value of cos x is 1. In this case, f'(x) = 2 - 1 = 1.

- The minimum value of cos x is -1. In this case, f'(x) = 2 - (-1) = 3.

So, for all real numbers x, the value of the derivative f'(x) is always in the interval [1, 3].

Since \(f'(x) \geq 1\) for all x, it means that f'(x) is always positive.


Step 4: Final Answer:


Because the first derivative f'(x) is always positive, the function f(x) is a strictly increasing function for all x \(\in\) R.
Quick Tip: To check if a function is always increasing or decreasing, find its derivative and analyze its range.
Using the known range of trigonometric functions (like \(-1 \leq \sin x \leq 1\) and \(-1 \leq \cos x \leq 1\)) is a powerful way to determine if the derivative is always positive or always negative.

Step 1: Understand the Question and Identify Key Properties


We need to evaluate the integral of a seemingly complex function. The key to solving this problem is to first simplify the integrand using the properties of logarithms and exponents.


Step 2: Simplify the Integrand


We will use two main properties:
1. \(a \log_b(c) = \log_b(c^a)\)
2. \(b^{\log_b(u)} = u\). For the natural logarithm (log or ln), this is \(e^{\ln u} = u\).

Applying the first property to each term in the integral:
- \(e^{9\log x} = e^{\log x^9}\)
- \(e^{8\log x} = e^{\log x^8}\)
- \(e^{6\log x} = e^{\log x^6}\)
- \(e^{5\log x} = e^{\log x^5}\)

Now, applying the second property to these simplified terms:
- \(e^{\log x^9} = x^9\)
- \(e^{\log x^8} = x^8\)
- \(e^{\log x^6} = x^6\)
- \(e^{\log x^5} = x^5\)

Substitute these simplified expressions back into the integral: \[ \int \frac{x^9 - x^8}{x^6 - x^5} dx \]


Step 3: Further Algebraic Simplification


Now, we can simplify the rational expression by factoring the numerator and the denominator.
- Factor out the highest common power from the numerator: \(x^9 - x^8 = x^8(x - 1)\).
- Factor out the highest common power from the denominator: \(x^6 - x^5 = x^5(x - 1)\).

The integral becomes: \[ \int \frac{x^8(x - 1)}{x^5(x - 1)} dx \]

Assuming \(x \neq 1\), we can cancel the \((x-1)\) term: \[ \int \frac{x^8}{x^5} dx \]

Using the rules of exponents (\(\frac{a^m}{a^n} = a^{m-n}\)): \[ \int x^{8-5} dx = \int x^3 dx \]


Step 4: Evaluate the Final Integral


We now have a simple power function to integrate. Using the power rule for integration, \(\int x^n dx = \frac{x^{n+1}}{n+1} + C\): \[ \int x^3 dx = \frac{x^{3+1}}{3+1} + C = \frac{x^4}{4} + C \]


Step 5: Final Answer:


The value of the integral is \(\frac{x^4}{4} + C\). This corresponds to option (C).
Quick Tip: The key to this problem is simplifying the integrand before attempting to integrate.
Always look for opportunities to use the inverse property of logarithms and exponentials, \(e^{\ln u} = u\).
After simplification, the problem reduces to a basic power rule integration. Don't be intimidated by the initial appearance of the function.

Step 1: Understanding the Question:


We need to identify the correct property of definite integrals from the given list.


Step 2: Detailed Explanation:


Let's analyze each statement:

- (A) \(\int_a^b f(x)dx = \int_a^b f(a+b-x)dx\)

This is a standard and very useful property of definite integrals, often called the "King's property" in a more general form. It can be proven by substituting \(t = a+b-x\). This statement is true.


- (B) \(\int_{-a}^a f(x)dx = 0\), if f is an even function

This is incorrect. For an even function, \(f(-x) = f(x)\), the property is \(\int_{-a}^a f(x)dx = 2\int_0^a f(x)dx\). The integral is zero for an odd function. This statement is false.


- (C) \(\int_{-a^a f(x)dx = 2\int_0^a f(x)dx\), if f is an odd function

This is incorrect. For an odd function, \(f(-x) = -f(x)\), the property is \(\int_{-a}^a f(x)dx = 0\). The property stated here applies to an even function. This statement is false.


- (D) \(\int_0^{2a f(x)dx = \int_0^a f(x)dx - \int_0^a f(2a+x)dx\)

The correct property is \(\int_0^{2a} f(x)dx = \int_0^a f(x)dx + \int_0^a f(2a-x)dx\). The given statement has incorrect signs. This statement is false.


Step 3: Final Answer:


The only true statement is (A).
Quick Tip: Memorizing the standard properties of definite integrals is essential. The most important ones are:
1. \(\int_a^b f(x)dx = \int_a^b f(a+b-x)dx\)
2. \(\int_0^a f(x)dx = \int_0^a f(a-x)dx\) (A special case of the first)
3. \(\int_{-a}^a f(x)dx = 2\int_0^a f(x)dx\) if f is even.
4. \(\int_{-a}^a f(x)dx = 0\) if f is odd.
5. \(\int_0^{2a} f(x)dx = \int_0^a f(x)dx + \int_0^a f(2a-x)dx\)

Step 1: Understanding the Question:


We need to evaluate the integral. The structure of the denominator, \(\sqrt{a^2 - u^2}\), suggests an inverse trigonometric function, specifically \(\sin^{-1}\).


Step 2: Key Formula or Approach:


We will use the method of substitution. The standard integral form is:
\[ \int \frac{du}{\sqrt{a^2 - u^2}} = \sin^{-1}\left(\frac{u}{a}\right) + C \]

We need to choose a substitution that transforms our integral into this form.


Step 3: Detailed Explanation:


Let the integral be I.
\[ I = \int \frac{e^x}{\sqrt{4 - (e^x)^2}} dx \]

Let's make the substitution \(u = e^x\).

Then, the derivative is \(\frac{du}{dx} = e^x\), which means \(du = e^x dx\).

The numerator \(e^x dx\) is exactly our \(du\).

The integral transforms to:
\[ I = \int \frac{du}{\sqrt{4 - u^2}} \]

This matches the standard form with \(a^2 = 4\), so \(a=2\).

Applying the formula:
\[ I = \sin^{-1}\left(\frac{u}{2}\right) + C \]

Now, substitute back \(u = e^x\):
\[ I = \sin^{-1}\left(\frac{e^x}{2}\right) + C \]


Step 4: Final Answer:


The value of the integral is \(\sin^{-1}(\frac{e^x}{2}) + C\).
Quick Tip: When you see an integrand with a square root in the denominator, look for these patterns:
- \(\sqrt{a^2 - u^2}\) suggests a \(\sin^{-1}\) substitution (\(u = a\sin\theta\)).
- \(\sqrt{a^2 + u^2}\) suggests a \(\tan^{-1}\) substitution (\(u = a\tan\theta\)).
- \(\sqrt{u^2 - a^2}\) suggests a \(\sec^{-1}\) substitution (\(u = a\sec\theta\)).
The key is to choose 'u' such that 'du' also appears in the integrand.

Step 1: Understanding the Question:


The ropes are parallel, which means their direction vectors must be parallel.

Two vectors are parallel if one is a scalar multiple of the other.


Step 2: Key Formula or Approach:


Let the two vectors be \(\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}\) and \(\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}\).

If \(\vec{a}\) and \(\vec{b}\) are parallel, then their corresponding components must be in proportion.
\[ \frac{a_1}{b_1} = \frac{a_2}{b_2} = \frac{a_3}{b_3} \]


Step 3: Detailed Explanation:


Let \(\vec{a} = 3\hat{i} + 15\hat{j} + 6\hat{k}\) and \(\vec{b} = 2\hat{i} + 10\hat{j} + \lambda\hat{k}\).

Since they are parallel, we set up the ratio of their components:
\[ \frac{3}{2} = \frac{15}{10} = \frac{6}{\lambda} \]

Let's check the first two parts of the ratio to find the constant of proportionality:
\[ \frac{3}{2} = 1.5 \]
\[ \frac{15}{10} = 1.5 \]

The ratios are equal, which confirms the vectors are parallel. Now we use this ratio to find \(\lambda\).
\[ \frac{3}{2} = \frac{6}{\lambda} \]

Cross-multiply to solve for \(\lambda\):
\[ 3\lambda = 2 \times 6 = 12 \]
\[ \lambda = \frac{12}{3} = 4 \]


Step 4: Final Answer:


The value of \(\lambda\) is 4.
Quick Tip: Two vectors are parallel if their components are proportional.
Two vectors are perpendicular if their dot product is zero.
These are the two most fundamental conditions for relating vectors in 3D space.

Step 1: Understanding the Question:


We are given an equality involving the magnitudes of the sum and difference of two vectors. We need to determine the geometric relationship between these vectors.


Step 2: Key Formula or Approach:


The magnitude squared of a vector \(\vec{v}\) is equal to the dot product of the vector with itself: \(|\vec{v}|^2 = \vec{v} \cdot \vec{v}\).

We will square both sides of the given equation and use this property to expand the terms.


Step 3: Detailed Explanation:


Given: \(|\vec{a} + \vec{b}| = |\vec{a} - \vec{b}|\).

Square both sides:
\[ |\vec{a} + \vec{b}|^2 = |\vec{a} - \vec{b}|^2 \]

Use the property \(|\vec{v}|^2 = \vec{v} \cdot \vec{v}\):
\[ (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = (\vec{a} - \vec{b}) \cdot (\vec{a} - \vec{b}) \]

Expand both sides using the distributive property of the dot product:
\[ \vec{a}\cdot\vec{a} + \vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{a} + \vec{b}\cdot\vec{b} = \vec{a}\cdot\vec{a} - \vec{a}\cdot\vec{b} - \vec{b}\cdot\vec{a} + \vec{b}\cdot\vec{b} \]

Using \(|\vec{v}|^2 = \vec{v}\cdot\vec{v}\) and the commutative property \(\vec{a}\cdot\vec{b} = \vec{b}\cdot\vec{a}\):
\[ |\vec{a}|^2 + 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2 = |\vec{a}|^2 - 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2 \]

Cancel the \(|\vec{a}|^2\) and \(|\vec{b}|^2\) terms from both sides:
\[ 2(\vec{a} \cdot \vec{b}) = -2(\vec{a} \cdot \vec{b}) \]
\[ 4(\vec{a} \cdot \vec{b}) = 0 \]
\[ \vec{a} \cdot \vec{b} = 0 \]


Step 4: Final Answer:


The dot product of two non-zero vectors is zero if and only if they are orthogonal (perpendicular) to each other.

Therefore, \(\vec{a}\) and \(\vec{b}\) are orthogonal vectors.
Quick Tip: The condition \(|\vec{a} + \vec{b}| = |\vec{a} - \vec{b}|\) has a clear geometric interpretation.
The vectors \(\vec{a} + \vec{b}\) and \(\vec{a} - \vec{b}\) represent the diagonals of the parallelogram formed by vectors \(\vec{a}\) and \(\vec{b}\).
The diagonals of a parallelogram are equal in length if and only if the parallelogram is a rectangle. This means the adjacent sides, \(\vec{a}\) and \(\vec{b}\), must be orthogonal.

Step 1: Understanding the Question:


We need to find the conditional probability of the complement of A given B, i.e., P(A' | B).


Step 2: Key Formula or Approach:


We use the formula for conditional probability:
\[ P(\bar{A}|B) = \frac{P(\bar{A} \cap B)}{P(B)} \]

The event \(\bar{A} \cap B\) means "B happens but A does not happen". The probability for this is given by \(P(B) - P(A \cap B)\).

Alternatively, we can use the property \(P(\bar{A}|B) = 1 - P(A|B)\). This is often faster.


Step 3: Detailed Explanation (Method 2)


First, let's calculate P(A|B):
\[ P(A|B) = \frac{P(A \cap B)}{P(B)} \]

Substitute the given values:
\[ P(A|B) = \frac{4/7}{5/7} = \frac{4}{5} \]

Now, use the complement property for conditional probability:
\[ P(\bar{A}|B) = 1 - P(A|B) \]
\[ P(\bar{A}|B) = 1 - \frac{4}{5} = \frac{1}{5} \]


Step 4: Final Answer:


The value of P(\(\bar{A}\) | B) is \(\frac{1}{5}\).
Quick Tip: The property \(P(A'|B) = 1 - P(A|B)\) is very useful and can save you from having to calculate \(P(A' \cap B)\) directly.
It holds because, given that B has occurred, either A occurs or A' occurs, so the sum of their conditional probabilities must be 1.

Step 1: Understanding the Question:


We need to find the probability of two independent events both occurring: getting a head on a coin toss AND getting a face card from a deck.


Step 2: Key Formula or Approach:


For two independent events E1 and E2, the probability that both occur is the product of their individual probabilities:

P(E1 and E2) = P(E1) \(\times\) P(E2).


Step 3: Detailed Explanation:


Let E1 be the event of getting a head on a coin toss.

For a fair coin, P(E1) = P(Head) = \(\frac{1}{2}\).


Let E2 be the event of drawing a face card from a standard 52-card deck.


A standard deck has 12 face cards (Jack, Queen, King in each of the 4 suits).


The total number of cards is 52.


So, P(E2) = P(Face Card) = \(\frac{12}{52}\).


Simplify this fraction by dividing by 4: \(P(Face Card) = \frac{3}{13}\).


Since the events are independent, the probability of both happening is:


P(Head and Face Card) = P(Head) \(\times\) P(Face Card)

\[ = \frac{1}{2} \times \frac{3}{13} = \frac{3}{26} \]


Step 4: Final Answer:


The probability of getting a head and a face card is \(\frac{3}{26}\).
Quick Tip: Remember to identify if events are independent or dependent.
Tossing a coin and drawing a card from a separate deck are classic examples of independent events.
For independent events, the "and" rule means you multiply the probabilities.

Step 1: Understanding the Question:


We need to check the continuity of the function f(x) at x=0 using the concept of limits, and determine if the Reason correctly explains the Assertion.


Step 2: Analyze the Reason


The Reason (R) states that as x approaches 0, sin(1/x) is a finite value between -1 and 1.

The range of the sine function is always [-1, 1], regardless of its input. So, for any non-zero x, \(-1 \leq \sin(1/x) \leq 1\).

The statement is correct. Reason (R) is true.


Step 3: Verify the Assertion using the Squeeze Theorem


For f(x) to be continuous at x=0, we need \(\lim_{x \to 0} f(x) = f(0)\).

We are given f(0) = 0. We need to evaluate the limit:
\[ \lim_{x \to 0} x \sin(1/x) \]

We know from the Reason that \(-1 \leq \sin(1/x) \leq 1\).

Let's multiply this inequality by x. We must consider two cases:


- If x \(>\) 0: \(-x \leq x\sin(1/x) \leq x\).


- If x \(<\) 0: \(-x \geq x\sin(1/x) \geq x\), which is the same as \(x \leq x\sin(1/x) \leq -x\).


In both cases, as x \(\rightarrow\) 0, the lower bound (\(\pm x\)) and the upper bound (\(\mp x\)) both approach 0.

By the Squeeze (or Sandwich) Theorem, the function in the middle must also approach 0.

\[ \lim_{x \to 0} x \sin(1/x) = 0 \]


Since the limit (0) is equal to the function's value f(0) (which is 0), the function is continuous at x=0. Assertion (A) is true.


The Reason explains the key property of \(\sin(1/x)\) that allows the Squeeze Theorem to work and prove the limit. Therefore, the Reason is the correct explanation.


Step 4: Final Answer:


Both A and R are true, and R is the correct explanation of A.
Quick Tip: The Squeeze Theorem is the standard method for finding limits of functions like \(x^n \sin(1/x)\) or \(x^n \cos(1/x)\) as x approaches 0.
The key is that \(\sin(1/x)\) and \(\cos(1/x)\) are bounded (between -1 and 1), while the \(x^n\) term goes to zero, "squeezing" the whole expression to zero.

Step 1: Analyze the Reason


The Reason (R) states the domain of the inverse secant function. The domain of sec\(^{-1}\)x is indeed the set of all real numbers x such that \(|x| \geq 1\).

This can be written as x \(\in\) (-\(\infty\), -1] \(\cup\) [1, \(\infty\)).

The set R - (-1, 1) is exactly this domain. Therefore, Reason (R) is true.


Step 2: Verify the Assertion using the Reason


The Assertion (A) is about the value of sec\(^{-1}(\frac{\sqrt{3}}{2})\).

Let's evaluate the input value: \( \frac{\sqrt{3}}{2} \approx \frac{1.732}{2} = 0.866 \).

This value, 0.866, lies within the interval (-1, 1).

According to the Reason, the function sec\(^{-1}\)x is NOT defined for values inside the interval (-1, 1).

Since \(\frac{\sqrt{3}}{2}\) is not in the domain of sec\(^{-1}\)x, there is no value for sec\(^{-1}(\frac{\sqrt{3}}{2})\).

This means the set of its possible values is the empty set (or null set). Therefore, Assertion (A) is true.

The Reason provides the domain rule that directly leads to the conclusion in the Assertion.


Step 3: Final Answer:


Both A and R are true, and R is the correct explanation of A.
Quick Tip: Domain and Range are fundamental to understanding inverse trigonometric functions.
- The domains of \(\sin^{-1}x\) and \(\cos^{-1}x\) are [-1, 1].
- The domains of \(\sec^{-1}x\) and \(\csc^{-1}x\) are the "outside" of that: \(|x| \geq 1\).
- The domains of \(\tan^{-1}x\) and \(\cot^{-1}x\) are all real numbers.

Step 1: Understanding Bijectivity


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

- One-one (Injective): \(f(x_1) = f(x_2) \implies x_1 = x_2\).

- Onto (Surjective): For every y in the codomain B, there is an x in the domain A such that f(x) = y.


Step 2: Check for One-one (Injective)


Let \(f(x_1) = f(x_2)\) for \(x_1, x_2 \in A\).
\[ \frac{x_1 - 2}{x_1 - 3} = \frac{x_2 - 2}{x_2 - 3} \]

Cross-multiply:
\[ (x_1 - 2)(x_2 - 3) = (x_2 - 2)(x_1 - 3) \]
\[ x_1x_2 - 3x_1 - 2x_2 + 6 = x_1x_2 - 3x_2 - 2x_1 + 6 \]

Cancel common terms (\(x_1x_2\) and 6) from both sides:
\[ -3x_1 - 2x_2 = -3x_2 - 2x_1 \]

Rearrange to group \(x_1\) and \(x_2\) terms:
\[ 3x_2 - 2x_2 = 3x_1 - 2x_1 \]
\[ x_2 = x_1 \]

Since \(f(x_1) = f(x_2)\) implies \(x_1 = x_2\), the function is one-one.


Step 3: Check for Onto (Surjective)


Let y be an arbitrary element in the codomain B (so \(y \in R - \{1\}\)). We need to find if there is a pre-image x in the domain A.

Set \(y = f(x)\) and solve for x:
\[ y = \frac{x-2}{x-3} \]
\[ y(x-3) = x-2 \]
\[ yx - 3y = x - 2 \]

Group terms with x:
\[ yx - x = 3y - 2 \]
\[ x(y - 1) = 3y - 2 \]
\[ x = \frac{3y - 2}{y - 1} \]

For x to be defined, the denominator cannot be zero, so \(y-1 \neq 0\), which means \(y \neq 1\). This is guaranteed because our codomain B is R - \{1\.

We also need to check if the calculated x can ever be 3 (the value excluded from the domain A).

If \(x=3\), then \(3 = \frac{3y-2}{y-1} \implies 3y-3 = 3y-2 \implies -3=-2\), which is a contradiction.

So, x is never 3.

Thus, for every y in the codomain B, we can find a valid x in the domain A. The function is onto.


Step 4: Final Answer:


Since the function is both one-one and onto, it is a bijection.
Quick Tip: For rational functions \(f(x) = \frac{ax+b}{cx+d}\), they are typically one-one on their domain.
To check for surjectivity, solve for x in terms of y. The function is onto its range. If the given codomain matches the calculated range, the function is surjective. The range of this function is R - \{a/c\}, which is R - \{1\} in this case.

Step 1: Understanding the Question:


We need to verify that the matrix A satisfies the given polynomial equation. This is an application of the Cayley-Hamilton theorem, which states that every square matrix satisfies its own characteristic equation.


Step 2: Calculate A\(^2\)

\[ A^2 = A \cdot A = \begin{bmatrix} 2 & 3
-1 & 2 \end{bmatrix} \begin{bmatrix} 2 & 3
-1 & 2 \end{bmatrix} \]
\[ A^2 = \begin{bmatrix} (2)(2)+(3)(-1) & (2)(3)+(3)(2)
(-1)(2)+(2)(-1) & (-1)(3)+(2)(2) \end{bmatrix} \]
\[ A^2 = \begin{bmatrix} 4-3 & 6+6
-2-2 & -3+4 \end{bmatrix} = \begin{bmatrix} 1 & 12
-4 & 1 \end{bmatrix} \]


Step 3: Calculate 4A

\[ 4A = 4 \begin{bmatrix} 2 & 3
-1 & 2 \end{bmatrix} = \begin{bmatrix} 8 & 12
-4 & 8 \end{bmatrix} \]


Step 4: Calculate 7I


I is the 2x2 identity matrix.
\[ 7I = 7 \begin{bmatrix} 1 & 0
0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & 0
0 & 7 \end{bmatrix} \]


Step 5: Evaluate the Expression A\(^2\) - 4A + 7I

\[ A^2 - 4A + 7I = \begin{bmatrix} 1 & 12
-4 & 1 \end{bmatrix} - \begin{bmatrix} 8 & 12
-4 & 8 \end{bmatrix} + \begin{bmatrix} 7 & 0
0 & 7 \end{bmatrix} \]

Perform the subtraction and addition element-wise:
\[ = \begin{bmatrix} 1-8+7 & 12-12+0
-4-(-4)+0 & 1-8+7 \end{bmatrix} \]
\[ = \begin{bmatrix} 0 & 0
0 & 0 \end{bmatrix} = O \]


Step 6: Final Answer:


The calculation shows that \(A^2 - 4A + 7I\) is indeed the zero matrix, O.
Quick Tip: For a 2x2 matrix \(A = \begin{bmatrix} a & b
c & d \end{bmatrix}\), its characteristic equation is always \( \lambda^2 - tr(A)\lambda + \det(A) = 0 \).
The Cayley-Hamilton theorem states that the matrix itself satisfies this: \( A^2 - tr(A)A + \det(A)I = O \).
Here, trace(A) = a+d = 2+2 = 4.
det(A) = ad-bc = (2)(2) - (3)(-1) = 4+3 = 7.
So the equation must be \(A^2 - 4A + 7I = O\), which matches the question. This is a great way to verify the result quickly.

Step 1: Understanding the Question:


We need to find the derivative of a function that is a quotient of an exponential function and a power function.


Step 2: Key Formula or Approach:


We will use the quotient rule for differentiation:
\[ \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \]

Here, \(u = 5^x\) and \(v = x^5\).

We also need the derivative of an exponential function: \(\frac{d}{dx}(a^x) = a^x \ln(a)\).


Step 3: Detailed Explanation:


Let \(y = \frac{5^x}{x^5}\).

First, find the derivatives of u and v:

- \(u' = \frac{d}{dx}(5^x) = 5^x \ln(5)\).

- \(v' = \frac{d}{dx}(x^5) = 5x^4\).

Now, apply the quotient rule:
\[ \frac{dy}{dx} = \frac{(5^x \ln 5)(x^5) - (5^x)(5x^4)}{(x^5)^2} \]

Factor out the common terms \(5^x\) and \(x^4\) from the numerator:
\[ \frac{dy}{dx} = \frac{5^x x^4 (\ln(5) \cdot x - 5)}{x^{10}} \]

Cancel \(x^4\) from the numerator and denominator:
\[ \frac{dy}{dx} = \frac{5^x (\ln(5) \cdot x - 5)}{x^6} \]


Step 4: Final Answer:


The derivative is \(\frac{5^x (x \ln 5 - 5)}{x^6}\).
Quick Tip: When using the quotient rule, remember the order: "low d-high minus high d-low, over the square of what's below".
Don't forget the derivative of \(a^x\) is \(a^x \ln a\), which is different from the derivative of \(x^n\) (power rule) and \(e^x\).

Step 1: Understanding the Question:


We need to find the derivative of y with respect to x for an implicitly defined function.


Step 2: Key Formula or Approach:


We will differentiate each term of the equation with respect to x, treating y as a function of x. This requires using the product rule for the `-5xy` term and the chain rule for the `y^3` term.


Step 3: Detailed Explanation:


The equation is \(-2x^2 – 5xy + y^3 = 76\).

Differentiate term by term with respect to x:
\[ \frac{d}{dx}(-2x^2) - \frac{d}{dx}(5xy) + \frac{d}{dx}(y^3) = \frac{d}{dx}(76) \]

- \(\frac{d}{dx}(-2x^2) = -4x\)

- For the middle term, use the product rule: \(\frac{d}{dx}(5xy) = 5\left( (1)y + x\frac{dy}{dx} \right) = 5y + 5x\frac{dy}{dx}\)

- For the y-term, use the chain rule: \(\frac{d}{dx}(y^3) = 3y^2 \frac{dy}{dx}\)

- The derivative of the constant 76 is 0.

Putting it all together:
\[ -4x - \left(5y + 5x\frac{dy}{dx}\right) + 3y^2\frac{dy}{dx} = 0 \]
\[ -4x - 5y - 5x\frac{dy}{dx} + 3y^2\frac{dy}{dx} = 0 \]


Step 4: Isolate \(\frac{dy}{dx}\)


Group the terms with \(\frac{dy}{dx}\) on one side and the other terms on the other side.
\[ 3y^2\frac{dy}{dx} - 5x\frac{dy}{dx} = 4x + 5y \]

Factor out \(\frac{dy}{dx}\):
\[ \frac{dy}{dx}(3y^2 - 5x) = 4x + 5y \]

Solve for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{4x + 5y}{3y^2 - 5x} \]


Step 5: Final Answer:


The derivative is \(\frac{dy}{dx} = \frac{4x+5y}{3y^2-5x}\).
Quick Tip: For implicit differentiation, go term by term.
Remember that whenever you differentiate a term containing 'y', you must multiply by \(\frac{dy}{dx}\) because of the chain rule.
Be especially careful with the product rule for terms that mix x and y, like 'xy'.

Step 1: Understanding the Problem


We need to graphically solve a maximization LPP. This involves plotting the constraints, identifying the feasible region, finding the corner points, and then testing these points in the objective function.


Step 2: Plot the Constraints and Find the Feasible Region


First, plot the boundary lines for each constraint.

1. 3x + y = 6: This line passes through (2, 0) and (0, 6). Since the inequality is \(\leq\), the region is on and below this line.

2. x = 1: This is a vertical line at x = 1. Since the inequality is \(\leq\), the region is to the left of this line.

3. x \(\geq\) 0 and y \(\geq\) 0: This restricts the region to the first quadrant.

The feasible region is the area that satisfies all these conditions simultaneously. It will be a bounded polygon.


Step 3: Identify the Corner Points


The corner points are the vertices of the feasible region.

- O (Origin): The intersection of x=0 and y=0, which is (0, 0).

- A: The intersection of the x-axis (y=0) and the line x=1. This gives the point (1, 0).

- B: The intersection of the lines x=1 and 3x + y = 6.

Substitute x=1 into the second equation: \(3(1) + y = 6 \implies 3 + y = 6 \implies y = 3\). This gives the point (1, 3).

- C: The intersection of the y-axis (x=0) and the line 3x + y = 6.

Substitute x=0: \(3(0) + y = 6 \implies y = 6\). This gives the point (0, 6).

The corner points are O(0, 0), A(1, 0), B(1, 3), and C(0, 6).


Step 4: Evaluate the Objective Function at Corner Points


The objective function is Z = 5x + 4y.

- At O(0, 0): \(Z = 5(0) + 4(0) = 0\)

- At A(1, 0): \(Z = 5(1) + 4(0) = 5\)

- At B(1, 3): \(Z = 5(1) + 4(3) = 5 + 12 = 17\) (Maximum)

- At C(0, 6): \(Z = 5(0) + 4(6) = 24\).

Wait, let me re-check. `Z(C) = 24`. This is greater than 17.
Let's re-check the constraints.
`3x+y <= 6`. `x <= 1`.
Let's check C(0,6). x=0, y=6. `3(0)+6 = 6`. `6 <= 6` is true. `0 <= 1` is true. So C(0,6) is a valid corner point.
Let's check B(1,3). x=1, y=3. `3(1)+3 = 6`. `6 <= 6` is true. `1 <= 1` is true. So B(1,3) is a valid corner point.
Calculation of Z:
Z(B) = 5(1)+4(3) = 17.
Z(C) = 5(0)+4(6) = 24.
The maximum value is 24 at C(0,6).

Final Answer:


The corner points are O(0,0), A(1,0), B(1,3), and C(0,6). The maximum value of Z is 24, which occurs at the point C(0, 6).


\begin{figure[h!]
\centering
\begin{tikzpicture[scale=1]
\draw[->, thick] (-1,0) -- (4,0) node[right] {x;
\draw[->, thick] (0,-1) -- (0,7) node[above] {y;
\node[below left] at (0,0) {O;

\foreach \x in {1,2,3 \draw (\x,1pt) -- (\x,-1pt) node[below] {\x;
\foreach \y in {1,2,3,4,5,6 \draw (1pt,\y) -- (-1pt,\y) node[left] {\y;

% Define corner points
\coordinate (O) at (0,0);
\coordinate (A) at (1,0);
\coordinate (B) at (1,3);
\coordinate (C) at (0,6);

% Shade the feasible region
\fill[cyan!20, opacity=0.7] (O) -- (A) -- (B) -- (C) -- cycle;

% Draw boundary lines
\draw[red, thick] (-0.5, 7.5) -- (2.5, -1.5) node[pos=0.8, below, sloped] {\(3x+y=6\);
\draw[blue, thick] (1,-1) -- (1,7) node[right] {\(x=1\);

% Label corner points
\filldraw[black] (O) circle (2pt) node[below left] {O(0,0);
\filldraw[black] (A) circle (2pt) node[below right] {A(1,0);
\filldraw[black] (B) circle (2pt) node[above right] {B(1,3);
\filldraw[black] (C) circle (2pt) node[above left] {C(0,6);

\node[align=center, font=\bfseries] at (0.5, 2) {Feasible
Region;
\end{tikzpicture
\end{figure Quick Tip: Graphing an LPP is a systematic process:
1. Plot each constraint line using its intercepts.
2. Shade the correct side of each line based on the inequality sign (\(\leq\) or \(\geq\)). A good way is to test the origin (0,0).
3. Identify the overlapping (feasible) region.
4. Find the coordinates of all vertices (corner points) of this region.
5. Test each vertex in the objective function to find the maximum or minimum.

Step 1: Understand the Random Variable and Outcomes


The random variable X represents the number on a randomly drawn block.

There are 10 blocks in total.

The possible values (outcomes) for X are {0, 1, 2, 3.


Step 2: Find the Probability Distribution of X


The probability of each outcome is the number of blocks with that number divided by the total number of blocks (10).

- Probability of X = 0:

There are 2 blocks marked with '0'.

\(P(X=0) = \frac{2}{10} = \frac{1}{5}\).


- Probability of X = 1:

There are 3 blocks marked with '1'.

\(P(X=1) = \frac{3}{10}\).


- Probability of X = 2:

There are 4 blocks marked with '2'.

\(P(X=2) = \frac{4}{10} = \frac{2}{5}\).


- Probability of X = 3:

There is 1 block marked with '3'.

\(P(X=3) = \frac{1}{10}\).


The probability distribution can be presented in a table:


\begin{tabular{|c|c|c|c|c|
\hline
X = x\(_i\) & 0 & 1 & 2 & 3

\hline
P(X = x\(_i\)) & 2/10 & 3/10 & 4/10 & 1/10

\hline
\end{tabular


*(As a check, the sum of probabilities is \(\frac{2}{10} + \frac{3}{10} + \frac{4}{10} + \frac{1}{10} = \frac{10}{10} = 1\).)*


Step 3: Calculate the Mean of the Distribution


The mean (\(\mu\) or E[X]) of a discrete random variable is calculated using the formula \(\mu = E[X] = \sum x_i \cdot P(X=x_i)\).
\[ \mu = (0)\left(\frac{2}{10}\right) + (1)\left(\frac{3}{10}\right) + (2)\left(\frac{4}{10}\right) + (3)\left(\frac{1}{10}\right) \]
\[ \mu = 0 + \frac{3}{10} + \frac{8}{10} + \frac{3}{10} \]
\[ \mu = \frac{3 + 8 + 3}{10} = \frac{14}{10} = 1.4 \]


Step 4: Final Answer:


The probability distribution of X is given by the table above.

The mean of the distribution is 1.4.
Quick Tip: A probability distribution lists all possible outcomes of a random variable and their corresponding probabilities.
Always check that the sum of all probabilities in your distribution equals 1, as this confirms you haven't missed any outcomes or made a calculation error.
The mean, or expected value, is a weighted average of the outcomes, where each outcome is weighted by its probability of occurring.

Step 1: Understand the Question and Define Events


We need to find the probability of the union of two events: P(Woman \(\cup\) Works Outside).

Let W be the event that the chosen person is a woman.

Let O be the event that the chosen person works outside the village.


Step 2: Key Formula or Approach


We will use the addition rule for probability:
\[ P(W \cup O) = P(W) + P(O) - P(W \cap O) \]


Step 3: Calculate the Probabilities of Individual and Intersection Events


Total number of people = 8000.

- P(W): Probability of being a woman.


Number of women = 4000.


\( P(W) = \frac{4000}{8000} = \frac{1}{2} \).


- P(O): Probability of working outside.


Number of people working outside = 3000.


\( P(O) = \frac{3000}{8000} = \frac{3}{8} \).


- P(W \(\cap\) O): Probability of being a woman AND working outside.


We are told 30% of women work outside.


Number of women working outside = 30% of 4000 = 0.30 \(\times\) 4000 = 1200.


\( P(W \cap O) = \frac{1200}{8000} = \frac{12}{80} = \frac{3}{20} \).


Step 4: Apply the Addition Rule

\[ P(W \cup O) = P(W) + P(O) - P(W \cap O) \]

\[ P(W \cup O) = \frac{1}{2} + \frac{3}{8} - \frac{3}{20} \]

Find a common denominator, which is 40.
\[ P(W \cup O) = \frac{20}{40} + \frac{15}{40} - \frac{6}{40} = \frac{20 + 15 - 6}{40} = \frac{29}{40} \]


Step 5: Final Answer:


The probability that the person is either a woman or works outside is \(\frac{29}{40}\).
Quick Tip: For "A or B" probability questions, the addition rule \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\) is fundamental.
The most common mistake is forgetting to subtract the intersection, which accounts for the overlap (in this case, the women who work outside).
Using a Venn diagram or a contingency table can help organize the numbers.

Step 1: Understanding Bijectivity


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


Step 2: Proving One-one (Injective)


Let \(f(x_1) = f(x_2)\) for any \(x_1, x_2 \in R\).
\[ 4x_1^3 - 5 = 4x_2^3 - 5 \]

Add 5 to both sides:
\[ 4x_1^3 = 4x_2^3 \]

Divide by 4:
\[ x_1^3 = x_2^3 \]

Take the cube root of both sides. Since the cube root function is a well-defined one-to-one function for all real numbers:
\[ x_1 = x_2 \]

Since \(f(x_1) = f(x_2)\) implies \(x_1 = x_2\), the function is one-one.


Step 3: Proving Onto (Surjective)


Let y be an arbitrary element in the codomain R. We need to show that there exists a pre-image x in the domain R such that f(x) = y.

Set \(y = f(x)\) and solve for x:
\[ y = 4x^3 - 5 \]
\[ y + 5 = 4x^3 \]
\[ x^3 = \frac{y+5}{4} \]
\[ x = \sqrt[3]{\frac{y+5}{4}} \]

For any real number y, the expression \(\frac{y+5}{4}\) is also a real number. The cube root of any real number is a well-defined real number.

Therefore, for every y in the codomain R, we can find a corresponding x in the domain R.

The function is onto.


Step 4: Final Answer:


Since the function f(x) is both one-one and onto, it is a bijection.
Quick Tip: An alternative method to prove injectivity for differentiable functions is to check the derivative.
Here, \(f'(x) = 12x^2\). Since \(f'(x) \geq 0\) for all x, the function is monotonically non-decreasing. It is strictly increasing for \(x \neq 0\). This implies it is one-one.
Odd-degree polynomial functions are often good candidates for being bijections from R to R.

Step 1: Understanding the Relation


The relation R on the set of natural numbers N is defined as (x, y) \(\in\) R if their product, xy, is a perfect square.

To be an equivalence relation, R must be reflexive, symmetric, and transitive.


Step 2: Checking for Reflexivity


A relation is reflexive if (x, x) \(\in\) R for all x \(\in\) N.

We need to check if \(x \cdot x\) is a perfect square.
\(x \cdot x = x^2\), which is by definition the square of a natural number x.

So, (x, x) \(\in\) R for all x \(\in\) N. The relation is reflexive.


Step 3: Checking for Symmetry


A relation is symmetric if (x, y) \(\in\) R implies (y, x) \(\in\) R.

Assume (x, y) \(\in\) R. This means \(xy = k^2\) for some natural number k.

We need to check if yx is a perfect square.

Since multiplication is commutative, \(yx = xy\).

Therefore, \(yx = k^2\), which is a perfect square.

So, (y, x) \(\in\) R. The relation is symmetric.


Step 4: Checking for Transitivity


A relation is transitive if (x, y) \(\in\) R and (y, z) \(\in\) R implies (x, z) \(\in\) R.

Assume (x, y) \(\in\) R, which means \(xy = k^2\) for some \(k \in N\).

Assume (y, z) \(\in\) R, which means \(yz = m^2\) for some \(m \in N\).

We need to check if xz is a perfect square.

Let's multiply the two equations:
\[ (xy)(yz) = k^2 m^2 \]
\[ xz \cdot y^2 = (km)^2 \]
\[ xz = \frac{(km)^2}{y^2} = \left(\frac{km}{y}\right)^2 \]

Now, we need to be sure that \(\frac{km}{y}\) is a natural number.

We know \(x=k^2/y\) and \(z=m^2/y\). Since x and z are natural numbers, y must be a divisor of both \(k^2\) and \(m^2\).
This implies that the prime factorization of y contains prime factors with powers less than or equal to their powers in \(k^2\) and \(m^2\).
This does not guarantee that y divides km.
Let's use a prime factorization approach. A number is a perfect square if all the exponents in its prime factorization are even.

Let the prime factorization of x, y, z be \(x = p_1^{a_1} p_2^{a_2} \dots\), \(y = p_1^{b_1} p_2^{b_2} \dots\), \(z = p_1^{c_1} p_2^{c_2} \dots\).

- (x,y) \(\in\) R means \(a_i + b_i\) is even for all i.

- (y,z) \(\in\) R means \(b_i + c_i\) is even for all i.

We need to check if \(a_i + c_i\) is even.

If \(b_i\) is even, then for \(a_i+b_i\) to be even, \(a_i\) must be even. For \(b_i+c_i\) to be even, \(c_i\) must be even. Then \(a_i+c_i\) (even+even) is even.

If \(b_i\) is odd, then for \(a_i+b_i\) to be even, \(a_i\) must be odd. For \(b_i+c_i\) to be even, \(c_i\) must be odd. Then \(a_i+c_i\) (odd+odd) is even.

In both cases, \(a_i + c_i\) is even for all i. This means xz is a perfect square.

The relation is transitive.


Step 5: Final Answer:


Since the relation R is reflexive, symmetric, and transitive, it is an equivalence relation.
Quick Tip: For relations involving properties like "is a multiple of", "is a factor of", or "is a perfect square", thinking in terms of prime factorization can be a very powerful and rigorous way to prove or disprove transitivity.
A number 'n' is a perfect square if and only if every exponent in its prime factorization is an even number.

Step 1: Understanding the Question:


We need to find the point of intersection of two lines. The problem specifies that we must use the matrix method.

This involves setting up the system of linear equations in matrix form (AX = B) and solving for X using the inverse of matrix A (\(X = A^{-1}B\)).


Step 2: Set Up the Matrix Equation


First, rewrite the equations in standard form:
\(2x + 5y = 1\)
\(3x + 2y = 7\)

Now, express this system in the matrix form AX = B:
\[ \begin{bmatrix} 2 & 5
3 & 2 \end{bmatrix} \begin{bmatrix} x
y \end{bmatrix} = \begin{bmatrix} 1
7 \end{bmatrix} \]

Where \( A = \begin{bmatrix} 2 & 5
3 & 2 \end{bmatrix} \), \( X = \begin{bmatrix} x
y \end{bmatrix} \), and \( B = \begin{bmatrix} 1
7 \end{bmatrix} \).


Step 3: Find the Inverse of Matrix A


First, calculate the determinant of A:
\(|A| = (2)(2) - (5)(3) = 4 - 15 = -11\).

Since \(|A| \neq 0\), the inverse exists.

Next, find the adjugate of A. For a 2x2 matrix \(\begin{bmatrix} a & b
c & d \end{bmatrix}\), the adjugate is \(\begin{bmatrix} d & -b
-c & a \end{bmatrix}\).
\[ adj(A) = \begin{bmatrix} 2 & -5
-3 & 2 \end{bmatrix} \]

The inverse is \(A^{-1} = \frac{1}{|A|} adj(A)\):
\[ A^{-1} = \frac{1}{-11} \begin{bmatrix} 2 & -5
-3 & 2 \end{bmatrix} \]


Step 4: Solve for X


Now, use the formula \(X = A^{-1}B\):
\[ \begin{bmatrix} x
y \end{bmatrix} = \frac{1}{-11} \begin{bmatrix} 2 & -5
-3 & 2 \end{bmatrix} \begin{bmatrix} 1
7 \end{bmatrix} \]
\[ \begin{bmatrix} x
y \end{bmatrix} = \frac{1}{-11} \begin{bmatrix} (2)(1) + (-5)(7)
(-3)(1) + (2)(7) \end{bmatrix} \]
\[ \begin{bmatrix} x
y \end{bmatrix} = \frac{1}{-11} \begin{bmatrix} 2 - 35
-3 + 14 \end{bmatrix} = \frac{1}{-11} \begin{bmatrix} -33
11 \end{bmatrix} \]
\[ \begin{bmatrix} x
y \end{bmatrix} = \begin{bmatrix} -33 / -11
11 / -11 \end{bmatrix} = \begin{bmatrix} 3
-1 \end{bmatrix} \]


Step 5: Final Answer:


The common point to the paths is (x, y) = (3, -1).
Quick Tip: The matrix method is a systematic way to solve systems of linear equations.
For a 2x2 system, the key steps are:
1. Write the system as AX = B.
2. Find the determinant and inverse of A.
3. The solution is always given by \(X = A^{-1}B\).
This method is particularly powerful for larger systems.

Step 1: Understanding the Question:


We are asked to perform two tasks:

1. Calculate the total sales revenue over two days using matrix multiplication.

2. Calculate the profit by subtracting the given total cost from the total sales revenue.


Step 2: Set Up the Matrices


We can represent the sales data in a 2 \(\times\) 3 matrix (let's call it S), where the rows represent Day I and Day II, and the columns represent the number of Chemistry, Physics, and Maths books sold.
\[ S = \begin{bmatrix} 50 & 60 & 35
40 & 45 & 50 \end{bmatrix} \]

We can represent the selling price of each book in a 3 \(\times\) 1 column matrix (let's call it P).
\[ P = \begin{bmatrix} 150
175
180 \end{bmatrix} \]


Step 3: Calculate Total Sales using Matrix Multiplication


To find the sales revenue for each day, we multiply the Sales matrix (S) by the Price matrix (P). The product SP will be a 2 \(\times\) 1 matrix, where the first row is the total sale for Day I and the second row is the total sale for Day II.
\[ Sales per day = SP = \begin{bmatrix} 50 & 60 & 35
40 & 45 & 50 \end{bmatrix} \begin{bmatrix} 150
175
180 \end{bmatrix} \]
\[ = \begin{bmatrix} (50)(150) + (60)(175) + (35)(180)
(40)(150) + (45)(175) + (50)(180) \end{bmatrix} \]
\[ = \begin{bmatrix} 7500 + 10500 + 6300
6000 + 7875 + 9000 \end{bmatrix} = \begin{bmatrix} 24300
22875 \end{bmatrix} \]

This matrix shows that the sale on Day I was Rs 24,300 and the sale on Day II was Rs 22,875.

The total sale over two days is the sum of these values:

Total Sale = 24,300 + 22,875 = Rs 47,175.


Step 4: Calculate the Profit


The profit is the difference between the total sales and the total cost.

Total Selling Price = Rs 47,175.

Total Cost Price = Rs 35,000 (given).
\[ Profit = Total Selling Price - Total Cost Price \]
\[ Profit = 47,175 - 35,000 = 12,175 \]


Step 5: Final Answer:


Using the matrix method, the total sale in two days is Rs 47,175.

The profit earned after the sale of two days is Rs 12,175.
Quick Tip: Matrix multiplication is a powerful tool for organizing and calculating data in business and economics problems.
When setting up the matrices, ensure the dimensions are compatible for multiplication. To get the total revenue per day, a (Days \(\times\) Items) matrix should be multiplied by an (Items \(\times\) Price) matrix.
The resulting (Days \(\times\) Price) matrix provides a clear breakdown of revenue for each day.

Step 1: Understanding the Question:


We need to differentiate a composite function. This will require the chain rule to be applied multiple times.

The function can be seen as layers: log(u), where u = sin(v), and v = \(x^3/3 - 1\).


Step 2: Key Formula or Approach:


We will use the chain rule: \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dx}\).


The derivatives we need are:


- \(\frac{d}{du}(\log u) = \frac{1}{u}\)


- \(\frac{d}{dv}(\sin v) = \cos v\)


- \(\frac{d}{dx}\left(\frac{x^3}{3} - 1\right) = x^2\)


Step 3: Detailed Explanation:


Let's apply the chain rule step-by-step from the outermost function to the innermost.
\[ \frac{dy}{dx} = \frac{1}{\sin\left(\frac{x^3}{3} - 1\right)} \cdot \frac{d}{dx}\left[\sin\left(\frac{x^3}{3} - 1\right)\right] \]

Now, differentiate the sine function:
\[ \frac{dy}{dx} = \frac{1}{\sin\left(\frac{x^3}{3} - 1\right)} \cdot \cos\left(\frac{x^3}{3} - 1\right) \cdot \frac{d}{dx}\left(\frac{x^3}{3} - 1\right) \]

Finally, differentiate the innermost expression:
\[ \frac{d}{dx}\left(\frac{x^3}{3} - 1\right) = \frac{3x^2}{3} - 0 = x^2 \]

Putting it all together:
\[ \frac{dy}{dx} = \frac{1}{\sin\left(\frac{x^3}{3} - 1\right)} \cdot \cos\left(\frac{x^3}{3} - 1\right) \cdot x^2 \]

We can simplify this using the definition of cotangent, \(\cot\theta = \frac{\cos\theta}{\sin\theta}\):
\[ \frac{dy}{dx} = x^2 \cot\left(\frac{x^3}{3} - 1\right) \]


Step 4: Final Answer:


The derivative is \(\frac{dy}{dx} = x^2 \cot\left(\frac{x^3}{3} - 1\right)\).
Quick Tip: For nested functions, the chain rule is applied from the "outside-in".
Differentiate the outermost function, keeping the inside the same, then multiply by the derivative of the next function inside, and so on, until you reach the innermost variable.
Here: \( (\log(\dots))' \rightarrow (\sin(\dots))' \rightarrow (x^3/3-1)' \).

Step 1: Understanding the Question:


We are looking for two positive integers, let's call them x and y, that satisfy two conditions:

1. Their product is 289: \(xy = 289\).

2. Their sum, S = x + y, is minimized.


Step 2: Key Formula or Approach:


This is an optimization problem. We can express the sum S as a function of a single variable and find its minimum.

From the product condition, we can write \(y = \frac{289}{x}\).

Substitute this into the sum: \(S(x) = x + \frac{289}{x}\).

To find the minimum, we find the derivative \(S'(x)\) and set it to zero.


Step 3: Detailed Explanation:


The function to minimize is \(S(x) = x + 289x^{-1}\).

Find the derivative:
\[ S'(x) = 1 - 289x^{-2} = 1 - \frac{289}{x^2} \]

Set the derivative to zero to find critical points:
\[ 1 - \frac{289}{x^2} = 0 \implies 1 = \frac{289}{x^2} \implies x^2 = 289 \]

Since x must be a positive integer, we take the positive square root:
\[ x = \sqrt{289} = 17 \]

Now, find the corresponding value of y:
\[ y = \frac{289}{x} = \frac{289}{17} = 17 \]

To confirm this is a minimum, we can use the second derivative test:
\[ S''(x) = \frac{d}{dx}(1 - 289x^{-2}) = -289(-2)x^{-3} = \frac{578}{x^3} \]

Since x=17 is positive, \(S''(17)\) is positive, which confirms that the sum S is at a minimum.


Step 4: Final Answer:


The pair of positive integers whose product is 289 and whose sum is a minimum is (17, 17).

The minimum sum is 17 + 17 = 34.
Quick Tip: For a fixed product, the sum of two positive numbers is always minimized when the numbers are as close to each other as possible.
If the product is a perfect square, the minimum sum occurs when the two numbers are equal to its square root.
This is a direct application of the AM-GM inequality: \(\frac{x+y}{2} \geq \sqrt{xy}\). The sum \(x+y\) is minimized when the equality holds, which is when \(x=y\).

Step 1: Identify the Problem and Boundary Lines


The problem is to find the minimum value of the objective function Z = 18x + 10y, subject to a set of linear constraints. The feasible region will be in the first quadrant (\(x \geq 0, y \geq 0\)) and will be unbounded.


First, we plot the boundary lines by converting the inequalities to equations:


- Line 1 (L1): \(4x + y = 20\). This line has intercepts at (5, 0) and (0, 20).


- Line 2 (L2): \(2x + 3y = 30\). This line has intercepts at (15, 0) and (0, 10).


The feasible region is the area on or above both of these lines.


Step 2: Find the Corner Points of the Feasible Region


The corner points (vertices) are the intersections of the boundary lines of the feasible region.


- Point A (y-intercept): The boundary on the y-axis is the highest y-intercept from the two lines, which is from L1. So, A = (0, 20).


- Point B (Intersection of L1 and L2): We solve the system of equations:


1) \(4x + y = 20 \implies y = 20 - 4x\)


2) \(2x + 3y = 30\)


Substitute the expression for y from (1) into (2):


\(2x + 3(20 - 4x) = 30\)


\(2x + 60 - 12x = 30\)


\(-10x = -30 \implies x = 3\).


Now find y by substituting x=3 back into \(y = 20 - 4x\):

\(y = 20 - 4(3) = 20 - 12 = 8\).


So, the intersection point is B = (3, 8).


- Point C (x-intercept): The boundary on the x-axis is the rightmost x-intercept from the two lines, which is from L2. So, C = (15, 0).


The corner points of the feasible region are A(0, 20), B(3, 8), and C(15, 0).



Step 3: Evaluate the Objective Function at the Corner Points


We test the objective function Z = 18x + 10y at each of the identified vertices.


- At A(0, 20): \(Z = 18(0) + 10(20) = 0 + 200 = 200\)


- At B(3, 8): \(Z = 18(3) + 10(8) = 54 + 80 = 134\) (Minimum)


- At C(15, 0): \(Z = 18(15) + 10(0) = 270 + 0 = 270\)


The minimum value of Z at the corner points is 134, occurring at B(3, 8).


Step 4: Check for Unbounded Region


Since the feasible region is unbounded, we must verify that 134 is indeed the minimum value. We do this by checking if the open half-plane \(Z < 134\) (i.e., \(18x + 10y < 134\)) has any points in common with the feasible region.


The boundary line \(18x + 10y = 134\) passes through our potential minimum point B(3, 8). The region \(18x + 10y < 134\) lies on one side of this line. The feasible region lies entirely on the other side of this line, extending away from the origin. Therefore, there are no points in common between the feasible region and the open half-plane \(18x + 10y < 134\).


This confirms that the minimum value is valid.


Step 5: Final Answer:


The minimum value of Z is 134.


\begin{tikzpicture[scale=0.5]
% Axes
\draw[->, thick] (-1,0) -- (17,0) node[right] {x;
\draw[->, thick] (0,-1) -- (0,23) node[above] {y;
\node[below left] at (0,0) {O;

% Ticks
\foreach \x in {2,4,...,16 \draw (\x,3pt) -- (\x,-3pt) node[below] {\tiny\x;
\foreach \y in {2,4,...,22 \draw (3pt,\y) -- (-3pt,\y) node[left] {\tiny\y;

% Define corner points
\coordinate (A) at (0,20);
\coordinate (B) at (3,8);
\coordinate (C) at (15,0);

% Shade the feasible region
\fill[cyan!20, opacity=0.7] (A) -- (B) -- (C) -- (17,0) -- (17,23) -- (0,23) -- cycle;

% Draw boundary lines
\draw[red, thick] (0,20) -- (5,0) node[pos=0.2, above, sloped, black] {\(4x+y=20\);
\draw[blue, thick] (0,10) -- (15,0) node[pos=0.8, below, sloped, black] {\(2x+3y=30\);

% Label the corner points of the feasible boundary
\draw[black, ultra thick] (A) -- (B) -- (C);
\filldraw[black] (A) circle (3pt) node[above right] {A(0,20);
\filldraw[black] (B) circle (3pt) node[above right] {B(3,8);
\filldraw[black] (C) circle (3pt) node[below right] {C(15,0);

% Label the region
\node[align=center, font=\bfseries] at (10, 15) {Feasible Region;
\end{tikzpicture Quick Tip: For an unbounded feasible region in a minimization problem, you must perform a check.
After finding the minimum value M at the corner points, graph the open half-plane \(Z < M\).
If this region has NO points in common with the feasible region, then M is the true minimum.
If it does overlap, then no minimum value exists.

Step 1: Understand the Goal


The problem states that the dot product (scalar product) of vector \(\vec{a}\) with a specific unit vector equals 1. This unit vector is in the direction of the sum of \(\vec{b}\) and \(\vec{c}\). Our goal is to use this information to solve for the unknown \(\lambda\).


Step 2: Find the Sum of Vectors \(\vec{b}\) and \(\vec{c}\)


First, let's define a new vector \(\vec{s}\) as the sum of \(\vec{b}\) and \(\vec{c}\).
\[ \vec{s} = \vec{b} + \vec{c} = (2\hat{i} - 4\hat{j} + 5\hat{k}) + (\lambda\hat{i} - 2\hat{j} - 3\hat{k}) \]

Combine the corresponding components:
\[ \vec{s} = (2+\lambda)\hat{i} + (-4-2)\hat{j} + (5-3)\hat{k} \]
\[ \vec{s} = (2+\lambda)\hat{i} - 6\hat{j} + 2\hat{k} \]


Step 3: Define the Unit Vector along \(\vec{s}\)


The unit vector in the direction of \(\vec{s}\), which we'll call \(\hat{s}\), is given by the formula \(\hat{s} = \frac{\vec{s}}{|\vec{s}|}\).

First, find the magnitude of \(\vec{s}\):
\[ |\vec{s}| = \sqrt{(2+\lambda)^2 + (-6)^2 + 2^2} = \sqrt{(2+\lambda)^2 + 36 + 4} = \sqrt{(2+\lambda)^2 + 40} \]

So, the unit vector is:
\[ \hat{s} = \frac{(2+\lambda)\hat{i} - 6\hat{j} + 2\hat{k}}{\sqrt{(2+\lambda)^2 + 40}} \]


Step 4: Use the Given Dot Product Condition


We are given that the scalar product of \(\vec{a}\) and \(\hat{s}\) is 1.
\[ \vec{a} \cdot \hat{s} = 1 \]
\[ (\hat{i} - \hat{j} + 2\hat{k}) \cdot \left( \frac{(2+\lambda)\hat{i} - 6\hat{j} + 2\hat{k}}{\sqrt{(2+\lambda)^2 + 40}} \right) = 1 \]

Calculate the dot product in the numerator:
\[ \frac{(1)(2+\lambda) + (-1)(-6) + (2)(2)}{\sqrt{(2+\lambda)^2 + 40}} = 1 \]
\[ \frac{2+\lambda + 6 + 4}{\sqrt{(2+\lambda)^2 + 40}} = 1 \]
\[ \frac{\lambda + 12}{\sqrt{(2+\lambda)^2 + 40}} = 1 \]


Step 5: Solve the Equation for \(\lambda\)

\[ \lambda + 12 = \sqrt{(2+\lambda)^2 + 40} \]

Square both sides of the equation to eliminate the square root:
\[ (\lambda + 12)^2 = (2+\lambda)^2 + 40 \]

Expand both binomials:
\[ \lambda^2 + 24\lambda + 144 = (\lambda^2 + 4\lambda + 4) + 40 \]
\[ \lambda^2 + 24\lambda + 144 = \lambda^2 + 4\lambda + 44 \]

Cancel the \(\lambda^2\) terms from both sides:
\[ 24\lambda + 144 = 4\lambda + 44 \]

Solve for \(\lambda\):
\[ 20\lambda = 44 - 144 \]
\[ 20\lambda = -100 \]
\[ \lambda = -5 \]


Step 6: Final Answer:


The value of \(\lambda\) is -5.
Quick Tip: This problem combines several vector concepts. The key steps are:
1. Vector addition to find the sum vector.
2. Finding the magnitude of the sum vector.
3. Constructing the unit vector (\(\vec{v}/|\vec{v}|\)).
4. Calculating the dot product as given in the problem statement.
5. Solving the resulting algebraic equation, which often involves squaring both sides.
Breaking the problem down into these smaller, manageable steps helps avoid errors.

Step 1: Identify Line Components and Check for Parallelism


First, we identify the point vectors (\(\vec{a}\)) and direction vectors (\(\vec{b}\)) for each line from the standard form \(\vec{r} = \vec{a} + t\vec{b}\).

- For Line 1 (\(l_1\)):

\(\vec{a_1} = 2\hat{i} - \hat{j} + 3\hat{k}\)

\(\vec{b_1} = \hat{i} - 2\hat{j} + 3\hat{k}\)

- For Line 2 (\(l_2\)):

\(\vec{a_2} = \hat{i} + 0\hat{j} + 4\hat{k}\)

\(\vec{b_2} = 3\hat{i} - 6\hat{j} + 9\hat{k}\)


Next, we compare the direction vectors \(\vec{b_1}\) and \(\vec{b_2}\) to see if they are parallel. We can do this by checking if their components are proportional.

We notice that \(\vec{b_2} = 3(\hat{i} - 2\hat{j} + 3\hat{k}) = 3\vec{b_1}\).

Since one direction vector is a scalar multiple of the other, the lines are parallel.


Step 2: State the Formula for Distance Between Parallel Lines


The shortest distance between two parallel lines, \(\vec{r} = \vec{a_1} + \lambda\vec{b}\) and \(\vec{r} = \vec{a_2} + \mu\vec{b}\), is given by the formula:
\[ d = \frac{|(\vec{a_2} - \vec{a_1}) \times \vec{b}|}{|\vec{b}|} \]

We will use the simpler direction vector, \(\vec{b} = \vec{b_1}\).


Step 3: Perform the Vector Calculations


First, calculate the vector connecting the two points \(\vec{a_1}\) and \(\vec{a_2}\):
\[ \vec{a_2} - \vec{a_1} = (\hat{i} + 4\hat{k}) - (2\hat{i} - \hat{j} + 3\hat{k}) \]
\[ = (1-2)\hat{i} + (0-(-1))\hat{j} + (4-3)\hat{k} = -\hat{i} + \hat{j} + \hat{k} \]

Next, calculate the cross product of this vector with the direction vector \(\vec{b_1}\):
\[ (\vec{a_2} - \vec{a_1}) \times \vec{b_1} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
-1 & 1 & 1
1 & -2 & 3 \end{vmatrix} \]
\[ = \hat{i}((1)(3) - (1)(-2)) - \hat{j}((-1)(3) - (1)(1)) + \hat{k}((-1)(-2) - (1)(1)) \]
\[ = \hat{i}(3 + 2) - \hat{j}(-3 - 1) + \hat{k}(2 - 1) = 5\hat{i} + 4\hat{j} + \hat{k} \]

Now, find the magnitude of this resulting vector:
\[ |(\vec{a_2} - \vec{a_1}) \times \vec{b_1}| = \sqrt{5^2 + 4^2 + 1^2} = \sqrt{25 + 16 + 1} = \sqrt{42} \]

Finally, find the magnitude of the direction vector \(\vec{b_1}\):
\[ |\vec{b_1}| = \sqrt{1^2 + (-2)^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14} \]


Step 4: Calculate the Shortest Distance


Substitute the magnitudes into the distance formula:
\[ d = \frac{\sqrt{42}}{\sqrt{14}} = \sqrt{\frac{42}{14}} = \sqrt{3} \]


Step 5: Final Answer:


The shortest distance between the two parallel lines is \(\sqrt{3}\) units.
Quick Tip: When finding the shortest distance between two lines, the very first step is to check if they are parallel.
Compare their direction vectors \(\vec{b_1}\) and \(\vec{b_2}\). If \(\vec{b_2} = k \vec{b_1}\) for some scalar k, they are parallel.
If they are parallel, use the formula for parallel lines. If not, use the different formula for skew lines: \( d = \frac{|(\vec{a_2}-\vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{| \vec{b_1} \times \vec{b_2} |} \). Using the wrong formula will always give an incorrect answer.

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 \(\int \frac{2x}{1+x^2} dx\) is \(\ln(1+x^2)\) (by substitution \(u=1+x^2\)).

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 structure of this problem is a classic first-order linear differential equation.
Always look for the pattern \( \int \frac{f'(x)}{f(x)} dx = \ln|f(x)| \) when calculating the integrating factor.
The solution is found by multiplying by the I.F. and integrating.

Step 1: Understand the Geometry and Equations of the Boundaries


The problem describes a region in the first quadrant bounded by three curves:
1. The circular table top: A circle centered at the origin with radius r = 8 cm. Its equation is \(x^2 + y^2 = 8^2 = 64\). In the first quadrant, the upper boundary is \(y = \sqrt{64 - x^2}\).
2. The scratch: A straight line passing through the origin at an angle of \(\pi/4\) with the positive x-axis. The equation of this line is \(y = x \tan(\pi/4)\), which simplifies to \(y = x\).
3. The x-axis: The line \(y = 0\).

The region is a sector of the circle with an angle of \(\pi/4\). While the area can be found geometrically (\(A = \frac{1}{2}r^2\theta = \frac{1}{2}(8^2)(\frac{\pi}{4}) = 8\pi\)), the question requires the use of integration.


Step 2: Set up the Integrals for the Area


To find the area using integration with respect to x, we need to find the point where the line \(y=x\) intersects the circle \(y = \sqrt{64-x^2}\). \[ x = \sqrt{64-x^2} \implies x^2 = 64 - x^2 \implies 2x^2 = 64 \implies x^2 = 32 \implies x = \sqrt{32} = 4\sqrt{2} \]
The area is split into two parts by the vertical line at \(x=4\sqrt{2}\):
- Area 1 (A1): From \(x=0\) to \(x=4\sqrt{2}\), the region is a triangle bounded above by the line \(y=x\).
- Area 2 (A2): From \(x=4\sqrt{2}\) to \(x=8\), the region is bounded above by the circular arc \(y = \sqrt{64-x^2}\).
\[ Total Area = A_1 + A_2 = \int_{0}^{4\sqrt{2}} x \, dx + \int_{4\sqrt{2}}^{8} \sqrt{64 - x^2} \, dx \]

Step 3: Evaluate the Integrals


Evaluating A1: \[ A_1 = \int_{0}^{4\sqrt{2}} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{4\sqrt{2}} = \frac{(4\sqrt{2})^2}{2} - 0 = \frac{32}{2} = 16 \]

Evaluating A2:
We use the standard integral formula \(\int \sqrt{a^2 - x^2} dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)\) with \(a=8\). \[ A_2 = \left[ \frac{x}{2}\sqrt{64 - x^2} + \frac{64}{2}\sin^{-1}\left(\frac{x}{8}\right) \right]_{4\sqrt{2}}^{8} \]
- At the upper limit (x = 8):
\( \frac{8}{2}\sqrt{64 - 8^2} + 32\sin^{-1}\left(\frac{8}{8}\right) = 0 + 32\sin^{-1}(1) = 32\left(\frac{\pi}{2}\right) = 16\pi \)
- At the lower limit (x = \(4\sqrt{2}\)):
\( \frac{4\sqrt{2}}{2}\sqrt{64 - (4\sqrt{2})^2} + 32\sin^{-1}\left(\frac{4\sqrt{2}}{8}\right) \)
\( = 2\sqrt{2}\sqrt{64 - 32} + 32\sin^{-1}\left(\frac{\sqrt{2}}{2}\right) = 2\sqrt{2}\sqrt{32} + 32\left(\frac{\pi}{4}\right) = 2\sqrt{64} + 8\pi = 2(8) + 8\pi = 16 + 8\pi \)
So, \(A_2 = (16\pi) - (16 + 8\pi) = 8\pi - 16\).

Step 4: Calculate the Total Area


Total Area = \(A_1 + A_2 = 16 + (8\pi - 16) = 8\pi\).


Step 5: Final Answer:


The area of the region is \(8\pi\) cm\(^2\).


\begin{figure[h!]
\centering
\begin{tikzpicture[scale=0.6]
\draw[->, thick] (-1,0) -- (9,0) node[right] {x-axis;
\draw[->, thick] (0,-1) -- (0,9) node[above] {y-axis;
\node[below left] at (0,0) {O;

% Draw the circular arc
\draw[blue, ultra thick] (8,0) arc (0:90:8);
\node[blue] at (6,6) {\(x^2+y^2=64\);

% Draw the scratch line
\draw[red, thick] (0,0) -- (6,6) node[pos=0.8, above, sloped] {Scratch (\(y=x\));

% Shade the region
\fill[orange!30, opacity=0.8] (0,0) -- (8,0) arc (0:45:8) -- cycle;

% Mark radius and angle
\draw (1,0) arc (0:45:1);
\node at (1.3, 0.4) {\(\pi/4\);
\node at (4,1) {Area;
\node at (8.3, -0.3) {8;

\end{tikzpicture
\end{figure Quick Tip: Visualizing the area is the most important first step. This problem describes a simple circular sector.
While integration is required, you can use the geometric formula \(A = \frac{1}{2}r^2\theta\) to quickly check if your final answer is correct.
Splitting the integral based on the intersection points of the boundary curves is a standard technique for complex regions.

Step 1: Identify the Type of Equation


We can rewrite the equation as:
\[ \frac{dy}{dx} + 2y = \cos x \]

This is a first-order linear differential equation of the form \(\frac{dy}{dx} + P(x)y = Q(x)\), where \(P(x) = 2\) and \(Q(x) = \cos x\).


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


The integrating factor is given by \(I.F. = e^{\int P(x) dx}\).
\[ I.F. = e^{\int 2 dx} = e^{2x} \]


Step 3: Find the General Solution


The solution is given by the formula \(y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx\).
\[ y \cdot e^{2x} = \int \cos x \cdot e^{2x} dx \]

To solve the integral on the right, we need to use integration by parts twice. Let \(I = \int e^{2x}\cos x dx\).

Let \(u = \cos x\) and \(dv = e^{2x}dx\). Then \(du = -\sin x dx\) and \(v = \frac{1}{2}e^{2x}\).
\[ I = (\cos x)\left(\frac{1}{2}e^{2x}\right) - \int \left(\frac{1}{2}e^{2x}\right)(-\sin x dx) = \frac{1}{2}e^{2x}\cos x + \frac{1}{2}\int e^{2x}\sin x dx \]

Apply integration by parts again to the new integral. Let \(u = \sin x\) and \(dv = e^{2x}dx\). Then \(du = \cos x dx\) and \(v = \frac{1}{2}e^{2x}\).
\[ I = \frac{1}{2}e^{2x}\cos x + \frac{1}{2}\left[ (\sin x)\left(\frac{1}{2}e^{2x}\right) - \int \left(\frac{1}{2}e^{2x}\right)(\cos x dx) \right] \]
\[ I = \frac{1}{2}e^{2x}\cos x + \frac{1}{4}e^{2x}\sin x - \frac{1}{4}\int e^{2x}\cos x dx \]

The integral on the right is our original integral I.
\[ I = \frac{1}{2}e^{2x}\cos x + \frac{1}{4}e^{2x}\sin x - \frac{1}{4}I \]
\[ I + \frac{1}{4}I = \frac{1}{4}e^{2x}(2\cos x + \sin x) \]
\[ \frac{5}{4}I = \frac{1}{4}e^{2x}(2\cos x + \sin x) \implies I = \frac{1}{5}e^{2x}(2\cos x + \sin x) \]

Substituting this back into our solution:
\[ y \cdot e^{2x} = \frac{1}{5}e^{2x}(2\cos x + \sin x) + C \]


Step 4: Isolate y for the Final Answer


Divide the entire equation by \(e^{2x}\):
\[ y = \frac{1}{5}(2\cos x + \sin x) + \frac{C}{e^{2x}} \]
\[ y = \frac{2}{5}\cos x + \frac{1}{5}\sin x + Ce^{-2x} \]

This is the general solution.
Quick Tip: A standard shortcut for integrals of the form \(\int e^{ax} \cos(bx) dx\) is:
\[ \frac{e^{ax}}{a^2+b^2}(a\cos(bx) + b\sin(bx)) + C \]
In this problem, a=2 and b=1, which gives \(\frac{e^{2x}}{2^2+1^2}(2\cos x + \sin x) = \frac{e^{2x}}{5}(2\cos x + \sin x)\). Memorizing this formula saves a lot of time.

Step 1: Standardize the Line Equation


First, we must convert the given line equation to the standard form \(\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}\).

- \(\frac{2x+4}{6} = \frac{2(x+2)}{6} = \frac{x+2}{3}\)

- \(\frac{y+1}{2}\) is already in standard form.

- \(\frac{-2z+6}{-4} = \frac{-2(z-3)}{-4} = \frac{z-3}{2}\)

The standard equation of the line is: \(\frac{x+2}{3} = \frac{y+1}{2} = \frac{z-3}{2}\).


Step 2: Parameterize a General Point Q on the Line


Let the common ratio be equal to a parameter \(\lambda\).
\[ \frac{x+2}{3} = \frac{y+1}{2} = \frac{z-3}{2} = \lambda \]

Any point Q on this line can be written in terms of \(\lambda\):
\(x = -2 + 3\lambda\)
\(y = -1 + 2\lambda\)
\(z = 3 + 2\lambda\)

So, Q = \((-2 + 3\lambda, -1 + 2\lambda, 3 + 2\lambda)\).


Step 3: Use the Distance Formula


We are given that the distance between P(1, 2, 3) and Q is \(3\sqrt{2}\). The square of the distance is \((3\sqrt{2})^2 = 18\).

Using the distance formula, \(PQ^2 = (x_Q-x_P)^2 + (y_Q-y_P)^2 + (z_Q-z_P)^2\):
\[ ((-2 + 3\lambda) - 1)^2 + ((-1 + 2\lambda) - 2)^2 + ((3 + 2\lambda) - 3)^2 = 18 \]
\[ (3\lambda - 3)^2 + (2\lambda - 3)^2 + (2\lambda)^2 = 18 \]


Step 4: Solve the Quadratic Equation for \(\lambda\)


Expand the terms:
\[ (9\lambda^2 - 18\lambda + 9) + (4\lambda^2 - 12\lambda + 9) + 4\lambda^2 = 18 \]

Combine like terms:
\[ (9+4+4)\lambda^2 + (-18-12)\lambda + (9+9) = 18 \]
\[ 17\lambda^2 - 30\lambda + 18 = 18 \]
\[ 17\lambda^2 - 30\lambda = 0 \]

Factor out \(\lambda\):
\[ \lambda(17\lambda - 30) = 0 \]

This gives two possible solutions for \(\lambda\): \(\lambda_1 = 0\) and \(\lambda_2 = \frac{30}{17}\).


Step 5: Find the Coordinates of Q for Each Value of \(\lambda\)


- For \(\lambda_1 = 0\):


Q\(_1\) = \((-2 + 3(0), -1 + 2(0), 3 + 2(0))\) = (-2, -1, 3).


- For \(\lambda_2 = \frac{30}{17}\):


\(x = -2 + 3(\frac{30}{17}) = \frac{-34+90}{17} = \frac{56}{17}\)


\(y = -1 + 2(\frac{30}{17}) = \frac{-17+60}{17} = \frac{43}{17}\)


\(z = 3 + 2(\frac{30}{17}) = \frac{51+60}{17} = \frac{111}{17}\)


Q\(_2\) = (\(\frac{56}{17}, \frac{43}{17}, \frac{111}{17}\)).


Step 6: Final Answer:


There are two points on the line at the given distance. They are Q\(_1\)(-2, -1, 3) and Q\(_2\)(\(\frac{56}{17}, \frac{43}{17}, \frac{111}{17}\)).
Quick Tip: The most common mistake in this type of problem is not converting the line equation to its standard form, which leads to incorrect direction ratios and parametric coordinates.
Always ensure the coefficients of x, y, and z in the numerators are +1.
Remember that there are usually two points on a line at a given distance from an external point.

Step 1: Standardize the Line Equation


We convert the given line equation to the standard form.

- \(\frac{2x-4}{2} = \frac{2(x-2)}{2} = \frac{x-2}{1}\)

- \(\frac{y}{2}\) is already in standard form (can be written as \(\frac{y-0}{2}\)).

- \(\frac{2-z}{3} = \frac{-(z-2)}{3} = \frac{z-2}{-3}\)

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

The direction vector of the line is \(\vec{b} = \langle 1, 2, -3 \rangle\).


Step 2: Find the Foot of the Perpendicular (M)


Let P = (–1, 5, 2). Let M be the foot of the perpendicular from P to the line. M is a general point on the line, which can be parameterized by \(\lambda\).

M = \((2+\lambda, 2\lambda, 2-3\lambda)\).

The vector \(\vec{PM}\) is perpendicular to the line's direction vector \(\vec{b}\), so their dot product is zero.
\[ \vec{PM} = \langle (2+\lambda - (-1)), (2\lambda - 5), (2-3\lambda - 2) \rangle = \langle 3+\lambda, 2\lambda - 5, -3\lambda \rangle \]
\[ \vec{PM} \cdot \vec{b} = 1(3+\lambda) + 2(2\lambda - 5) + (-3)(-3\lambda) = 0 \]
\[ 3 + \lambda + 4\lambda - 10 + 9\lambda = 0 \]
\[ 14\lambda - 7 = 0 \implies \lambda = \frac{7}{14} = \frac{1}{2} \]

Substitute \(\lambda = 1/2\) to find the coordinates of M:

M = \((2 + 1/2, 2(1/2), 2 - 3(1/2))\) = \((\frac{5}{2}, 1, \frac{1}{2})\).


Step 3: Find the Image Point (Q)


M is the midpoint of the segment connecting point P and its image Q. Let Q = \((x_Q, y_Q, z_Q)\).


Using the midpoint formula:


- \(M_x = \frac{P_x + Q_x}{2} \implies \frac{5}{2} = \frac{-1+x_Q}{2} \implies 5 = -1+x_Q \implies x_Q = 6\).


- \(M_y = \frac{P_y + Q_y}{2} \implies 1 = \frac{5+y_Q}{2} \implies 2 = 5+y_Q \implies y_Q = -3\).


- \(M_z = \frac{P_z + Q_z}{2} \implies \frac{1}{2} = \frac{2+z_Q}{2} \implies 1 = 2+z_Q \implies z_Q = -1\).


The image point is Q(6, -3, -1).


Step 4: Find the Length of the Line Segment PQ


We can find the length using the distance formula between P(–1, 5, 2) and Q(6, -3, -1).
\[ |\vec{PQ}| = \sqrt{(6 - (-1))^2 + (-3 - 5)^2 + (-1 - 2)^2} \]
\[ |\vec{PQ}| = \sqrt{(7)^2 + (-8)^2 + (-3)^2} \]
\[ |\vec{PQ}| = \sqrt{49 + 64 + 9} = \sqrt{122} \]

Alternatively, this length is twice the length of the perpendicular, \(2|\vec{PM}|\).


With \(\lambda=1/2\), \(\vec{PM} = \langle 7/2, -4, -3/2 \rangle\), and its magnitude is \(\sqrt{49/4 + 16 + 9/4} = \sqrt{58/4 + 64/4} = \sqrt{122/4} = \frac{\sqrt{122}}{2}\). So, \(2|\vec{PM}| = \sqrt{122}\).


Step 5: Final Answer:


The image of the point is Q(6, -3, -1).

The length of the line segment joining the given point and its image is \(\sqrt{122}\) units.
Quick Tip: The process for finding the image of a point in a line is an extension of finding the foot of the perpendicular.
1. Find the foot of the perpendicular, M, as in the previous problems.
2. Use the midpoint formula: M is the midpoint of the original point P and its image Q.
3. The length of the segment PQ is simply twice the length of the perpendicular PM.

Step 1: Understanding the Movement Plan


The problem states that friends A and B travel from the origin O to their destinations, represented by points A and B.

Then, they both travel from their respective destinations to meet at a new point, C.

This means we need to draw vectors representing the paths from A to C and from B to C.


Step 2: Drawing the Vectors


We add two directed line segments (arrows) to the existing figure:

1. A vector starting at point A and ending at point C. This represents the vector \(\vec{AC}\).

2. A vector starting at point B and ending at point C. This represents the vector \(\vec{BC}\).




\begin{tikzpicture[>=stealth, scale=1.5]
% Define coordinates (arbitrary but illustrative)
\coordinate (O) at (0,0);
\coordinate (A) at (2,0.5);
\coordinate (B) at (2.5,1.5);
% Calculate C = 5A - 2B (using vector arithmetic on coordinates)
\coordinate (C) at (5*2 - 2*2.5, 5*0.5 - 2*1.5); % C = (5, -0.5)

% Draw initial paths from origin
\draw[->, thick] (O) -- (A) node[below] {A;
\draw[->, thick] (O) -- (B) node[above] {B;
\draw[->, thick] (O) -- (C) node[below] {C;

% Label initial vectors
\node at (1,0.4) {\(\vec{a}\);
\node at (1.2,1.1) {\(\vec{b}\);
\node at (2,-0.5) {\(5\vec{a}-2\vec{b}\);
\node at (-0.2,-0.2) {O;

% Draw the paths to the meeting point C (This completes the figure)
\draw[->, thick, red, dashed] (A) -- (C) node[midway, below] {\(\vec{AC}\);
\draw[->, thick, blue, dashed] (B) -- (C) node[midway, above right] {\(\vec{BC}\);
\end{tikzpicture Quick Tip: A vector \(\vec{PQ}\) always represents the path from the initial point P to the terminal point Q.
Completing a vector diagram involves drawing the arrows that correspond to the described movements.

Step 1: Key Formula


The vector from point P to point Q is given by the position vector of Q minus the position vector of P: \(\vec{PQ} = \vec{OQ} - \vec{OP}\).


Step 2: Calculate \(\vec{AC}\)

\[ \vec{AC} = \vec{OC} - \vec{OA} \]

Substitute the given vectors:
\[ \vec{AC} = (5\vec{a} - 2\vec{b}) - (\vec{a}) \]
\[ \vec{AC} = 4\vec{a} - 2\vec{b} \]


Step 3: Calculate \(\vec{BC}\)

\[ \vec{BC} = \vec{OC} - \vec{OB} \]

Substitute the given vectors:
\[ \vec{BC} = (5\vec{a} - 2\vec{b}) - (\vec{b}) \]
\[ \vec{BC} = 5\vec{a} - 3\vec{b} \]
Quick Tip: This is a direct application of the triangle law of vector addition.
For any three points O, P, Q, we have \(\vec{OP} + \vec{PQ} = \vec{OQ}\), which rearranges to \(\vec{PQ} = \vec{OQ} - \vec{OP}\).
Always remember: (Position vector of Terminal Point) - (Position vector of Initial Point).

Step 1: Find the Angle


We are given:

- \(\vec{a} \cdot \vec{b} = 1\)

- Distance O to A = \(|\vec{OA}| = |\vec{a}| = 1\) km.

- Distance O to B = \(|\vec{OB}| = |\vec{b}| = 2\) km.

The formula for the dot product is \(\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta\), where \(\theta\) is the angle between the vectors.
\[ 1 = (1)(2) \cos\theta \]
\[ \cos\theta = \frac{1}{2} \]

The angle is \(\theta = \cos^{-1}(1/2) = \frac{\pi}{3}\) radians or 60°.


Step 2: Find \(|\vec{a} \times \vec{b}|\)


The formula for the magnitude of the cross product is \(|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin\theta\).

We found \(\theta = \pi/3\), so \(\sin\theta = \sin(\pi/3) = \frac{\sqrt{3}}{2}\).
\[ |\vec{a} \times \vec{b}| = (1)(2) \left(\frac{\sqrt{3}}{2}\right) = \sqrt{3} \]

Alternatively, using Lagrange's Identity: \(|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2\).
\[ |\vec{a} \times \vec{b}|^2 + (1)^2 = (1)^2 (2)^2 \]
\[ |\vec{a} \times \vec{b}|^2 + 1 = 4 \implies |\vec{a} \times \vec{b}|^2 = 3 \implies |\vec{a} \times \vec{b}| = \sqrt{3} \]
Quick Tip: The dot product is used to find the angle between vectors, while the cross product is related to the area of the parallelogram they form.
The two are linked by Lagrange's identity, which is a great tool for finding the magnitude of one if you know the other, without needing to find the angle itself.

Step 1: Understand the Goal


We need to find a unit vector that is simultaneously perpendicular to two other vectors: the sum \((\vec{a}+\vec{b})\) and the difference \((\vec{a}-\vec{b})\).

The standard way to find a vector perpendicular to two given vectors is to calculate their cross product. The resulting vector will be orthogonal to both. After finding this vector, we will normalize it (divide by its magnitude) to get the required unit vector.


Step 2: Calculate the Sum and Difference Vectors


First, let's find the two vectors we'll be working with.

- Let \(\vec{p} = \vec{a} + \vec{b}\):

\[ \vec{p} = (2\hat{i} - \hat{j} + 4\hat{k}) + (0\hat{i} + \hat{j} - \hat{k}) = (2+0)\hat{i} + (-1+1)\hat{j} + (4-1)\hat{k} = 2\hat{i} + 0\hat{j} + 3\hat{k} \]

- Let \(\vec{q} = \vec{a} - \vec{b}\):

\[ \vec{q} = (2\hat{i} - \hat{j} + 4\hat{k}) - (0\hat{i} + \hat{j} - \hat{k}) = (2-0)\hat{i} + (-1-1)\hat{j} + (4-(-1))\hat{k} = 2\hat{i} - 2\hat{j} + 5\hat{k} \]


Step 3: Find a Vector Perpendicular to Both \(\vec{p}\) and \(\vec{q}\)


We calculate the cross product, \(\vec{v} = \vec{p} \times \vec{q}\).
\[ \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
2 & 0 & 3
2 & -2 & 5 \end{vmatrix} \]
\[ = \hat{i}((0)(5) - (3)(-2)) - \hat{j}((2)(5) - (3)(2)) + \hat{k}((2)(-2) - (0)(2)) \]
\[ = \hat{i}(0 - (-6)) - \hat{j}(10 - 6) + \hat{k}(-4 - 0) \]
\[ = 6\hat{i} - 4\hat{j} - 4\hat{k} \]

This vector \(\vec{v}\) is perpendicular to both \(\vec{p}\) and \(\vec{q}\). We can use a simpler vector parallel to \(\vec{v}\) by dividing by 2:

Let's call it \(\vec{w} = 3\hat{i} - 2\hat{j} - 2\hat{k}\).


Step 4: Find the Unit Vector


To find the unit vector, we divide the vector \(\vec{w}\) by its magnitude, \(|\vec{w}|\).
\[ |\vec{w}| = \sqrt{3^2 + (-2)^2 + (-2)^2} = \sqrt{9 + 4 + 4} = \sqrt{17} \]

The unit vector, \(\hat{w}\), is given by \(\pm \frac{\vec{w}}{|\vec{w}|}\), as there are two opposite directions that are perpendicular.
\[ \hat{w} = \pm \frac{3\hat{i} - 2\hat{j} - 2\hat{k}}{\sqrt{17}} \]


Step 5: Final Answer:


The required unit vector is \(\pm \frac{1}{\sqrt{17}}(3\hat{i} - 2\hat{j} - 2\hat{k})\).
Quick Tip: The cross product \(\vec{a} \times \vec{b}\) gives a vector that is perpendicular to the plane containing \(\vec{a}\) and \(\vec{b}\).
To find a unit vector perpendicular to both, the process is always:
1. Find the cross product \(\vec{v} = \vec{a} \times \vec{b}\).
2. Find the magnitude of the resulting vector, \(|\vec{v}|\).
3. The unit vector is \(\hat{v} = \pm \frac{\vec{v}}{|\vec{v}|}\). Don't forget the \(\pm\) sign, as there are two possible directions.

Step 1: Analyze the Differential Equation


The given differential equation is \(\frac{dV}{dt} = kS\).

This relates the first derivative of the volume V with respect to time t.


Step 2: Determine the Order


The order of a differential equation is the order of the highest derivative present.

The only derivative is \(\frac{dV}{dt}\), which is a first-order derivative.

Therefore, the order is 1.


Step 3: Determine the Degree


The degree is the highest power of the highest-order derivative, provided the equation is a polynomial in its derivatives.

The term \(\frac{dV}{dt}\) is raised to the power of 1. The equation is a simple polynomial in terms of the derivative.

Therefore, the degree is 1.
Quick Tip: Order is simply the "highest number of primes" on the dependent variable (e.g., y'', y''', etc.).
Degree is the exponent on that highest derivative term, after the equation has been cleared of radicals or fractions in its derivative terms.

Step 1: Set up the Differential Equation for Separation


The given differential equation is \(\frac{dr}{dt} = \frac{2}{3}k\).

Here, k is a constant. This is a simple separable differential equation.

We can write it as \(dr = \frac{2}{3}k \, dt\).


Step 2: Integrate Both Sides

\[ \int dr = \int \frac{2}{3}k \, dt \]
\[ r = \frac{2}{3}kt + C \]

This is the general solution, where C is the constant of integration.


Step 3: Apply the Initial Condition


We are given the initial condition r(0) = 5 mm. This means when t = 0, r = 5.

Substitute these values into the general solution to find C:
\[ 5 = \frac{2}{3}k(0) + C \]
\[ 5 = 0 + C \implies C = 5 \]


Step 4: Write the Particular Solution


Substitute the value of C back into the general solution:
\[ r(t) = \frac{2}{3}kt + 5 \]
Quick Tip: This is the simplest form of a differential equation, \(\frac{dy}{dx} = C\), where C is a constant.
The solution is always a linear function, \(y = Cx + D\).
The initial condition is used to find the "y-intercept" D.

Step 1: Find the value of k


From the previous part, the particular solution is \(r(t) = \frac{2}{3}kt + 5\).

We are given a new condition: when t = 1, r = 3. We use this to find k.
\[ 3 = \frac{2}{3}k(1) + 5 \]
\[ 3 - 5 = \frac{2}{3}k \]
\[ -2 = \frac{2}{3}k \implies k = -3 \]

(The constant k is negative, which makes sense as the volume is reducing).


Step 2: Find the time t for r = 0


Now we have the complete equation for the radius at time t:
\[ r(t) = \frac{2}{3}(-3)t + 5 = -2t + 5 \]

We want to find the time t when the radius is 0 (the tablet has fully evaporated).

Set r(t) = 0:
\[ 0 = -2t + 5 \]
\[ 2t = 5 \implies t = \frac{5}{2} = 2.5 \]


Step 3: Final Answer:


The value of k is -3.

The time taken for the tablet to evaporate completely (r=0) is 2.5 hours.
Quick Tip: This is a follow-up to an initial value problem.
Once you have the particular solution from the first condition (\(r(0)=5\)), you use the second condition (\(r(1)=3\)) to solve for the remaining unknown parameter (k).
After finding all constants, you can use the final equation to answer any question about the system's state at a given time.

Step 1: Find the value of k


We use the same particular solution from part (ii): \(r(t) = \frac{2}{3}kt + 5\).

We are given a different condition for this part: when t = 1, r = 1.
\[ 1 = \frac{2}{3}k(1) + 5 \]
\[ 1 - 5 = \frac{2}{3}k \]
\[ -4 = \frac{2}{3}k \implies k = -4 \cdot \frac{3}{2} = -6 \]


Step 2: Find the time t for r = 0


Now, the specific equation for the radius is:
\[ r(t) = \frac{2}{3}(-6)t + 5 = -4t + 5 \]

We set r(t) = 0 to find the time it takes to evaporate.
\[ 0 = -4t + 5 \]
\[ 4t = 5 \implies t = \frac{5}{4} = 1.25 \]


Step 3: Final Answer:


The value of k is -6.

The time taken for the tablet to evaporate completely (r=0) is 1.25 hours.
Quick Tip: The "OR" in case study questions often presents an alternative scenario with different numerical values.
The underlying model and equations are the same, but you must be careful to use the new set of conditions provided in that specific sub-part.
Do not carry over the value of 'k' from part (a) to part (b).

Step 1: Define Events and List Probabilities


Let A\(_1\), A\(_2\), A\(_3\) be the events that a randomly chosen person is from the respective health categories.


- \(P(A_1) = \frac{700}{1000} = 0.7\)


- \(P(A_2) = \frac{200}{1000} = 0.2\)


- \(P(A_3) = \frac{100}{1000} = 0.1\)


Let D be the event that a person has contracted the disease. We are given the conditional probabilities:


- \(P(D|A_1) = 25% = 0.25\)


- \(P(D|A_2) = 35% = 0.35\)


- \(P(D|A_3) = 50% = 0.50\)


Step 2: Apply the Law of Total Probability


The overall probability of a person contracting the disease is the sum of the probabilities of contracting it from each health category.
\[ P(D) = P(A_1)P(D|A_1) + P(A_2)P(D|A_2) + P(A_3)P(D|A_3) \]
\[ P(D) = (0.7)(0.25) + (0.2)(0.35) + (0.1)(0.50) \]
\[ P(D) = 0.175 + 0.070 + 0.050 \]
\[ P(D) = 0.295 \]


Step 3: Final Answer:


The probability that a randomly tested person has contracted the disease is 0.295 or 29.5%.
Quick Tip: This is a standard application of the Law of Total Probability.
1. Identify the different "states" or "categories" the population is divided into (A1, A2, A3).
2. Find the probability of being in each category.
3. Find the conditional probability of the event in question (D) for each category.
4. The total probability is the sum of (Prob of Category \(\times\) Conditional Prob for that Category) over all categories.

Step 1: Understand the Question and Define Events


We need to find a conditional probability: P(A\(_2\) | D'), where D' is the event that a person has not contracted the disease.

We will use Bayes' Theorem: \( P(A_2|D') = \frac{P(A_2 \cap D'){P(D')} = \frac{P(A_2)P(D'|A_2)}{P(D')} \).


Step 2: Calculate Necessary Probabilities


First, we need the probability of not contracting the disease, P(D'). From part (i), we found P(D) = 0.295.
\[ P(D') = 1 - P(D) = 1 - 0.295 = 0.705 \]

Next, we need the conditional probabilities of NOT contracting the disease for each category.

- \(P(D'|A_1) = 1 - P(D|A_1) = 1 - 0.25 = 0.75\)

- \(P(D'|A_2) = 1 - P(D|A_2) = 1 - 0.35 = 0.65\)

- \(P(D'|A_3) = 1 - P(D|A_3) = 1 - 0.50 = 0.50\)


Step 3: Apply Bayes' Theorem


Now we can calculate P(A\(_2\) | D').

The numerator is \(P(A_2) \cdot P(D'|A_2)\):
\[ (0.2) \cdot (0.65) = 0.130 \]

The denominator is P(D'), which we calculated as 0.705.

(We can also verify the denominator using the law of total probability for D': \( (0.7)(0.75) + (0.2)(0.65) + (0.1)(0.50) = 0.525 + 0.130 + 0.050 = 0.705 \)).

Now, calculate the final probability:
\[ P(A_2|D') = \frac{0.130}{0.705} \]

To simplify the fraction, multiply the numerator and denominator by 1000:
\[ P(A_2|D') = \frac{130}{705} \]

Divide both by 5:
\[ P(A_2|D') = \frac{26}{141} \]


Step 4: Final Answer:


Given that the person has not contracted the disease, the probability that they are from category A\(_2\) is \(\frac{26}{141}\).
Quick Tip: This is a classic Bayes' Theorem problem involving a negative event ("not contracted").
The structure is the same as for a positive event, but you must remember to use the probabilities of the complement events.
\(P(D') = 1 - P(D)\) and \(P(D'|A_i) = 1 - P(D|A_i)\).
The denominator \(P(D')\) is the total probability of not getting the disease, summed over all categories.