Step 1: Understanding the Concept:

The principal value of an inverse sine function, \( \sin^{-1}(x) \), is the value \( \theta \) that lies in the range \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \). We need to find an angle in this range that has the same sine value as \( \frac{10\pi}{3} \).


Step 2: Detailed Explanation:

First, we simplify the angle inside the sine function.
\[ \sin\left(\frac{10\pi}{3}\right) \]
We can rewrite \( \frac{10\pi}{3} \) as a multiple of \( \pi \) to simplify it.
\[ \frac{10\pi}{3} = \frac{9\pi + \pi}{3} = 3\pi + \frac{\pi}{3} \]
Now, we find the value of \( \sin\left(3\pi + \frac{\pi}{3}\right) \).

Using the property \( \sin(\pi + \theta) = -\sin(\theta) \), we can write:
\[ \sin\left(3\pi + \frac{\pi}{3}\right) = \sin\left(2\pi + \pi + \frac{\pi}{3}\right) = \sin\left(\pi + \frac{\pi}{3}\right) = -\sin\left(\frac{\pi}{3}\right) \]
We know that \( \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \).

Therefore,
\[ \sin\left(\frac{10\pi}{3}\right) = -\frac{\sqrt{3}}{2} \]
The original expression becomes:
\[ \sin^{-1}\left(\sin\frac{10\pi}{3}\right) = \sin^{-1}\left(-\frac{\sqrt{3}}{2}\right) \]
Now we need to find the principal value of \( \sin^{-1}\left(-\frac{\sqrt{3}}{2}\right) \). Let this value be \( y \).

So, \( \sin(y) = -\frac{\sqrt{3}}{2} \), where \( y \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \).

Since \( \sin(y) \) is negative, \( y \) must be in the fourth quadrant of the principal value range, i.e., \( y \in \left[-\frac{\pi}{2}, 0\right] \).

We know \( \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \). Using the identity \( \sin(-\theta) = -\sin(\theta) \), we get:
\[ \sin\left(-\frac{\pi}{3}\right) = -\sin\left(\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2} \]
The angle \( -\frac{\pi}{3} \) lies in the principal value range \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \).


Step 3: Final Answer:

Thus, the principal value is \( -\frac{\pi}{3} \).
Quick Tip: For any inverse trigonometric function, first check its principal value range. For \( \sin^{-1}(\sin x) \), the answer is not always \( x \). You must find an angle \( \theta \) within the range \( [-\pi/2, \pi/2] \) such that \( \sin \theta = \sin x \).

Step 1: Understanding the Concept:

We are given two relations for square matrices A and B: \( AB = A \) and \( BA = B \). We need to use these relations to simplify the expression \( A^2 + B^2 \). Matrices that satisfy \( M^2 = M \) are called idempotent matrices.


Step 2: Detailed Explanation:

Let's compute \( A^2 \) first.
\[ A^2 = A \cdot A \]
We are given \( AB = A \). Substitute this into the expression for \( A^2 \):
\[ A^2 = (AB)A \]
Using the associative property of matrix multiplication, \( (XY)Z = X(YZ) \):
\[ A^2 = A(BA) \]
Now, we are given \( BA = B \). Substitute this into the equation:
\[ A^2 = A(B) = AB \]
Finally, since \( AB = A \), we have:
\[ A^2 = A \]
Now let's compute \( B^2 \).
\[ B^2 = B \cdot B \]
We are given \( BA = B \). Substitute this into the expression for \( B^2 \):
\[ B^2 = (BA)B \]
Using the associative property:
\[ B^2 = B(AB) \]
Now, we are given \( AB = A \). Substitute this into the equation:
\[ B^2 = B(A) = BA \]
Finally, since \( BA = B \), we have:
\[ B^2 = B \]
So, we have found that \( A^2 = A \) and \( B^2 = B \).


Step 3: Final Answer:

Now we can find \( A^2 + B^2 \):
\[ A^2 + B^2 = A + B \]
The correct option is (A).
Quick Tip: When manipulating matrix equations, always look for opportunities to substitute the given relations. The associative property of multiplication is your most powerful tool. The conditions given imply that A and B are idempotent, which is a key observation.

Step 1: Understanding the Concept:

We need to determine if the function \( f(x) = x^3 + 5x + 1 \) is one-one (injective) and onto (surjective) over the set of real numbers R.

A function is one-one if every distinct element in the domain maps to a distinct element in the co-domain. For a differentiable function, if its derivative is strictly positive or strictly negative, it is one-one.

A function is onto if its range is equal to its co-domain.


Step 2: Checking for One-one (Injectivity):

To check if the function is one-one, we can examine its derivative, \( f'(x) \).
\[ f(x) = x^3 + 5x + 1 \] \[ f'(x) = \frac{d}{dx}(x^3 + 5x + 1) = 3x^2 + 5 \]
Now, let's analyze the sign of \( f'(x) \). For any real number \( x \), \( x^2 \ge 0 \).

Therefore, \( 3x^2 \ge 0 \).

This implies \( f'(x) = 3x^2 + 5 \ge 5 \).

Since \( f'(x) > 0 \) for all \( x \in R \), the function \( f(x) \) is strictly increasing. A strictly monotonic function is always one-one.

So, f is one-one.


Step 3: Checking for Onto (Surjectivity):

The function \( f(x) \) is a polynomial of degree 3 (an odd degree). The domain and co-domain are both R.

For any polynomial function with an odd degree and real coefficients, the range is the set of all real numbers, \( (-\infty, \infty) \).

We can also verify this by checking the limits as \( x \) approaches \( \pm\infty \).
\[ \lim_{x \to \infty} f(x) = \lim_{x \to \infty} (x^3 + 5x + 1) = \infty \] \[ \lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} (x^3 + 5x + 1) = -\infty \]
Since the function is continuous and extends from \( -\infty \) to \( \infty \), its range is R. The co-domain is also R.

Therefore, the function is onto.


Step 4: Final Answer:

Since the function is both one-one and onto, it is a bijective function on R. The correct option is (C).
Quick Tip: A quick way to solve this: 1. \textbf{One-one}: Find the derivative. If \( f'(x) \) is always positive or always negative, the function is one-one. 2. \textbf{Onto}: If \( f(x) \) is a polynomial of an odd degree from R to R, it is always onto.

Step 1: Understanding the Concept:

We are asked to find an expression for the second derivative of \( y = \sin^{-1} x \) multiplied by \( (1-x^2) \). This involves finding the first and second derivatives of the given function.


Step 2: Key Formula or Approach:

We will first find \( \frac{dy}{dx} \). Then, instead of directly finding the second derivative, we will rearrange the first derivative equation to simplify the differentiation process.


Step 3: Detailed Explanation:

Given the function:
\[ y = \sin^{-1} x \]
Differentiating with respect to x, we get the first derivative:
\[ \frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}} \]
To avoid a complicated quotient/chain rule for the second derivative, we can rearrange this equation:
\[ \sqrt{1-x^2} \frac{dy}{dx} = 1 \]
Squaring both sides to eliminate the square root:
\[ (1-x^2) \left(\frac{dy}{dx}\right)^2 = 1^2 = 1 \]
Now, differentiate both sides of this equation with respect to x, using the product rule on the left side. The product rule is \( (uv)' = u'v + uv' \).

Let \( u = (1-x^2) \) and \( v = \left(\frac{dy}{dx}\right)^2 \).
\[ \frac{d}{dx}(1-x^2) \cdot \left(\frac{dy}{dx}\right)^2 + (1-x^2) \cdot \frac{d}{dx}\left(\left(\frac{dy}{dx}\right)^2\right) = \frac{d}{dx}(1) \] \[ (-2x) \left(\frac{dy}{dx}\right)^2 + (1-x^2) \cdot \left(2 \frac{dy}{dx} \cdot \frac{d^2y}{dx^2}\right) = 0 \]
We can divide the entire equation by \( 2 \frac{dy}{dx} \) (since \( \frac{dy}{dx} \neq 0 \) for \( |x| < 1 \)).
\[ -x \frac{dy}{dx} + (1-x^2) \frac{d^2y}{dx^2} = 0 \]
Rearranging the terms to find the required expression:
\[ (1-x^2) \frac{d^2y}{dx^2} = x \frac{dy}{dx} \]

Step 4: Final Answer:

The expression \( (1-x^2) \frac{d^2y}{dx^2} \) is equal to \( x \frac{dy}{dx} \). The correct option is (A).
Quick Tip: When dealing with second derivatives involving square roots, it's often a good strategy to rearrange the first derivative equation to eliminate the root (usually by squaring) and then differentiate implicitly. This avoids complex differentiation and leads to the answer more quickly.

Step 1: Understanding the Concept:

A function \( f(x) \) is decreasing for all real values of x if its derivative, \( f'(x) \), is less than or equal to zero for all \( x \in R \). That is, \( f'(x) \le 0 \).


Step 2: Key Formula or Approach:

We will find the derivative \( f'(x) \), set up the inequality \( f'(x) \le 0 \), and then use the property that the maximum value of \( a\sin x + b\cos x \) is \( \sqrt{a^2 + b^2} \).


Step 3: Detailed Explanation:

The given function is:
\[ f(x) = \sin x - \cos x - \lambda x + C \]
First, we find its derivative with respect to x:
\[ f'(x) = \frac{d}{dx}(\sin x - \cos x - \lambda x + C) \] \[ f'(x) = \cos x - (-\sin x) - \lambda \] \[ f'(x) = \sin x + \cos x - \lambda \]
For the function \( f(x) \) to be decreasing for all x, we must have \( f'(x) \le 0 \) for all \( x \in R \).
\[ \sin x + \cos x - \lambda \le 0 \] \[ \sin x + \cos x \le \lambda \]
This inequality must hold true for all real values of x. This means that \( \lambda \) must be greater than or equal to the maximum possible value of the expression \( \sin x + \cos x \).

We know that for an expression of the form \( a\sin x + b\cos x \), the maximum value is \( \sqrt{a^2+b^2} \).

In our case, \( a=1 \) and \( b=1 \).

So, the maximum value of \( \sin x + \cos x \) is \( \sqrt{1^2 + 1^2} = \sqrt{2} \).

Therefore, for the inequality \( \sin x + \cos x \le \lambda \) to be true for all x, \( \lambda \) must be greater than or equal to this maximum value.
\[ \lambda \ge \sqrt{2} \]

Step 4: Final Answer:

The values of \( \lambda \) for which the function decreases for all real x are given by \( \lambda \ge \sqrt{2} \). The correct option is (C).
Quick Tip: Remember the range of the function \( f(x) = a\sin x + b\cos x \) is \( [-\sqrt{a^2+b^2}, \sqrt{a^2+b^2}] \). This is extremely useful for solving problems involving monotonicity or finding the range of trigonometric expressions.

Step 1: Understanding the Concept:

We can find the coordinates of a point P that divides a line segment in a certain ratio using the section formula. We are given one coordinate of P, which allows us to find the ratio. Then, we can use this ratio to find the unknown z-coordinate.


Step 2: Key Formula or Approach:

Let the two points be \( A(x_1, y_1, z_1) \) and \( B(x_2, y_2, z_2) \). If point P divides AB in the ratio \( k:1 \), its coordinates are given by the section formula:
\[ P = \left( \frac{kx_2+x_1}{k+1}, \frac{ky_2+y_1}{k+1}, \frac{kz_2+z_1}{k+1} \right) \]

Step 3: Detailed Explanation:

Let the given points be A(3, 6, -1) and B(6, 2, -2). Let the point P divide the line segment AB in the ratio \( k:1 \).

The coordinates of P are:
\[ P_x = \frac{k(6)+1(3)}{k+1} = \frac{6k+3}{k+1} \] \[ P_y = \frac{k(2)+1(6)}{k+1} = \frac{2k+6}{k+1} \] \[ P_z = \frac{k(-2)+1(-1)}{k+1} = \frac{-2k-1}{k+1} \]
We are given that the y-coordinate of P is 4. So, \( P_y = 4 \).
\[ \frac{2k+6}{k+1} = 4 \]
Now, we solve for k:
\[ 2k+6 = 4(k+1) \] \[ 2k+6 = 4k+4 \] \[ 6-4 = 4k-2k \] \[ 2 = 2k \] \[ k=1 \]
A ratio of \( k:1 = 1:1 \) means that P is the midpoint of the line segment AB.

Now we can find the z-coordinate of P by substituting \( k=1 \) into the expression for \( P_z \).
\[ P_z = \frac{-2(1)-1}{1+1} = \frac{-2-1}{2} = -\frac{3}{2} \]

Step 4: Final Answer:

The z-coordinate of point P is \( -\frac{3}{2} \). The correct option is (A).
Quick Tip: For problems involving a point dividing a line segment, always use the section formula. If one coordinate is given, use it to find the ratio \(k\). Once \(k\) is known, you can find any other unknown coordinate. If \(k=1\), the point is simply the midpoint.

Step 1: Understanding the Concept:

We are given an equation involving matrices M and N and need to find the determinant of N. We will use the properties of determinants to solve this problem.


Step 2: Key Formula or Approach:

The key properties of determinants we will use are:
1. \( \det(AB) = \det(A) \det(B) \)
2. \( \det(kA) = k^n \det(A) \), where A is a square matrix of order n and k is a scalar.

3. \( \det(I) = 1 \), where I is the identity matrix.


Step 3: Detailed Explanation:

We are given the matrix equation:
\[ MN = mI \]
We are also given that M and N are square matrices of order 3, and \( \det(M) = m \).

Taking the determinant of both sides of the equation:
\[ \det(MN) = \det(mI) \]
Using the property \( \det(AB) = \det(A)\det(B) \), the left side becomes:
\[ \det(M)\det(N) = \det(mI) \]
For the right side, we use the property \( \det(kA) = k^n \det(A) \). Here, the scalar is \( k=m \), the matrix is \( A=I \), and the order is \( n=3 \).
\[ \det(mI) = m^3 \det(I) \]
Since \( \det(I) = 1 \), we have:
\[ \det(mI) = m^3 \cdot 1 = m^3 \]
Now substitute these back into the main equation:
\[ \det(M)\det(N) = m^3 \]
We are given that \( \det(M) = m \). Substituting this value:
\[ m \cdot \det(N) = m^3 \]
Assuming \( m \neq 0 \) (if \( m=0 \), the problem changes, but the options suggest a unique answer), we can divide both sides by m:
\[ \det(N) = \frac{m^3}{m} = m^2 \]

Step 4: Final Answer:

The determinant of N is \( m^2 \). The correct option is (D).
Quick Tip: When you see a matrix equation and are asked about determinants, immediately think of taking the determinant of both sides. Remember the rule for scalars inside a determinant: \( \det(kA) = k^n \det(A) \), where n is the order of the matrix. This is a common point of error.

Step 1: Understanding the Concept:

A function is continuous at a point \( x=c \) if the left-hand limit (LHL), the right-hand limit (RHL), and the value of the function at that point are all equal. That is, \( \lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c) \). For a piecewise function to be continuous over an interval, it must be continuous at the points where the definition of the function changes.


Step 2: Detailed Explanation:

The given function is: \[ f(x) = \begin{cases} 3x-2, & 0 < x \le 1
2x^2+ax, & 1 < x < 2 \end{cases} \]
The function is composed of two polynomials, which are continuous everywhere on their defined open intervals. The only potential point of discontinuity in the interval \( (0, 2) \) is at \( x=1 \), where the function definition changes.

For \( f(x) \) to be continuous at \( x=1 \), the following condition must be met: \[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) \]
Let's calculate the Left-Hand Limit (LHL):
As \( x \) approaches 1 from the left (\( x < 1 \)), we use the definition \( f(x) = 3x - 2 \). \[ LHL = \lim_{x \to 1^-} (3x-2) = 3(1) - 2 = 1 \]
Now, let's calculate the Right-Hand Limit (RHL):
As \( x \) approaches 1 from the right (\( x > 1 \)), we use the definition \( f(x) = 2x^2 + ax \). \[ RHL = \lim_{x \to 1^+} (2x^2+ax) = 2(1)^2 + a(1) = 2+a \]
For continuity, LHL must be equal to RHL. \[ 1 = 2+a \]
Solving for a: \[ a = 1 - 2 = -1 \]
We should also check \( f(1) \). From the definition, for \( x=1 \), \( f(x) = 3x-2 \), so \( f(1) = 3(1)-2=1 \). Since LHL = f(1) = 1, we only needed to equate LHL and RHL.


Step 3: Final Answer:

The value of a for which the function is continuous is -1. The correct option is (D).
Quick Tip: For piecewise functions, the first step to check for continuity is to identify the "break points" where the definition changes. Equate the left-hand and right-hand limits at these points to find the values of any unknown constants.

Step 1: Understanding the Concept:

We need to analyze the given function \( f: N \to W \), where N = {1, 2, 3, ... is the set of natural numbers (domain) and W = {0, 1, 2, ... is the set of whole numbers (co-domain).

- Injective (One-to-one): A function is injective if \( f(n_1) = f(n_2) \implies n_1 = n_2 \).
- Surjective (Onto): A function is surjective if for every element \( w \) in the co-domain W, there exists at least one element \( n \) in the domain N such that \( f(n) = w \). This means the range of the function must be equal to the co-domain.


Step 2: Checking for Injectivity:

Let's test a few inputs.
For odd inputs: \( f(1) = 0 \) \( f(3) = 0 \) \( f(5) = 0 \)
We can see that \( f(1) = f(3) \), but \( 1 \neq 3 \). Since different inputs (1 and 3) map to the same output (0), the function is not injective.


Step 3: Checking for Surjectivity:

We need to check if the range of f is equal to the co-domain W = {0, 1, 2, 3, ....
Let's find the range by checking the outputs for different types of inputs.
If n is odd, \( f(n) = 0 \). So, 0 is in the range.
If n is even, \( f(n) = \frac{n-2}{2} \).
Let's check the outputs for some even numbers: \( f(2) = \frac{2-2}{2} = 0 \) \( f(4) = \frac{4-2}{2} = 1 \) \( f(6) = \frac{6-2}{2} = 2 \) \( f(8) = \frac{8-2}{2} = 3 \)
It appears that for any whole number \( w \in W \), we can find a pre-image in N.
Let's try to find a pre-image for an arbitrary \( w \in W \). We need to find an \( n \in N \) such that \( f(n) = w \).
We can set \( \frac{n-2}{2} = w \), assuming n is even. \[ n-2 = 2w \] \[ n = 2w + 2 \]
For any \( w \in \{0, 1, 2, ...\} \), \( n = 2w+2 \) will be an even natural number. For example:
- if \( w=0 \), \( n = 2(0)+2 = 2 \). \( f(2)=0 \).
- if \( w=10 \), \( n = 2(10)+2 = 22 \). \( f(22)=10 \).
Since we can find a pre-image \( n = 2w+2 \) for every whole number \( w \), the range of the function is {0, 1, 2, ... which is W.
Therefore, the function is surjective.


Step 4: Final Answer:

The function is surjective but not injective. The correct option is (B).
Quick Tip: To test for injectivity, try to find a counterexample: two different inputs giving the same output. To test for surjectivity, take an arbitrary element 'y' from the co-domain and try to solve the equation f(x) = y for 'x'. If you can always find at least one 'x' in the domain, the function is surjective.

Step 1: Understanding the Concept:

We need to identify the type of the given matrix by checking the definitions of diagonal, symmetric, skew-symmetric, and scalar matrices.
- Diagonal Matrix: All non-diagonal elements are zero.
- Scalar Matrix: A diagonal matrix where all diagonal elements are equal.
- Symmetric Matrix: A square matrix A such that \( A^T = A \), where \( A^T \) is the transpose of A. This means \( a_{ij} = a_{ji} \) for all i, j.
- Skew-Symmetric Matrix: A square matrix A such that \( A^T = -A \). This means \( a_{ij} = -a_{ji} \) for all i, j. A consequence is that all diagonal elements must be zero.


Step 2: Detailed Explanation:

Let the given matrix be A. \[ A = \begin{pmatrix} 0 & 1 & -2
-1 & 0 & -7
2 & 7 & 0 \end{pmatrix} \]
1. It is not a diagonal matrix because non-diagonal elements like \( a_{12}=1 \) are not zero.
2. It is not a scalar matrix because it's not a diagonal matrix.
3. Let's check if it is a symmetric matrix. For this, we need \( A^T = A \). Let's find the transpose of A.
\[ A^T = \begin{pmatrix} 0 & -1 & 2
1 & 0 & 7
-2 & -7 & 0 \end{pmatrix} \]
Clearly, \( A^T \neq A \). For example, \( a_{12}^T = -1 \) while \( a_{12} = 1 \). So, it is not symmetric.
4. Let's check if it is a skew-symmetric matrix. For this, we need \( A^T = -A \).
Let's compute -A.
\[ -A = -1 \cdot \begin{pmatrix} 0 & 1 & -2
-1 & 0 & -7
2 & 7 & 0 \end{pmatrix} = \begin{pmatrix} 0 & -1 & 2
1 & 0 & 7
-2 & -7 & 0 \end{pmatrix} \]
Comparing \( A^T \) and \( -A \), we see that they are identical.
\[ A^T = -A \]
Therefore, the matrix A is a skew-symmetric matrix.


Step 3: Final Answer:

The given matrix is a skew-symmetric matrix. The correct option is (C).
Quick Tip: There are two quick checks for a skew-symmetric matrix: 1. All elements on the main diagonal must be zero. 2. The elements across the main diagonal must be negatives of each other (i.e., \( a_{ij} = -a_{ji} \)). If both conditions are met, the matrix is skew-symmetric.

Step 1: Understanding the Concept:

The median of a triangle from a vertex A is the line segment that connects A to the midpoint of the opposite side BC. We are given the vectors for the sides \( \vec{AB} \) and \( \vec{AC} \). The vector for the median from A can be found using these two vectors. The length of the median is the magnitude of this median vector.


Step 2: Key Formula or Approach:

If D is the midpoint of BC, the vector representing the median from A is \( \vec{AD} \). By the parallelogram law of vector addition, or by finding the position vector of the midpoint, the median vector \( \vec{AD} \) is given by: \[ \vec{AD} = \frac{\vec{AB} + \vec{AC}}{2} \]
The length of the median is then \( |\vec{AD}| \).


Step 3: Detailed Explanation:

Let the given vectors be: \[ \vec{AB} = \hat{j} + \hat{k} = 0\hat{i} + 1\hat{j} + 1\hat{k} \] \[ \vec{AC} = 3\hat{i} - \hat{j} + 4\hat{k} \]
Let D be the midpoint of BC. The vector for the median from A is \( \vec{AD} \).
Using the formula for the median vector: \[ \vec{AD} = \frac{\vec{AB} + \vec{AC}}{2} \] \[ \vec{AD} = \frac{(0\hat{i} + 1\hat{j} + 1\hat{k}) + (3\hat{i} - 1\hat{j} + 4\hat{k})}{2} \]
Combine the corresponding components: \[ \vec{AD} = \frac{(0+3)\hat{i} + (1-1)\hat{j} + (1+4)\hat{k}}{2} \] \[ \vec{AD} = \frac{3\hat{i} + 0\hat{j} + 5\hat{k}}{2} = \frac{3}{2}\hat{i} + \frac{5}{2}\hat{k} \]
Now, we find the length of this median, which is the magnitude of the vector \( \vec{AD} \). \[ |\vec{AD}| = \sqrt{\left(\frac{3}{2}\right)^2 + (0)^2 + \left(\frac{5}{2}\right)^2} \] \[ |\vec{AD}| = \sqrt{\frac{9}{4} + 0 + \frac{25}{4}} \] \[ |\vec{AD}| = \sqrt{\frac{9+25}{4}} = \sqrt{\frac{34}{4}} \] \[ |\vec{AD}| = \frac{\sqrt{34}}{\sqrt{4}} = \frac{\sqrt{34}}{2} \]

Step 4: Final Answer:

The length of the median through A is \( \frac{\sqrt{34}}{2} \) units. The correct option is (C).
Quick Tip: A very useful formula to remember: In a triangle ABC, the vector of the median from vertex A to side BC is the average of the vectors of the two sides originating from A, i.e., \( \vec{AD} = \frac{1}{2}(\vec{AB} + \vec{AC}) \). This saves you from having to find the vector BC and its midpoint explicitly.

Step 1: Understanding the Concept:

A function is continuous at a point \( x=c \) if the limit of the function as x approaches c exists and is equal to the function's value at c. For piecewise functions, we must check for continuity at the points where the function's definition changes. This involves comparing the left-hand limit (LHL), right-hand limit (RHL), and the function's value at that point.


Step 2: Detailed Explanation:

The given piecewise function is: \[ f(x) = \begin{cases} x, & if x \le 1
5, & if x > 1 \end{cases} \]
The function is defined by \( y=x \) (a line) and \( y=5 \) (a constant function). Both of these are continuous everywhere. The only point where a discontinuity might occur is at \( x=1 \), where the rule changes. Let's check the conditions for continuity at \( x=1 \).


1. Left-Hand Limit (LHL): We consider values of x approaching 1 from the left, so \( x < 1 \). For these values, \( f(x) = x \).
\[ LHL = \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} x = 1 \]

2. Right-Hand Limit (RHL): We consider values of x approaching 1 from the right, so \( x > 1 \). For these values, \( f(x) = 5 \).
\[ RHL = \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} 5 = 5 \]

3. Value of the function at x=1: For \( x=1 \), the rule is \( f(x) = x \).
\[ f(1) = 1 \]

For the function to be continuous at \( x=1 \), we must have LHL = RHL = f(1).
Here, we have LHL = 1 and RHL = 5.
Since \( LHL \neq RHL \), the limit \( \lim_{x \to 1} f(x) \) does not exist. Therefore, the function is not continuous at \( x=1 \).


At any other point \( c \neq 1 \), the function is continuous. For \( c < 1 \), \( f(x)=x \) is continuous. For \( c > 1 \), \( f(x)=5 \) is continuous.


Step 3: Final Answer:

The function f is not continuous at \( x=1 \). The correct option is (B).
Quick Tip: For piecewise functions, always focus your continuity check on the "boundary" points where the function definition splits. A jump in the graph, where the left and right limits do not match, indicates a "jump discontinuity".

Step 1: Understanding the Concept:

To determine if a function is increasing, decreasing, or has maxima/minima, we analyze its first derivative, \( f'(x) \).
- If \( f'(x) > 0 \) for all x in an interval, the function is increasing on that interval.
- If \( f'(x) < 0 \) for all x in an interval, the function is decreasing on that interval.
- Critical points (where maxima/minima can occur) are found where \( f'(x) = 0 \) or is undefined.


Step 2: Detailed Explanation:

The given function is: \[ f(x) = 2x + \cos(x) \]
First, we find the derivative of the function with respect to x: \[ f'(x) = \frac{d}{dx}(2x + \cos(x)) \] \[ f'(x) = 2 - \sin(x) \]
Now, we need to analyze the sign of \( f'(x) \) for all real numbers x.
We know that the range of the sine function is \( -1 \le \sin(x) \le 1 \).
Let's use this to find the range of \( f'(x) \).
The maximum value of \( \sin(x) \) is 1. When \( \sin(x) = 1 \): \[ f'(x) = 2 - 1 = 1 \]
The minimum value of \( \sin(x) \) is -1. When \( \sin(x) = -1 \): \[ f'(x) = 2 - (-1) = 3 \]
So, for any real value of x, the value of \( f'(x) \) will be between 1 and 3, inclusive. \[ 1 \le f'(x) \le 3 \]
Since \( f'(x) \) is always greater than 0 (specifically, \( f'(x) \ge 1 \)) for all \( x \in R \), the function \( f(x) \) is strictly increasing for all real numbers.

Because the function is always increasing, it does not have any local maxima or minima.


Step 3: Final Answer:

The function \( f(x) = 2x + \cos(x) \) is an increasing function. The correct option is (C).
Quick Tip: When analyzing the derivative, use your knowledge of the range of basic functions like sin(x) and cos(x). This can quickly tell you if the derivative is always positive or always negative, allowing you to determine the function's monotonic behavior without solving for critical points.

Step 1: Understanding the Concept:

The problem requires finding the indefinite integral of a trigonometric function. The key to solving this is to simplify the integrand using trigonometric identities before performing the integration.


Step 2: Key Formula or Approach:

We will use the double angle identity for cosine:
\[ \cos 2\theta = 2\cos^2 \theta - 1 \]
We will also use the difference of squares factorization: \( a^2 - b^2 = (a - b)(a + b) \).


Step 3: Detailed Explanation:

Let the given integral be \( I \).
\[ I = \int \frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha} dx \]
Apply the double angle formula to the numerator:
\[ \cos 2x = 2\cos^2 x - 1 \] \[ \cos 2\alpha = 2\cos^2 \alpha - 1 \]
Substitute these into the numerator:
\[ Numerator = (2\cos^2 x - 1) - (2\cos^2 \alpha - 1) = 2\cos^2 x - 1 - 2\cos^2 \alpha + 1 = 2(\cos^2 x - \cos^2 \alpha) \]
Now, factor the numerator using the difference of squares formula:
\[ 2(\cos^2 x - \cos^2 \alpha) = 2(\cos x - \cos \alpha)(\cos x + \cos \alpha) \]
Substitute this simplified numerator back into the integral:
\[ I = \int \frac{2(\cos x - \cos \alpha)(\cos x + \cos \alpha)}{\cos x - \cos \alpha} dx \]
Cancel the common term \( (\cos x - \cos \alpha) \) from the numerator and denominator, assuming \( \cos x \neq \cos \alpha \).
\[ I = \int 2(\cos x + \cos \alpha) dx \]
Now, we can integrate term by term. Note that \( \cos \alpha \) is a constant with respect to \( x \).
\[ I = 2 \int \cos x \,dx + 2 \int \cos \alpha \,dx \] \[ I = 2 \sin x + 2 (\cos \alpha) \int 1 \,dx \] \[ I = 2 \sin x + 2x \cos \alpha + C \]
Factoring out the 2, we get:
\[ I = 2(\sin x + x \cos \alpha) + C \]
This matches option (A).


Step 4: Final Answer:

The value of the integral is \( 2(\sin x + x \cos \alpha) + C \).
Quick Tip: When faced with integrals involving trigonometric functions, always look for identities that can simplify the expression. The double angle formulas and sum-to-product/product-to-sum formulas are particularly useful for simplifying fractions.

Step 1: Understanding the Concept:

This problem involves evaluating a definite integral of a function with exponential terms. The strategy is to manipulate the integrand into a standard form that can be integrated, often using substitution.


Step 2: Key Formula or Approach:

The key is to rewrite the integrand to fit the standard integral form:
\[ \int \frac{du}{a^2 + u^2} = \frac{1}{a} \tan^{-1}\left(\frac{u}{a}\right) + C \]
In this case, we will aim for the form \( \int \frac{du}{1 + u^2} = \tan^{-1}(u) + C \).


Step 3: Detailed Explanation:

Let the integral be \( I \).
\[ I = \int_{0}^{1} \frac{dx}{e^x + e^{-x}} \]
First, simplify the integrand by multiplying the numerator and denominator by \( e^x \):
\[ I = \int_{0}^{1} \frac{e^x \cdot dx}{e^x(e^x + e^{-x})} = \int_{0}^{1} \frac{e^x}{e^{2x} + e^{0}} dx = \int_{0}^{1} \frac{e^x}{(e^x)^2 + 1} dx \]
Now, we can use u-substitution. Let \( u = e^x \).

Then, \( du = e^x dx \).

We also need to change the limits of integration from \( x \) to \( u \):

When \( x = 0 \), \( u = e^0 = 1 \).

When \( x = 1 \), \( u = e^1 = e \).

Substitute \( u \) and \( du \) and the new limits into the integral:
\[ I = \int_{1}^{e} \frac{du}{u^2 + 1} \]
This is a standard integral form. The integral of \( \frac{1}{u^2+1} \) is \( \tan^{-1}(u) \).

Now, evaluate the definite integral:
\[ I = [\tan^{-1}(u)]_{1}^{e} \] \[ I = \tan^{-1}(e) - \tan^{-1}(1) \]
We know that \( \tan^{-1}(1) = \frac{\pi}{4} \).
\[ I = \tan^{-1}(e) - \frac{\pi}{4} \]
This matches option (C).


Step 4: Final Answer:

The value of the integral is \( \tan^{-1} e - \frac{\pi}{4} \).
Quick Tip: For integrals containing \(e^x\) and \(e^{-x}\), a common and effective trick is to multiply both the numerator and denominator by \(e^x\). This often converts the integrand into a form suitable for substitution, typically with \(u = e^x\).

Step 1: Understanding the Concept:

Order of a differential equation is the order of the highest derivative appearing in the equation.

Degree of a differential equation is the highest power (positive integer) of the highest order derivative, after the equation has been cleared of radicals and fractions in its derivatives. A key condition is that the differential equation must be a polynomial equation in its derivatives.


Step 2: Detailed Explanation:

Let's analyze the given differential equation:
\[ \left(\frac{d^2y}{dx^2}\right)^2 + \left(\frac{dy}{dx}\right)^2 = x \sin\left(\frac{dy}{dx}\right) \]
Finding the Order:

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

The highest order derivative is \( \frac{d^2y}{dx^2} \), which is of order 2.

Therefore, the order of the differential equation is 2.


Finding the Degree:

For the degree to be defined, the differential equation must be expressible as a polynomial in its derivatives.

Consider the term on the right-hand side: \( x \sin\left(\frac{dy}{dx}\right) \).

The function \( \sin(\cdot) \) is a transcendental function. If we expand \( \sin\left(\frac{dy}{dx}\right) \) using its Taylor (Maclaurin) series, we get:
\[ \sin(z) = z - \frac{z^3}{3!} + \frac{z^5}{5!} - \dots \]
Substituting \( z = \frac{dy}{dx} \), the equation becomes:
\[ \left(\frac{d^2y}{dx^2}\right)^2 + \left(\frac{dy}{dx}\right)^2 = x \left[ \left(\frac{dy}{dx}\right) - \frac{1}{3!}\left(\frac{dy}{dx}\right)^3 + \frac{1}{5!}\left(\frac{dy}{dx}\right)^5 - \dots \right] \]
This is an infinite series in terms of \( \frac{dy}{dx} \). Because the equation cannot be written as a finite polynomial in its derivatives, the degree is not defined.


Step 3: Final Answer:

The order of the differential equation is 2, and the degree is not defined. This corresponds to option (C).
Quick Tip: A simple rule to remember: if any derivative in the differential equation appears as an argument of a transcendental function (like sin, cos, log, exponential), the degree of the equation is not defined. Always check for this first when determining the degree.

Step 1: Understanding the Concept:

The area of a region bounded by a curve \( y = f(x) \), the x-axis, and the vertical lines \( x = a \) and \( x = b \) is given by the definite integral of the function \( f(x) \) from \( a \) to \( b \).


Step 2: Key Formula or Approach:

The formula for the area \( A \) is:
\[ A = \int_{a}^{b} y \,dx \]
We will use the power rule for integration: \( \int x^n \,dx = \frac{x^{n+1}}{n+1} \).


Step 3: Detailed Explanation:

We are given the following boundaries:

The curve: \( y = \sqrt{x} = x^{1/2} \).

The vertical lines: \( x = 0 \) and \( x = 4 \).

The boundary line: the x-axis (\( y = 0 \)).

Here, \( f(x) = x^{1/2} \), \( a = 0 \), and \( b = 4 \). Since \( y = \sqrt{x} \) is non-negative for \( x \in [0, 4] \), the curve is above the x-axis in this interval.

Set up the definite integral for the area:
\[ A = \int_{0}^{4} x^{1/2} \,dx \]
Now, apply the power rule for integration:
\[ A = \left[ \frac{x^{(1/2) + 1}}{(1/2) + 1} \right]_{0}^{4} \] \[ A = \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{4} \] \[ A = \left[ \frac{2}{3} x^{3/2} \right]_{0}^{4} \]
Evaluate the integral at the upper and lower limits:
\[ A = \frac{2}{3} (4)^{3/2} - \frac{2}{3} (0)^{3/2} \]
Calculate \( (4)^{3/2} \):
\[ (4)^{3/2} = (\sqrt{4})^3 = (2)^3 = 8 \]
Now substitute this value back into the area equation:
\[ A = \frac{2}{3} (8) - 0 \] \[ A = \frac{16}{3} \]
The area is \( \frac{16}{3} \) square units. This matches option (C).


Step 4: Final Answer:

The area of the enclosed region is \( \frac{16}{3} \) sq. units.
Quick Tip: Before setting up the integral for an area problem, it's helpful to sketch the graph of the function and the boundaries. This helps to visualize the region and confirm that the function is above the x-axis (or to see if you need to split the integral).

Step 1: Understanding the Concept:

In a Linear Programming Problem (LPP), the optimal (maximum or minimum) value of the objective function, if it exists, must occur at one of the corner points (vertices) of the feasible region. If the optimal value occurs at two distinct corner points, it also occurs at every point on the line segment connecting these two points. A consequence of this is that the value of the objective function at these two corner points must be equal.


Step 2: Key Formula or Approach:

The objective function is given by \( Z = ax + by \).

We are told that the maximum value of \( Z \) occurs at two points: \( P_1(0, 2) \) and \( P_2(3, 0) \).

This implies that the value of \( Z \) at \( P_1 \) is equal to the value of \( Z \) at \( P_2 \).

Let \( Z_1 \) be the value of Z at \( P_1 \) and \( Z_2 \) be the value of Z at \( P_2 \). Then \( Z_1 = Z_2 \).


Step 3: Detailed Explanation:

Calculate the value of the objective function \( Z \) at the first point (0, 2):
\[ Z_1 = a(0) + b(2) = 2b \]
Calculate the value of the objective function \( Z \) at the second point (3, 0):
\[ Z_2 = a(3) + b(0) = 3a \]
Since the maximum value is obtained at both points, these two values must be equal:
\[ Z_1 = Z_2 \] \[ 2b = 3a \]
This gives the relation between \( a \) and \( b \). Rearranging, it is \( 3a = 2b \), which corresponds to option (D).

Let's verify that this value is indeed the maximum by checking other points with a relation like \(a=2k, b=3k\) for some \(k>0\).
Then \(Z = 2kx + 3ky\).
Z at (0,2) is \(6k\).
Z at (3,0) is \(6k\).
Z at (6,0) is \(12k\).
Z at (6,8) is \(12k+24k = 36k\).
Z at (0,5) is \(15k\).
The maximum is at (6,8). The question states the maximum is obtained at (0,2) and (3,0). There seems to be an inconsistency in the problem statement. However, based on the condition that "the maximum value of Z is obtained at (0, 2) and (3, 0)", it implicitly means that the values of Z at these two points are equal and also maximal among all corner points. This condition of equality is what is being tested. Thus, we proceed by equating the values.
\[ 2b = 3a \]

Step 4: Final Answer:

The relation between a and b is \( 3a = 2b \).
Quick Tip: In LPP, if an objective function has the same optimal value at two different corner points, it means the slope of the objective function line is the same as the slope of the line segment (edge) connecting those two points. The problem is simply asking for the condition where \(Z(0,2) = Z(3,0)\).

Step 1: Understanding the Concept:

We need to evaluate the truthfulness of both the Assertion and the Reason, and then determine if the Reason correctly explains the Assertion.

- Assertion (A) makes a claim about independent events based on the condition \( P(A \cap B) = 0 \).

- Reason (R) provides the definition of independent events.


Step 2: Key Formula or Approach:

The mathematical condition for two events A and B to be independent is \( P(A \cap B) = P(A) \cdot P(B) \).

The condition \( P(A \cap B) = 0 \) means that A and B are mutually exclusive events (they cannot occur simultaneously).


Step 3: Detailed Explanation:

Analysis of Reason (R):

The statement "Two events are independent if the occurrence of one does not effect the occurrence of the other" is the standard conceptual definition of independent events. Therefore, Reason (R) is true.


Analysis of Assertion (A):

Assertion (A) states that if \( P(A \cap B) = 0 \), then A and B are independent.

For A and B to be independent, the condition \( P(A \cap B) = P(A) \cdot P(B) \) must hold.

If we are given \( P(A \cap B) = 0 \), then for the events to be independent, we must have \( P(A) \cdot P(B) = 0 \). This is only true if \( P(A) = 0 \) or \( P(B) = 0 \) (or both).

However, the assertion claims this is true for any two events with \( P(A \cap B) = 0 \). Let's consider a counterexample.

Let's consider the experiment of rolling a single fair die.

Let event A be "the outcome is 2". Then \( P(A) = 1/6 \).

Let event B be "the outcome is 3". Then \( P(B) = 1/6 \).

Here, A and B are mutually exclusive, as you cannot get both 2 and 3 in a single roll. So, \( P(A \cap B) = P(getting 2 and 3) = 0 \).

Now, let's check the condition for independence:
\( P(A) \cdot P(B) = \frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36} \).

Since \( P(A \cap B) = 0 \) and \( P(A) \cdot P(B) = \frac{1}{36} \), we have \( P(A \cap B) \neq P(A) \cdot P(B) \).

Therefore, A and B are not independent.

This counterexample shows that having \( P(A \cap B) = 0 \) does not imply that events are independent (unless one of the events has zero probability). Thus, Assertion (A) is false.


Step 4: Final Answer:

Assertion (A) is false and Reason (R) is true. This corresponds to option (D).
Quick Tip: Do not confuse mutually exclusive events with independent events. - \textbf{Mutually Exclusive}: \( P(A \cap B) = 0 \). If one happens, the other cannot. They are highly dependent. - \textbf{Independent}: \( P(A \cap B) = P(A) \cdot P(B) \). The occurrence of one does not influence the probability of the other. Two events with non-zero probabilities cannot be both mutually exclusive and independent.

Step 1: Understanding the Concept:

This question tests the fundamental concepts of Linear Programming Problems (LPP). We need to understand what a feasible region is and what its implications are for the existence of a solution.


Step 2: Detailed Explanation:

Analysis of Reason (R):

The statement "A feasible region is defined as the region that satisfies all the constraints" is the correct and standard definition of a feasible region in an LPP. The feasible region represents the set of all possible points (or combinations of decision variables) that are valid according to the given constraints (e.g., \( x \ge 0, y \ge 0, x+y \le 10 \), etc.). So, Reason (R) is true.


Analysis of Assertion (A):

The statement "In a Linear Programming Problem, if the feasible region is empty, then the Linear Programming Problem has no solution" is also correct. A solution to an LPP must satisfy all of its constraints. The set of all such points is the feasible region. If this region is empty, it means there are no points that satisfy all the constraints simultaneously. Therefore, there is no valid solution to the problem. So, Assertion (A) is true.


Connecting Reason and Assertion:

The Reason (R) defines the feasible region as the set of all points satisfying the constraints. The Assertion (A) states a direct consequence of this definition: if this set of points is empty, then no solution exists. The logic flows directly from the reason to the assertion. Because the feasible region is the collection of all possible solutions, its emptiness logically implies the non-existence of a solution.

Therefore, Reason (R) is the correct explanation for Assertion (A).


Step 3: Final Answer:

Both Assertion (A) and Reason (R) are true, and Reason (R) correctly explains Assertion (A). This corresponds to option (A).
Quick Tip: In LPP, the existence of a solution depends entirely on the feasible region. - \textbf{Empty Feasible Region}: No solution. - \textbf{Bounded Feasible Region}: An optimal (maximum and minimum) solution always exists. - \textbf{Unbounded Feasible Region}: An optimal solution may or may not exist. If it exists, it will be at a corner point.

Step 1: Understanding the Concept:

We need to find the determinant of a product of matrices, where one of the terms is a scalar multiple of a matrix. This requires using the properties of determinants related to scalar multiplication and matrix products.


Step 2: Key Formula or Approach:

We will use the following two properties of determinants for square matrices of order 'n':
1. \( \det(kM) = k^n \det(M) \), where k is a scalar.
2. \( \det(MN) = \det(M) \det(N) \).


Step 3: Detailed Explanation:

We are asked to find the value of \( \det(-6AB) \).

The matrices A and B are of order 3, so \( n=3 \). The scalar is \( k=-6 \).

Using the first property, we can take the scalar \( -6 \) out of the determinant: \[ \det(-6AB) = (-6)^3 \det(AB) \]
Now, we calculate \( (-6)^3 \): \[ (-6)^3 = -216 \]
So the expression becomes: \[ \det(-6AB) = -216 \cdot \det(AB) \]
Next, we use the second property, \( \det(AB) = \det(A) \det(B) \): \[ \det(-6AB) = -216 \cdot (\det(A) \cdot \det(B)) \]
We are given the values \( \det(A) = 3 \) and \( \det(B) = -4 \). Substituting these values: \[ \det(-6AB) = -216 \cdot (3 \cdot (-4)) \] \[ \det(-6AB) = -216 \cdot (-12) \]
Now, we perform the final multiplication: \[ \det(-6AB) = 2592 \]

Step 4: Final Answer:

The value of \( \det(-6AB) \) is 2592.
Quick Tip: A common mistake is forgetting to raise the scalar to the power of the matrix order 'n'. Always remember that \( \det(kA) = k^n \det(A) \), not \( k \det(A) \). Pay close attention to the order of the matrices given in the problem.

Step 1: Understanding the Concept:

For a differentiable function \( f(x) \) to be increasing on a given interval, its first derivative, \( f'(x) \), must be greater than or equal to zero (\( f'(x) \ge 0 \)) for all x in that interval.


Step 2: Detailed Explanation:

The given function is \( f(x) = 2x^2 - ax + 3 \).

First, we find the derivative of \( f(x) \) with respect to x: \[ f'(x) = \frac{d}{dx}(2x^2 - ax + 3) = 4x - a \]
For \( f(x) \) to be an increasing function on the closed interval [2, 4], we must have \( f'(x) \ge 0 \) for all \( x \in [2, 4] \). \[ 4x - a \ge 0 \]
This inequality can be rearranged to find the condition on 'a': \[ a \le 4x \quad for all x \in [2, 4] \]
This means that 'a' must be less than or equal to every value that \( 4x \) takes in the interval [2, 4]. To satisfy this for the entire interval, 'a' must be less than or equal to the minimum value of \( 4x \) on this interval.

The expression \( g(x) = 4x \) is a simple linear function that increases as x increases. Therefore, its minimum value on the interval [2, 4] will occur at the smallest value of x, which is \( x=2 \).
Minimum value of \( 4x \) on [2, 4] is \( 4(2) = 8 \).

So, the condition on 'a' becomes: \[ a \le 8 \]
The set of all possible values for 'a' is the interval \( (-\infty, 8] \).


Step 3: Final Answer:

The question asks for the "least value of 'a'". The interval \( (-\infty, 8] \) does not have a least value (it extends to negative infinity). This suggests a possible typo in the question, which might have intended to ask for the "greatest value of 'a'". If that were the case, the answer would be 8. Based on the phrasing, the condition derived is \( a \le 8 \).
Quick Tip: When a condition like \( a \le g(x) \) must hold for an entire interval, it means 'a' must be less than or equal to the absolute minimum of \( g(x) \) on that interval. Conversely, if \( a \ge g(x) \) must hold, 'a' must be greater than or equal to the absolute maximum of \( g(x) \).

Step 1: Understanding the Concept:

To show that a function is increasing on an interval, we need to prove that its first derivative is non-negative (\( \ge 0 \)) for all values of x in that interval.


Step 2: Detailed Explanation:

The given function is: \[ f(x) = x + \frac{1}{x} \quad for x \ge 1 \]
First, we find the derivative of \( f(x) \) with respect to x. We can write \( f(x) = x + x^{-1} \). \[ f'(x) = \frac{d}{dx}(x + x^{-1}) = 1 - 1 \cdot x^{-2} = 1 - \frac{1}{x^2} \]
To simplify, we can write the derivative as a single fraction: \[ f'(x) = \frac{x^2 - 1}{x^2} \]
Now, we need to determine the sign of \( f'(x) \) for the given domain, \( x \ge 1 \).
Let's analyze the numerator and the denominator separately.

Denominator: The denominator is \( x^2 \). For any \( x \ge 1 \), \( x \) is positive, so \( x^2 \) will also be positive. Thus, \( x^2 > 0 \).

Numerator: The numerator is \( x^2 - 1 \). For the domain \( x \ge 1 \), we have \( x^2 \ge 1^2 \), which means \( x^2 \ge 1 \). Subtracting 1 from both sides gives \( x^2 - 1 \ge 0 \). So, the numerator is non-negative.

Since the numerator is non-negative (\( \ge 0 \)) and the denominator is positive (\( > 0 \)) for all \( x \ge 1 \), their quotient will be non-negative. \[ f'(x) = \frac{x^2 - 1}{x^2} \ge 0 \quad for all x \ge 1 \]

Step 3: Final Answer:

Since the first derivative \( f'(x) \) is greater than or equal to zero for all \( x \ge 1 \), the function \( f(x) = x + \frac{1}{x} \) is an increasing function on the interval \( [1, \infty) \).
Quick Tip: For showing monotonicity (increasing/decreasing), always analyze the sign of the derivative. It's often helpful to express the derivative as a single fraction and then analyze the sign of the numerator and denominator separately over the given domain.

Step 1: Understanding the Concept:

The expression inside the inverse sine function, \( \frac{x}{\sqrt{1+x^2}} \), resembles a trigonometric ratio. We can simplify this expression by making a suitable trigonometric substitution for x. The form \( \sqrt{1+x^2} \) suggests using a tangent substitution.


Step 2: Detailed Explanation:

Let \( x = \tan(\theta) \). This implies \( \theta = \tan^{-1}(x) \).

The principal value range for the inverse tangent function is \( -\frac{\pi}{2} < \theta < \frac{\pi}{2} \).

Now, substitute \( x = \tan(\theta) \) into the given expression: \[ \sin^{-1}\left(\frac{\tan(\theta)}{\sqrt{1+\tan^2(\theta)}}\right) \]
Using the trigonometric identity \( 1 + \tan^2(\theta) = \sec^2(\theta) \), the expression becomes: \[ \sin^{-1}\left(\frac{\tan(\theta)}{\sqrt{\sec^2(\theta)}}\right) = \sin^{-1}\left(\frac{\tan(\theta)}{|\sec(\theta)|}\right) \]
We need to consider the sign of \( \sec(\theta) \). Since \( \theta \) is in the interval \( (-\frac{\pi}{2}, \frac{\pi}{2}) \), \( \cos(\theta) \) is positive. As \( \sec(\theta) = 1/\cos(\theta) \), \( \sec(\theta) \) is also positive in this interval.

Therefore, \( |\sec(\theta)| = \sec(\theta) \).

The expression simplifies to: \[ \sin^{-1}\left(\frac{\tan(\theta)}{\sec(\theta)}\right) \]
Now, let's simplify the trigonometric part: \[ \frac{\tan(\theta)}{\sec(\theta)} = \frac{\sin(\theta)/\cos(\theta)}{1/\cos(\theta)} = \sin(\theta) \]
Substituting this back, we get: \[ \sin^{-1}(\sin(\theta)) \]
Since \( \theta \) lies in the interval \( (-\frac{\pi}{2}, \frac{\pi}{2}) \), which is the principal value range of the \( \sin^{-1} \) function, we can simplify \( \sin^{-1}(\sin(\theta)) \) to \( \theta \).

Finally, we substitute back the value of \( \theta \): \[ \theta = \tan^{-1}(x) \]

Step 3: Final Answer:

The simplified form of \( \sin^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right) \) is \( \tan^{-1}(x) \).
Quick Tip: Remember these common trigonometric substitutions for simplifying expressions involving inverse trigonometric functions: - If you see \( \sqrt{a^2 - x^2} \), try \( x = a\sin(\theta) \) or \( x = a\cos(\theta) \). - If you see \( \sqrt{a^2 + x^2} \), try \( x = a\tan(\theta) \). - If you see \( \sqrt{x^2 - a^2} \), try \( x = a\sec(\theta) \).

Step 1: Understanding the Concept:

To find the domain of a composite function like this, we must satisfy two conditions:
1. The expression inside the square root must be non-negative.
2. The argument of the inverse sine function must be within its domain, which is [-1, 1].


Step 2: Detailed Explanation:

Let the function be \( f(x) = \sin^{-1}\sqrt{x-1} \).


Condition 1: The argument of the square root must be non-negative.

For \( \sqrt{x-1} \) to be defined as a real number, the expression inside the root must be greater than or equal to zero. \[ x-1 \ge 0 \] \[ x \ge 1 \]

Condition 2: The argument of the inverse sine function must be in the interval [-1, 1].

The domain of \( \sin^{-1}(u) \) is \( -1 \le u \le 1 \). In our case, \( u = \sqrt{x-1} \).
So, we must have: \[ -1 \le \sqrt{x-1} \le 1 \]
This inequality can be split into two parts:
Part (i): \( \sqrt{x-1} \ge -1 \)

The square root of a real number, by definition, is always non-negative. Therefore, \( \sqrt{x-1} \ge 0 \). Since any non-negative number is always greater than -1, this part is true for all x for which the square root is defined (i.e., for \( x \ge 1 \)).

Part (ii): \( \sqrt{x-1} \le 1 \)

Since both sides of the inequality are non-negative, we can square both sides without changing the direction of the inequality sign. \[ (\sqrt{x-1})^2 \le 1^2 \] \[ x-1 \le 1 \] \[ x \le 2 \]

Combining the conditions:

From Condition 1, we have \( x \ge 1 \).

From Condition 2, we have \( x \le 2 \).

To find the overall domain, we must find the values of x that satisfy both conditions simultaneously. This is the intersection of the two sets of values. \[ x \ge 1 \quad and \quad x \le 2 \]
This can be written as \( 1 \le x \le 2 \).


Step 3: Final Answer:

The domain of the function \( \sin^{-1}\sqrt{x-1} \) is the closed interval [1, 2].
Quick Tip: When finding the domain of a composite function, work from the "outside in" to identify all constraints. For \( \sin^{-1}(\sqrt{g(x)}) \), the constraints are always \( g(x) \ge 0 \) (for the square root) and \( 0 \le \sqrt{g(x)} \le 1 \) (for the arcsin, simplified because square root is non-negative). This simplifies to \( 0 \le g(x) \le 1 \).

Step 1: Understanding the Concept:

The given equation represents an ellipse centered at the origin (0, 0) with a semi-major axis of \(a=3\) along the x-axis and a semi-minor axis of \(b=2\) along the y-axis. The region bounded by the ellipse and the x-axis corresponds to the upper half of the ellipse. The area can be found by integrating the function representing the upper semi-ellipse from one x-intercept to the other.


Step 2: Key Formula or Approach:

The area \(A\) under a curve \(y = f(x)\) from \(x = x_1\) to \(x = x_2\) is given by the definite integral:
\[ A = \int_{x_1}^{x_2} y \, dx \]
We will also use the standard integral formula:
\[ \int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) + C \]

Step 3: Detailed Explanation:

First, we express \(y\) in terms of \(x\) from the equation of the ellipse:
\[ \frac{x^2}{9} + \frac{y^2}{4} = 1 \] \[ \frac{y^2}{4} = 1 - \frac{x^2}{9} = \frac{9 - x^2}{9} \] \[ y^2 = \frac{4}{9}(9 - x^2) \]
Taking the square root, we get:
\[ y = \pm \frac{2}{3}\sqrt{9 - x^2} \]
Since we are interested in the area bounded by the curve and the x-axis, we consider the upper part of the ellipse, where \(y \geq 0\).
\[ y = \frac{2}{3}\sqrt{9 - x^2} \]
The ellipse intersects the x-axis when \(y=0\), which gives \(x^2/9 = 1\), so \(x = \pm 3\). Thus, our limits of integration are from -3 to 3.

The required area is:
\[ A = \int_{-3}^{3} \frac{2}{3}\sqrt{9 - x^2} \, dx \]
Due to the symmetry of the ellipse about the y-axis, we can calculate the area in the first quadrant (from 0 to 3) and multiply it by 2.
\[ A = 2 \int_{0}^{3} \frac{2}{3}\sqrt{9 - x^2} \, dx = \frac{4}{3} \int_{0}^{3} \sqrt{3^2 - x^2} \, dx \]
Using the standard integral formula with \(a=3\):
\[ A = \frac{4}{3} \left[ \frac{x}{2}\sqrt{9 - x^2} + \frac{9}{2}\sin^{-1}\left(\frac{x}{3}\right) \right]_{0}^{3} \]
Now, we evaluate the expression at the limits of integration.

At the upper limit, \(x=3\):
\[ \frac{3}{2}\sqrt{9 - 3^2} + \frac{9}{2}\sin^{-1}\left(\frac{3}{3}\right) = \frac{3}{2}\sqrt{0} + \frac{9}{2}\sin^{-1}(1) = 0 + \frac{9}{2}\left(\frac{\pi}{2}\right) = \frac{9\pi}{4} \]
At the lower limit, \(x=0\):
\[ \frac{0}{2}\sqrt{9 - 0^2} + \frac{9}{2}\sin^{-1}\left(\frac{0}{3}\right) = 0 + \frac{9}{2}\sin^{-1}(0) = 0 + 0 = 0 \]
So, the area is:
\[ A = \frac{4}{3} \left( \frac{9\pi}{4} - 0 \right) = \frac{4}{3} \cdot \frac{9\pi}{4} = 3\pi \]

Step 4: Final Answer:

The area of the region is \( 3\pi \) square units.
Quick Tip: For an ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), the total area is \( \pi ab \). The area of the region bounded by the ellipse and the x-axis will be half of the total area, which is \( \frac{1}{2}\pi ab \). In this case, \(a=3\) and \(b=2\), so the area is \( \frac{1}{2}\pi(3)(2) = 3\pi \). This is a great way to quickly verify your result from integration.

Step 1: Understanding the Concept:

This is a problem of related rates. We are given the equation of a curve and the rate at which the x-coordinate is changing. We need to find the rate at which the slope of the curve is changing at a specific point. The slope itself is a function of x, and its rate of change will depend on the rate of change of x.


Step 2: Key Formula or Approach:

1. Find the slope of the curve, \( m \), by differentiating \( y \) with respect to \( x \). So, \( m = \frac{dy}{dx} \).

2. Find the rate of change of the slope, \( \frac{dm}{dt} \), by differentiating \( m \) with respect to time \( t \), using the chain rule.

3. Substitute the given values of \( x \) and \( \frac{dx}{dt} \) to find the required rate.


Step 3: Detailed Explanation:

The equation of the curve is given by:
\[ y = 5x - 2x^3 \]
First, find the slope (m) of the curve. The slope is the first derivative of y with respect to x.
\[ m = \frac{dy}{dx} = \frac{d}{dx}(5x - 2x^3) \] \[ m = 5 - 6x^2 \]
Next, find the rate at which the slope is changing. We need to find \( \frac{dm}{dt} \). We differentiate the expression for \( m \) with respect to time \( t \).
\[ \frac{dm}{dt} = \frac{d}{dt}(5 - 6x^2) \]
Since 5 is a constant, its derivative is 0. For the term \( -6x^2 \), we use the chain rule because \( x \) is a function of \( t \).
\[ \frac{dm}{dt} = 0 - 6 \cdot (2x) \cdot \frac{dx}{dt} \] \[ \frac{dm}{dt} = -12x \frac{dx}{dt} \]
Finally, substitute the given values. We are given that \( x \) increases at a rate of 2 units/s, so \( \frac{dx}{dt} = 2 \) units/s. We need to find the rate of change of the slope when \( x = 2 \).
\[ \frac{dm}{dt} \bigg|_{x=2} = -12(2) \cdot (2) \] \[ \frac{dm}{dt} = -48 \]
The unit of the slope is unitless, so the rate of change of the slope is in units per second (units/s). The negative sign indicates that the slope is decreasing.


Step 4: Final Answer:

The slope of the curve is changing at a rate of -48 units/s, which means it is decreasing at 48 units/s.
Quick Tip: In related rates problems, it's crucial to identify what quantity's rate is given and what quantity's rate is required. Here, the rate of change of 'x' is given, and the rate of change of the 'slope' is required. Do not confuse the rate of change of 'y' (\( \frac{dy}{dt} \)) with the rate of change of the 'slope' (\( \frac{dm}{dt} \)).

Step 1: Understanding the Concept:

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

- Injective (One-to-one): A function \(f\) is injective if for every \(x_1, x_2\) in the domain, \(f(x_1) = f(x_2)\) implies \(x_1 = x_2\). In other words, different inputs produce different outputs.

- Surjective (Onto): A function \(f: A \to B\) is surjective if for every element \(y\) in the codomain \(B\), there exists at least one element \(x\) in the domain \(A\) such that \(f(x) = y\). In other words, the range of the function is equal to its codomain.


Step 2: Detailed Explanation:

The given function is \( f(x) = \log_a x \), with domain \( \mathbf{R}^+ \) (positive real numbers) and codomain \( \mathbf{R} \) (all real numbers).


Part 1: Proving f is Injective (One-to-one)

Let \(x_1, x_2\) be any two elements in the domain \( \mathbf{R}^+ \).

Assume that \( f(x_1) = f(x_2) \).
\[ \log_a x_1 = \log_a x_2 \]
By the fundamental property of logarithmic functions (which are one-to-one by definition), if the logarithms of two numbers to the same base are equal, the numbers themselves must be equal.

Therefore, \( x_1 = x_2 \).

Since \(f(x_1) = f(x_2)\) implies \(x_1 = x_2\), the function \(f(x) = \log_a x\) is injective.


Part 2: Proving f is Surjective (Onto)

Let \( y \) be an arbitrary element in the codomain \( \mathbf{R} \). We need to show that there exists an element \( x \) in the domain \( \mathbf{R}^+ \) such that \( f(x) = y \).
\[ f(x) = y \] \[ \log_a x = y \]
To find \(x\), we can rewrite this logarithmic equation in its equivalent exponential form:
\[ x = a^y \]
Now, we must check if this value of \(x\) is in the domain \( \mathbf{R}^+ \).

Since we are given that \( a > 0 \), for any real number \( y \), the value of \( a^y \) will always be a positive real number.

So, \( x = a^y \in \mathbf{R}^+ \).

Thus, for any \( y \in \mathbf{R} \), there exists an \( x = a^y \in \mathbf{R}^+ \) such that \( f(x) = f(a^y) = \log_a(a^y) = y \).

Therefore, the function \(f(x) = \log_a x\) is surjective.


Step 3: Final Answer:

Since the function \( f(x) = \log_a x \) is both injective and surjective, it is a bijection.
Quick Tip: The proof of bijectivity for standard functions like logarithms, exponentials, and linear functions often relies on their fundamental properties. For surjectivity, the key is to take an arbitrary element \(y\) from the codomain and explicitly find the corresponding \(x\) in the domain. For logarithms, this involves converting from log form to exponential form.

Step 1: Understanding the Concept:

A relation R from a set A to a set B is a subset of the Cartesian product A × B. A function is a special type of relation where every element in the domain (set A) is associated with exactly one element in the codomain (set B). The domain of a relation is the set of all first components of the ordered pairs, and the range is the set of all second components.


Step 2: Detailed Explanation:

We are given the sets A = \{1, 2, 3\ and B = \{4, 5, 6\.

The relation is defined by the rule \( x + y = 6 \), where \( x \in A \) and \( y \in B \).


(i) Write all elements of R.

We need to find all pairs (x, y) that satisfy the condition. We check each element of A:

- If \( x = 1 \): \( 1 + y = 6 \implies y = 5 \). Since \( 5 \in B \), the pair (1, 5) is in R.

- If \( x = 2 \): \( 2 + y = 6 \implies y = 4 \). Since \( 4 \in B \), the pair (2, 4) is in R.

- If \( x = 3 \): \( 3 + y = 6 \implies y = 3 \). Since \( 3 \notin B \), there is no pair in R with x = 3.

Therefore, the set of elements in R is {(1, 5), (2, 4)}.


(ii) Is R a function? Justify.

For R to be a function from A to B, every element of A must have exactly one image in B.

- The element 1 in A has one image, which is 5.

- The element 2 in A has one image, which is 4.

- The element 3 in A has no image in B under the relation R.

Since not every element of the set A is mapped to an element in set B, the relation R is not a function from A to B.

Justification: The domain of the relation R is \{1, 2\, which is not equal to the set A = \{1, 2, 3\. For a relation to be a function from A to B, its domain must be exactly A.


(iii) Determine domain and range of R.

- Domain of R: The set of all the first elements (x-values) in the ordered pairs of R.

The pairs are (1, 5) and (2, 4). The first elements are 1 and 2.

So, Domain(R) = {1, 2}.

- Range of R: The set of all the second elements (y-values) in the ordered pairs of R.

The pairs are (1, 5) and (2, 4). The second elements are 5 and 4.

So, Range(R) = {4, 5}.
Quick Tip: A common mistake is to confuse the definition of a relation with that of a function. A relation can connect any number of elements from the domain to the codomain (including none). A function has stricter rules: every single element in the domain must be connected to exactly one element in the codomain. Always check if all elements of the starting set (A) are used.

Step 1: Understanding the Concept:

For a function \( f(x) \) to be continuous at a point \( x = c \), three conditions must be met:

1. \( f(c) \) is defined.

2. The limit \( \lim_{x \to c} f(x) \) exists.

3. The limit equals the function's value: \( \lim_{x \to c} f(x) = f(c) \).

In this problem, we need to find the value of \(k\) that satisfies this third condition at \(c = -1\).


Step 2: Key Formula or Approach:

We will set the limit of the function as \(x\) approaches -1 equal to the value of the function at \(x=-1\).
\[ \lim_{x \to -1} \frac{x^2-2x-3}{x+1} = f(-1) \]

Step 3: Detailed Explanation:

From the definition of the function, the value at \(x = -1\) is:
\[ f(-1) = k \]
Now, we need to find the limit of \( f(x) \) as \( x \to -1 \). For the limit, we consider values of \(x\) close to -1 but not equal to -1, so we use the first expression:
\[ \lim_{x \to -1} f(x) = \lim_{x \to -1} \frac{x^2-2x-3}{x+1} \]
Substituting \(x = -1\) into the expression gives \( \frac{(-1)^2 - 2(-1) - 3}{-1 + 1} = \frac{1 + 2 - 3}{0} = \frac{0}{0} \), which is an indeterminate form. This indicates that we can simplify the expression.

We factor the numerator polynomial \( x^2 - 2x - 3 \):
\[ x^2 - 3x + x - 3 = x(x-3) + 1(x-3) = (x-3)(x+1) \]
Now, substitute the factored form back into the limit:
\[ \lim_{x \to -1} \frac{(x-3)(x+1)}{x+1} \]
Since \( x \to -1 \), \( x \neq -1 \), so we can cancel the \( (x+1) \) term from the numerator and denominator.
\[ \lim_{x \to -1} (x-3) \]
Now, we can substitute \( x = -1 \) into the simplified expression:
\[ -1 - 3 = -4 \]
For the function to be continuous at \( x = -1 \), the limit must equal \( f(-1) \).
\[ \lim_{x \to -1} f(x) = f(-1) \] \[ -4 = k \]

Step 4: Final Answer:

The value of \( k \) that makes the function continuous at \( x = -1 \) is -4.
Quick Tip: When evaluating limits for piecewise functions to check for continuity, if you encounter the indeterminate form 0/0, it's a strong hint that the numerator and denominator share a common factor. Factoring and canceling this common factor is the standard method to resolve the limit.

Step 1: Understanding the Concept:

A function \(f(x)\) is differentiable at a point \(x = c\) if its derivative exists at that point. This means that the Left-Hand Derivative (LHD) and the Right-Hand Derivative (RHD) at \(x = c\) both exist and are equal.

The LHD is defined as: \( Lf'(c) = \lim_{h \to 0^-} \frac{f(c+h) - f(c)}{h} \).

The RHD is defined as: \( Rf'(c) = \lim_{h \to 0^+} \frac{f(c+h) - f(c)}{h} \).


Step 2: Key Formula or Approach:

1. First, write \( f(x) = x|x| \) as a piecewise function to remove the absolute value.

2. Calculate the Left-Hand Derivative (LHD) at \(x = 0\).

3. Calculate the Right-Hand Derivative (RHD) at \(x = 0\).

4. Compare the LHD and RHD. If they are equal, the function is differentiable at \(x=0\). Otherwise, it is not.


Step 3: Detailed Explanation:

Step 1: Express f(x) as a piecewise function.

The definition of \(|x|\) is:
\( |x| = x \) if \( x \geq 0 \)
\( |x| = -x \) if \( x < 0 \)

So, the function \( f(x) = x|x| \) can be written as:
\[ f(x) = \begin{cases} x(x) = x^2, & if x \geq 0
x(-x) = -x^2, & if x < 0 \end{cases} \]
Also, the value of the function at \(x=0\) is \(f(0) = 0|0| = 0\).


Step 2: Calculate the Left-Hand Derivative (LHD) at x = 0.
\[ LHD = \lim_{h \to 0^-} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^-} \frac{f(h) - 0}{h} \]
Since \( h \to 0^- \), \(h\) is a small negative number (\(h < 0\)). So we use the definition \(f(h) = -h^2\).
\[ LHD = \lim_{h \to 0^-} \frac{-h^2}{h} = \lim_{h \to 0^-} (-h) \]
Substituting \(h=0\), we get:
\[ LHD = 0 \]

Step 3: Calculate the Right-Hand Derivative (RHD) at x = 0.
\[ RHD = \lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^+} \frac{f(h) - 0}{h} \]
Since \( h \to 0^+ \), \(h\) is a small positive number (\(h > 0\)). So we use the definition \(f(h) = h^2\).
\[ RHD = \lim_{h \to 0^+} \frac{h^2}{h} = \lim_{h \to 0^+} (h) \]
Substituting \(h=0\), we get:
\[ RHD = 0 \]

Step 4: Compare LHD and RHD.

We found that LHD = 0 and RHD = 0.

Since LHD = RHD, the function \(f(x) = x|x|\) is differentiable at \(x = 0\). The value of the derivative is \(f'(0) = 0\).


Step 5: Final Answer:

The function \( f(x) = x|x| \) is differentiable at \( x = 0 \).
Quick Tip: When checking for differentiability of functions involving absolute values like \(|x-c|\), it's always best to first rewrite the function as a piecewise function by splitting the definition at the point where the argument of the absolute value is zero (here, at x=0). This makes calculating the LHD and RHD much simpler.

Step 1: Understanding the Concept:

The integral is in a form that suggests using the property \( \int e^x (f(x) + f'(x)) dx = e^x f(x) + C \). To apply this, we need to manipulate the trigonometric expression \( \frac{1-\sin x}{1-\cos x} \) into the form \( f(x) + f'(x) \).


Step 2: Key Formula or Approach:

We will use the half-angle trigonometric identities to simplify the integrand: \[ 1 - \cos x = 2\sin^2\left(\frac{x}{2}\right) \] \[ \sin x = 2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right) \]

Step 3: Detailed Explanation:

Let \( I = \int_{\pi/2}^{\pi} e^x \left(\frac{1-\sin x}{1-\cos x}\right) dx \).

First, we simplify the term inside the parenthesis: \[ \frac{1-\sin x}{1-\cos x} = \frac{1 - 2\sin(x/2)\cos(x/2)}{2\sin^2(x/2)} \]
Now, split the fraction into two parts: \[ = \frac{1}{2\sin^2(x/2)} - \frac{2\sin(x/2)\cos(x/2)}{2\sin^2(x/2)} \] \[ = \frac{1}{2}\csc^2\left(\frac{x}{2}\right) - \cot\left(\frac{x}{2}\right) \]
Let's rearrange the terms: \[ = -\cot\left(\frac{x}{2}\right) + \frac{1}{2}\csc^2\left(\frac{x}{2}\right) \]
Now, let's check if this expression is in the form \( f(x) + f'(x) \).

Let \( f(x) = -\cot\left(\frac{x}{2}\right) \).

We find the derivative, \( f'(x) \): \[ f'(x) = \frac{d}{dx}\left(-\cot\left(\frac{x}{2}\right)\right) \]
Using the chain rule, the derivative of \( \cot(u) \) is \( -\csc^2(u) \cdot u' \). \[ f'(x) = -\left(-\csc^2\left(\frac{x}{2}\right) \cdot \frac{d}{dx}\left(\frac{x}{2}\right)\right) \] \[ f'(x) = \csc^2\left(\frac{x}{2}\right) \cdot \frac{1}{2} = \frac{1}{2}\csc^2\left(\frac{x}{2}\right) \]
We see that our expression is exactly \( f(x) + f'(x) \).

So, the integral becomes: \[ I = \int_{\pi/2}^{\pi} e^x \left(-\cot\left(\frac{x}{2}\right) + \frac{1}{2}\csc^2\left(\frac{x}{2}\right)\right) dx \]
Using the property \( \int e^x (f(x) + f'(x)) dx = e^x f(x) \), the integral evaluates to: \[ I = \left[ e^x \cdot \left(-\cot\left(\frac{x}{2}\right)\right) \right]_{\pi/2}^{\pi} \]
Now, we apply the limits of integration: \[ I = \left(-e^{\pi} \cot\left(\frac{\pi}{2}\right)\right) - \left(-e^{\pi/2} \cot\left(\frac{\pi/2}{2}\right)\right) \] \[ I = -e^{\pi} \cot\left(\frac{\pi}{2}\right) + e^{\pi/2} \cot\left(\frac{\pi}{4}\right) \]
We know that \( \cot(\pi/2) = 0 \) and \( \cot(\pi/4) = 1 \). \[ I = -e^{\pi}(0) + e^{\pi/2}(1) \] \[ I = 0 + e^{\pi/2} = e^{\pi/2} \]

Step 4: Final Answer:

The value of the integral is \( e^{\pi/2} \).
Quick Tip: Whenever you see an integral involving \( e^x \) multiplied by a complex function, immediately try to check if the function can be expressed in the form \( f(x) + f'(x) \). Using half-angle formulas is a very common technique for this manipulation with trigonometric functions.

Step 1: Understanding the Concept:

We need to find the probability for each possible number of boys in a family of three children. A probability distribution lists all possible outcomes (values of the random variable) and their corresponding probabilities. The random variable, let's call it X, is the "number of boys".


Step 2: Key Formula or Approach:

The probability of a boy (B) and a girl (G) is equal. \[ P(B) = \frac{1}{2} \quad and \quad P(G) = \frac{1}{2} \]
There are three children, so the total number of possible outcomes in the sample space is \( 2 \times 2 \times 2 = 8 \).
The number of boys, X, can take values {0, 1, 2, 3.


Step 3: Detailed Explanation:

Let's list the entire sample space S, where the order of children matters (e.g., BGB is different from BBG). \[ S = \{BBB, BBG, BGB, GBB, BGG, GBG, GGB, GGG\} \]
Each of these 8 outcomes is equally likely, with a probability of \( (\frac{1}{2})^3 = \frac{1}{8} \).


Now we calculate the probability for each value of X.

Case 1: X = 0 (No boys)

This corresponds to the outcome {GGG. There is only 1 such outcome. \[ P(X=0) = \frac{Number of outcomes with 0 boys}{Total number of outcomes} = \frac{1}{8} \]
Case 2: X = 1 (One boy)

This corresponds to the outcomes {BGG, GBG, GGB. There are 3 such outcomes. \[ P(X=1) = \frac{3}{8} \]
Case 3: X = 2 (Two boys)

This corresponds to the outcomes {BBG, BGB, GBB. There are 3 such outcomes. \[ P(X=2) = \frac{3}{8} \]
Case 4: X = 3 (Three boys)

This corresponds to the outcome {BBB. There is only 1 such outcome. \[ P(X=3) = \frac{1}{8} \]

Step 4: Final Answer:

The probability distribution of the number of boys (X) is:

\begin{tabular{|c|c|c|c|c|
\hline
X (Number of Boys) & 0 & 1 & 2 & 3

\hline
P(X) & \( \frac{1}{8} \) & \( \frac{3}{8} \) & \( \frac{3}{8} \) & \( \frac{1}{8} \)

\hline
\end{tabular

We can check that the sum of probabilities is \( \frac{1}{8} + \frac{3}{8} + \frac{3}{8} + \frac{1}{8} = \frac{8}{8} = 1 \).
Quick Tip: This is an example of a binomial distribution with \( n=3 \) trials (children) and probability of success \( p=1/2 \) (a child being a boy). The formula \( P(X=k) = \binom{n}{k}p^k(1-p)^{n-k} \) can be used to quickly find the probabilities without listing the sample space, which is especially useful for a larger number of trials.

Step 1: Understanding the Concept:

We first need to identify all possible outcomes of tossing a coin twice. Then, for each outcome, we calculate the value of the random variable X. Finally, we group the outcomes by the value of X to create the probability distribution and then calculate the mean (expected value).


Step 2: Key Formula or Approach:

The sample space for tossing a coin twice is S = {HH, HT, TH, TT.
The random variable is \( X = (Number of Heads) - (Number of Tails) \).
The mean (expected value) of a discrete random variable X is given by the formula \( E(X) = \sum x_i P(X=x_i) \).


Step 3: Detailed Explanation:

Part 1: Finding the Probability Distribution

The probability of each outcome in the sample space is \( \frac{1}{4} \).
Let's calculate X for each outcome:

For HH: Number of Heads = 2, Number of Tails = 0. \( \Rightarrow X = 2 - 0 = 2 \).
For HT: Number of Heads = 1, Number of Tails = 1. \( \Rightarrow X = 1 - 1 = 0 \).
For TH: Number of Heads = 1, Number of Tails = 1. \( \Rightarrow X = 1 - 1 = 0 \).
For TT: Number of Heads = 0, Number of Tails = 2. \( \Rightarrow X = 0 - 2 = -2 \).

The possible values for the random variable X are \{-2, 0, 2\.

Now, we find the probability for each value of X:

\( P(X = -2) = P(TT) = \frac{1}{4} \).
\( P(X = 0) = P(HT or TH) = P(HT) + P(TH) = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2} \).
\( P(X = 2) = P(HH) = \frac{1}{4} \).

The probability distribution of X is:

\begin{tabular{|c|c|c|c|
\hline
X & -2 & 0 & 2

\hline
P(X) & \( \frac{1}{4} \) & \( \frac{1}{2} \) & \( \frac{1}{4} \)

\hline
\end{tabular

Part 2: Finding the Mean of X

Using the formula for the mean, \( E(X) \): \[ E(X) = \sum x_i P(X=x_i) \] \[ E(X) = (-2) \cdot P(X=-2) + (0) \cdot P(X=0) + (2) \cdot P(X=2) \] \[ E(X) = (-2) \cdot \left(\frac{1}{4}\right) + (0) \cdot \left(\frac{1}{2}\right) + (2) \cdot \left(\frac{1}{4}\right) \] \[ E(X) = -\frac{2}{4} + 0 + \frac{2}{4} \] \[ E(X) = 0 \]

Step 4: Final Answer:

The probability distribution is given in the table above, and the mean of X is 0.
Quick Tip: To create a probability distribution, always start with the sample space. Systematically calculate the value of the random variable for each outcome. Then, group the outcomes and sum their probabilities for each unique value of the random variable. The mean is a weighted average of the values, where the weights are the probabilities.

Step 1: Understanding the Concept:

The problem requires a two-step process. First, we must find the coordinates of the point where the two given lines intersect. Second, we use the distance formula in 3D to find the distance between this intersection point and the given point (-1, -5, -10).


Step 2: Finding the Point of Intersection:

Let the first line be L1 and the second line be L2. We can write the coordinates of any point on these lines in parametric form.

For L1: \( \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4} = \lambda \) (say)
A general point on L1 is \( P_1(2\lambda + 1, 3\lambda + 2, 4\lambda + 3) \).


For L2: \( \frac{x-4}{5} = \frac{y-1}{2} = \frac{z-0}{1} = \mu \) (say)
A general point on L2 is \( P_2(5\mu + 4, 2\mu + 1, \mu) \).


If the lines intersect, there must be a point that is on both lines. Therefore, for some values of \( \lambda \) and \( \mu \), the coordinates of \( P_1 \) and \( P_2 \) must be identical.
Equating the coordinates:
1) \( 2\lambda + 1 = 5\mu + 4 \implies 2\lambda - 5\mu = 3 \)
2) \( 3\lambda + 2 = 2\mu + 1 \implies 3\lambda - 2\mu = -1 \)
3) \( 4\lambda + 3 = \mu \)

We can solve this system of linear equations. Let's substitute equation (3) into equation (2): \[ 3\lambda - 2(4\lambda + 3) = -1 \] \[ 3\lambda - 8\lambda - 6 = -1 \] \[ -5\lambda = 5 \implies \lambda = -1 \]
Now, substitute \( \lambda = -1 \) back into equation (3) to find \( \mu \): \[ \mu = 4(-1) + 3 = -1 \]
We should check if these values satisfy equation (1): \[ 2(-1) - 5(-1) = -2 + 5 = 3 \). This is correct.
Since the values of \( \lambda \) and \( \mu \) are consistent across all three equations, the lines intersect.
To find the intersection point, substitute \( \lambda = -1 \) into the coordinates for L1 (or \( \mu = -1 \) into L2): Intersection point Q: \( (2(-1)+1, 3(-1)+2, 4(-1)+3) = (-1, -1, -1) \).
\textbf{Step 3: Calculating the Distance:}
Now we need to find the distance between the given point P(-1, -5, -10) and the intersection point Q(-1, -1, -1). Using the 3D distance formula, \( d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2} \): \[ d = \sqrt{(-1 - (-1))^2 + (-1 - (-5))^2 + (-1 - (-10))^2} \] \[ d = \sqrt{(0)^2 + (4)^2 + (9)^2} \] \[ d = \sqrt{0 + 16 + 81} \] \[ d = \sqrt{97} \]

Step 4: Final Answer:

The distance is \( \sqrt{97} \) units.
Quick Tip: To find the intersection of two lines in 3D, always write their general points using two different parameters (e.g., \( \lambda \) and \( \mu \)). Equating the coordinates will give you a system of three equations. Solve any two to find the parameters, and then substitute into the third equation to verify that they intersect. If the third equation is not satisfied, the lines are skew.

Step 1: Understanding the Concept:

We need to find the values of x and y that maximize the objective function Z, while satisfying all the given linear inequalities (constraints). The graphical method involves plotting these constraints to identify a feasible region and then testing the corner points of this region in the objective function.


Step 2: Plotting the Constraints:

First, we treat each inequality as an equation to plot the boundary lines.
1. \( 3x + y = 600 \): Passes through (200, 0) and (0, 600).
2. \( x + y = 300 \): Passes through (300, 0) and (0, 300).
3. \( y = x + 200 \) (or \( -x + y = 200 \)): Passes through (0, 200) and intersects the x-axis at (-200, 0).
4. \( x \ge 0, y \ge 0 \): This restricts the solution to the first quadrant.

The inequalities \( \le \) mean that the feasible region will be below or on these lines.


Step 3: Identifying the Feasible Region and Corner Points:

The feasible region is the area in the first quadrant that is simultaneously below lines (1), (2), and (3). This forms a polygon. We need to find the coordinates of its vertices (corner points).

Point O: The origin, (0, 0).
Point A: The x-intercept of \( 3x+y=600 \), which is (200, 0).
Point D: The y-intercept of \( y=x+200 \), which is (0, 200).
Point B: Intersection of \( 3x+y=600 \) and \( x+y=300 \).
Subtracting the second from the first: \( (3x+y) - (x+y) = 600-300 \implies 2x = 300 \implies x=150 \).
Substituting \( x=150 \) into \( x+y=300 \implies 150+y=300 \implies y=150 \). So, B = (150, 150).
Point C: Intersection of \( x+y=300 \) and \( y=x+200 \).
Substituting \( y=x+200 \) into the first equation: \( x + (x+200) = 300 \implies 2x = 100 \implies x=50 \).
Then \( y = 50+200=250 \). So, C = (50, 250).

The corner points of the feasible region are O(0,0), A(200,0), B(150,150), C(50,250), and D(0,200).


Step 4: Evaluating Z at Corner Points:

The optimal solution must occur at one of these corner points. We evaluate Z = 100x + 50y at each point.

Z at O(0, 0): \( 100(0) + 50(0) = 0 \)
Z at A(200, 0): \( 100(200) + 50(0) = 20000 \)
Z at B(150, 150): \( 100(150) + 50(150) = 15000 + 7500 = 22500 \)
Z at C(50, 250): \( 100(50) + 50(250) = 5000 + 12500 = 17500 \)
Z at D(0, 200): \( 100(0) + 50(200) = 10000 \)


Step 5: Final Answer:

Comparing the values of Z, the maximum value is 22500, which occurs at the point B(150, 150).
Thus, the optimal solution is \( x = 150, y = 150 \), with a maximum Z value of 22500.
Quick Tip: In LPP, after plotting the lines, use a test point like (0,0) to determine which side of the line represents the inequality. For example, for \( 3x+y \le 600 \), plugging in (0,0) gives \( 0 \le 600 \), which is true. This means the feasible region is on the origin side of the line \( 3x+y=600 \).

Part 1: Proof

Step 1: Understanding the Concept:

To prove that \( \frac{1}{k}A^{-1} \) is the inverse of \( (kA) \), we need to show that their product is the identity matrix, I. That is, \( (kA) \left(\frac{1}{k}A^{-1}\right) = I \).


Step 2: Detailed Proof:

Let's compute the product: \[ (kA) \left(\frac{1}{k}A^{-1}\right) \]
Using the property of scalar multiplication with matrices, we can rearrange the scalars: \[ = \left(k \cdot \frac{1}{k}\right) (A \cdot A^{-1}) \]
Since \( k \ne 0 \), \( k \cdot \frac{1}{k} = 1 \).
Since A is an invertible matrix, by definition, \( A \cdot A^{-1} = I \).
Substituting these results back: \[ = (1)(I) = I \]
Since \( (kA) \left(\frac{1}{k}A^{-1}\right) = I \), it is proven that \( (kA)^{-1} = \frac{1}{k}A^{-1} \).


Part 2: Calculation

Step 1: Applying the Proven Property:

We need to calculate \( (3A)^{-1} \). Using the property we just proved with \( k=3 \), we get: \[ (3A)^{-1} = \frac{1}{3}A^{-1} \]
So, the task reduces to finding the inverse of A, \( A^{-1} \).


Step 2: Finding the Inverse of A, \( A^{-1} \):

The formula for the inverse is \( A^{-1} = \frac{1}{\det(A)} adj(A) \).

a) Calculate the determinant of A: \[ \det(A) = 2(2 \cdot 2 - (-1)(-1)) - (-1)(-1 \cdot 2 - (-1)(1)) + 1((-1)(-1) - 2 \cdot 1) \] \[ \det(A) = 2(4 - 1) + 1(-2 + 1) + 1(1 - 2) \] \[ \det(A) = 2(3) - 1 - 1 = 6 - 2 = 4 \]
b) Find the adjugate of A, adj(A):

The adjugate is the transpose of the cofactor matrix. Let's find the cofactors: \[ C_{11} = +(4-1) = 3 \] \[ C_{12} = -(-2-(-1)) = -(-1) = 1 \] \[ C_{13} = +(1-2) = -1 \] \[ C_{21} = -(-2-(-1)) = -(-1) = 1 \] \[ C_{22} = +(4-1) = 3 \] \[ C_{23} = -(-2-(-1)) = -(-1) = 1 \] \[ C_{31} = +(1-2) = -1 \] \[ C_{32} = -(-2-(-1)) = -(-1) = 1 \] \[ C_{33} = +(4-1) = 3 \]
The cofactor matrix is \( C = \begin{pmatrix} 3 & 1 & -1
1 & 3 & 1
-1 & 1 & 3 \end{pmatrix} \).

The adjugate matrix is \( adj(A) = C^T = \begin{pmatrix} 3 & 1 & -1
1 & 3 & 1
-1 & 1 & 3 \end{pmatrix} \). (The matrix C is symmetric, so \( C^T=C \)).

c) Form the inverse \( A^{-1} \): \[ A^{-1} = \frac{1}{4} \begin{pmatrix} 3 & 1 & -1
1 & 3 & 1
-1 & 1 & 3 \end{pmatrix} \]
Step 3: Calculate \( (3A)^{-1} \):
\[ (3A)^{-1} = \frac{1}{3} A^{-1} = \frac{1}{3} \left( \frac{1}{4} \begin{pmatrix} 3 & 1 & -1
1 & 3 & 1
-1 & 1 & 3 \end{pmatrix} \right) \] \[ (3A)^{-1} = \frac{1}{12} \begin{pmatrix} 3 & 1 & -1
1 & 3 & 1
-1 & 1 & 3 \end{pmatrix} \]

Step 4: Final Answer:

The final result is \( (3A)^{-1} = \frac{1}{12} \begin{pmatrix} 3 & 1 & -1
1 & 3 & 1
-1 & 1 & 3 \end{pmatrix} \).
Quick Tip: Remember the key properties of matrix operations: - Inverse: \( (AB)^{-1} = B^{-1}A^{-1} \) (order reverses) - Transpose: \( (AB)^T = B^T A^T \) (order reverses) - Scalar Inverse: \( (kA)^{-1} = \frac{1}{k}A^{-1} \) (scalar is inverted, order doesn't apply) These properties can significantly simplify complex matrix calculations.

The given equation for the height of the plant is \( y = 4x - \frac{1}{2}x^2 \).


(i) Find the rate of growth of the plant with respect to sunlight.

Step 1: Understanding the Concept:

The "rate of growth" is the instantaneous rate of change of the height (y) with respect to the number of days (x). This is found by calculating the first derivative of the height function, \( \frac{dy}{dx} \).


Step 2: Detailed Explanation:

We differentiate the height function \( y(x) \) with respect to x: \[ \frac{dy}{dx} = \frac{d}{dx}\left(4x - \frac{1}{2}x^2\right) \] \[ \frac{dy}{dx} = 4 - \frac{1}{2}(2x) \] \[ \frac{dy}{dx} = 4 - x \]

Step 3: Final Answer for (i):

The rate of growth of the plant with respect to sunlight is \( (4-x) \) cm/day.


(ii) In how many days will the plant attain its maximum height? What is the maximum height?

Step 1: Understanding the Concept:

The height of the plant will be maximum when its rate of growth is zero. This is a critical point of the function. We can find this point by setting the first derivative to zero and solving for x. Then, we substitute this value of x back into the original height equation to find the maximum height.


Step 2: Detailed Explanation:

To find the maximum height, we set the rate of growth, \( \frac{dy}{dx} \), to zero: \[ \frac{dy}{dx} = 0 \] \[ 4 - x = 0 \] \[ x = 4 \]
So, the plant attains its maximum height after 4 days.


To verify this is a maximum, we can use the second derivative test. \[ \frac{d^2y}{dx^2} = \frac{d}{dx}(4 - x) = -1 \]
Since \( \frac{d^2y}{dx^2} < 0 \), the height is indeed a maximum at \( x=4 \).


Now, to find the maximum height, we substitute \( x=4 \) into the original equation for y: \[ y_{max} = 4(4) - \frac{1}{2}(4)^2 \] \[ y_{max} = 16 - \frac{1}{2}(16) \] \[ y_{max} = 16 - 8 = 8 \]

Step 3: Final Answer for (ii):

The plant will attain its maximum height in 4 days. The maximum height is 8 cm.
Quick Tip: For any problem asking for a maximum or minimum value of a function, the standard procedure is: 1. Find the first derivative of the function. 2. Set the first derivative to zero and solve for the variable to find the critical points. 3. (Optional but good practice) Use the second derivative test to confirm if the point corresponds to a maximum (\( f''(x) < 0 \)) or a minimum (\( f''(x) > 0 \)). 4. Substitute the variable's value back into the original function to find the maximum or minimum value.

Step 1: Understanding the Concept:

The presence of \( \cos x \, dx \) in the numerator suggests a substitution involving \( \sin x \). We will first express the entire integrand in terms of \( \sin x \) and then perform the substitution, which will lead to an integral of a rational function that can be solved using partial fractions.


Step 2: Detailed Explanation:

Let \( I = \int \frac{\cos x}{(4+\sin^2 x)(5-4\cos^2 x)} dx \).

First, express the denominator solely in terms of \( \sin x \) using the identity \( \cos^2 x = 1 - \sin^2 x \). \[ 5-4\cos^2 x = 5 - 4(1-\sin^2 x) = 5 - 4 + 4\sin^2 x = 1 + 4\sin^2 x \]
So the integral becomes: \[ I = \int \frac{\cos x}{(4+\sin^2 x)(1+4\sin^2 x)} dx \]
Now, let's use the substitution \( u = \sin x \). Then \( du = \cos x \, dx \). \[ I = \int \frac{1}{(4+u^2)(1+4u^2)} du \]
We use partial fraction decomposition. Let \( v = u^2 \). \[ \frac{1}{(4+v)(1+4v)} = \frac{A}{4+v} + \frac{B}{1+4v} \] \[ 1 = A(1+4v) + B(4+v) \]
To find A, let \( v = -4 \): \( 1 = A(1 - 16) \implies 1 = -15A \implies A = -\frac{1}{15} \).

To find B, let \( v = -1/4 \): \( 1 = B(4 - 1/4) \implies 1 = B(\frac{15}{4}) \implies B = \frac{4}{15} \).

Substituting back \( v = u^2 \), our integrand is: \[ \frac{4/15}{1+4u^2} - \frac{1/15}{4+u^2} = \frac{1}{15} \left( \frac{4}{1+4u^2} - \frac{1}{4+u^2} \right) \]
The integral becomes: \[ I = \frac{1}{15} \int \left( \frac{4}{1+(2u)^2} - \frac{1}{2^2+u^2} \right) du \]
This splits into two standard integrals of the form \( \int \frac{1}{a^2+x^2}dx = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) \). \[ I = \frac{1}{15} \left[ 4 \int \frac{1}{1+(2u)^2} du - \int \frac{1}{2^2+u^2} du \right] \]
For the first integral, let \( w=2u, dw=2du \): \( \int \frac{1}{1+w^2} \frac{dw}{2} = \frac{1}{2}\tan^{-1}(w) = \frac{1}{2}\tan^{-1}(2u) \). \[ I = \frac{1}{15} \left[ 4 \left(\frac{1}{2}\tan^{-1}(2u)\right) - \frac{1}{2}\tan^{-1}\left(\frac{u}{2}\right) \right] + C \] \[ I = \frac{1}{15} \left[ 2\tan^{-1}(2u) - \frac{1}{2}\tan^{-1}\left(\frac{u}{2}\right) \right] + C \]
Finally, substitute back \( u = \sin x \).

Step 3: Final Answer:
\[ I = \frac{1}{15} \left( 2\tan^{-1}(2\sin x) - \frac{1}{2}\tan^{-1}\left(\frac{\sin x}{2}\right) \right) + C \] Quick Tip: When using partial fractions for expressions with only even powers of the variable (like \(u^2\) here), you can make a temporary substitution (like \(v=u^2\)) to simplify the algebra of finding the coefficients A, B, etc.

Step 1: Understanding the Concept:

This is a standard definite integral form. The integrand has a property of symmetry that allows us to simplify the integration interval. The presence of \( \sin^2 x \) and \( \cos^2 x \) suggests dividing by \( \cos^2 x \) to convert the integrand into a form involving \( \sec^2 x \) and \( \tan^2 x \), which is suitable for substitution.


Step 2: Detailed Explanation:

Let \( I = \int_0^\pi \frac{dx}{a^2\cos^2 x + b^2\sin^2 x} \).
Let \( f(x) = \frac{1}{a^2\cos^2 x + b^2\sin^2 x} \).
We check if \( f(\pi - x) = f(x) \). \[ f(\pi-x) = \frac{1}{a^2\cos^2(\pi-x) + b^2\sin^2(\pi-x)} = \frac{1}{a^2(-\cos x)^2 + b^2(\sin x)^2} = \frac{1}{a^2\cos^2 x + b^2\sin^2 x} = f(x) \]
Using the property \( \int_0^{2a} f(x) dx = 2\int_0^a f(x) dx \) if \( f(2a-x)=f(x) \), with \( 2a=\pi \), we get: \[ I = 2 \int_0^{\pi/2} \frac{dx}{a^2\cos^2 x + b^2\sin^2 x} \]
Now, divide the numerator and denominator by \( \cos^2 x \): \[ I = 2 \int_0^{\pi/2} \frac{\sec^2 x \, dx}{a^2 + b^2\tan^2 x} \]
Let's use the substitution \( t = \tan x \). Then \( dt = \sec^2 x \, dx \).
We must change the limits of integration:
When \( x=0 \), \( t = \tan(0) = 0 \).
When \( x=\pi/2 \), \( t = \tan(\pi/2) \to \infty \).
The integral becomes: \[ I = 2 \int_0^\infty \frac{dt}{a^2 + (bt)^2} \]
This is a standard integral of the form \( \int \frac{1}{c^2+u^2}du = \frac{1}{c}\tan^{-1}\left(\frac{u}{c}\right) \). \[ I = 2 \left[ \frac{1}{b} \cdot \frac{1}{a}\tan^{-1}\left(\frac{bt}{a}\right) \right]_0^\infty \] \[ I = \frac{2}{ab} \left[ \tan^{-1}\left(\frac{bt}{a}\right) \right]_0^\infty \]
Now, apply the limits: \[ I = \frac{2}{ab} \left( \lim_{t\to\infty} \tan^{-1}\left(\frac{bt}{a}\right) - \tan^{-1}(0) \right) \]
Assuming a, b > 0, as \( t \to \infty \), \( \frac{bt}{a} \to \infty \). \[ I = \frac{2}{ab} \left( \frac{\pi}{2} - 0 \right) \] \[ I = \frac{2}{ab} \cdot \frac{\pi}{2} = \frac{\pi}{ab} \]

Step 3: Final Answer:

The value of the integral is \( \frac{\pi}{ab} \).
Quick Tip: For definite integrals from 0 to \( \pi \) or 0 to \( 2\pi \) involving even powers of sin(x) and cos(x), always check the symmetry property \( f(2a-x)=f(x) \) to halve the interval of integration. This often avoids issues with discontinuities (like with tan(x) at \( \pi/2 \)) within the original interval.

Part 1: Proof

Step 1: Understanding the Concept:

The area of a parallelogram is given by the magnitude of the cross product of its adjacent sides. We are given the diagonals, so we must first express the adjacent sides in terms of the diagonals.


Step 2: Detailed Proof:

Let the adjacent sides of the parallelogram be represented by vectors \( \vec{p} \) and \( \vec{q} \).
The diagonals of the parallelogram are then given by \( \vec{d_1} = \vec{p} + \vec{q} \) and \( \vec{d_2} = \vec{p} - \vec{q} \).
We are given that the diagonals are \( \vec{a} \) and \( \vec{b} \). Let's set \( \vec{a} = \vec{p} + \vec{q} \) and \( \vec{b} = \vec{p} - \vec{q} \).
We need to find the vectors for the sides, \( \vec{p} \) and \( \vec{q} \), in terms of \( \vec{a} \) and \( \vec{b} \).
Adding the two diagonal equations: \[ \vec{a} + \vec{b} = (\vec{p} + \vec{q}) + (\vec{p} - \vec{q}) = 2\vec{p} \implies \vec{p} = \frac{1}{2}(\vec{a} + \vec{b}) \]
Subtracting the second from the first: \[ \vec{a} - \vec{b} = (\vec{p} + \vec{q}) - (\vec{p} - \vec{q}) = 2\vec{q} \implies \vec{q} = \frac{1}{2}(\vec{a} - \vec{b}) \]
The area of the parallelogram is given by \( |\vec{p} \times \vec{q}| \). \[ Area = \left| \frac{1}{2}(\vec{a} + \vec{b}) \times \frac{1}{2}(\vec{a} - \vec{b}) \right| \] \[ Area = \left| \frac{1}{4} [(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b})] \right| \]
Using the distributive property of the cross product: \[ (\vec{a} + \vec{b}) \times (\vec{a} - \vec{b}) = (\vec{a} \times \vec{a}) - (\vec{a} \times \vec{b}) + (\vec{b} \times \vec{a}) - (\vec{b} \times \vec{b}) \]
Since the cross product of a vector with itself is the zero vector (\( \vec{a} \times \vec{a} = \vec{0} \)) and \( \vec{b} \times \vec{a} = -(\vec{a} \times \vec{b}) \): \[ = \vec{0} - (\vec{a} \times \vec{b}) - (\vec{a} \times \vec{b}) - \vec{0} = -2(\vec{a} \times \vec{b}) \]
Substituting this back into the area formula: \[ Area = \left| \frac{1}{4}[-2(\vec{a} \times \vec{b})] \right| = \left| -\frac{1}{2}(\vec{a} \times \vec{b}) \right| \] \[ Area = \frac{1}{2}|\vec{a} \times \vec{b}| \]
This completes the proof.


Part 2: Calculation

Step 1: Applying the Formula:

We are given the diagonals \( \vec{d_1} = 2\hat{i} - \hat{j} + \hat{k} \) and \( \vec{d_2} = \hat{i} + 3\hat{j} - \hat{k} \).
The area is \( \frac{1}{2}|\vec{d_1} \times \vec{d_2}| \).

Step 2: Compute the Cross Product: \[ \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
2 & -1 & 1
1 & 3 & -1 \end{vmatrix} \] \[ = \hat{i}((-1)(-1) - (1)(3)) - \hat{j}((2)(-1) - (1)(1)) + \hat{k}((2)(3) - (-1)(1)) \] \[ = \hat{i}(1-3) - \hat{j}(-2-1) + \hat{k}(6+1) \] \[ = -2\hat{i} + 3\hat{j} + 7\hat{k} \]
Step 3: Find the Magnitude: \[ |\vec{d_1} \times \vec{d_2}| = \sqrt{(-2)^2 + 3^2 + 7^2} = \sqrt{4 + 9 + 49} = \sqrt{62} \]
Step 4: Final Answer: \[ Area = \frac{1}{2}|\vec{d_1} \times \vec{d_2}| = \frac{\sqrt{62}}{2} \]
The area of the parallelogram is \( \frac{\sqrt{62}}{2} \) square units.
Quick Tip: Be careful not to confuse the formulas for the area of a parallelogram based on adjacent sides versus diagonals. - Adjacent sides \( \vec{p}, \vec{q} \): Area = \( |\vec{p} \times \vec{q}| \) - Diagonals \( \vec{d_1}, \vec{d_2} \): Area = \( \frac{1}{2}|\vec{d_1} \times \vec{d_2}| \) The factor of 1/2 is crucial when using diagonals.

Step 1: Understanding the Concept:

The equation of a line requires a point on the line and a direction vector parallel to the line. We are given the point. Since the required line is perpendicular to two other lines, its direction vector will be perpendicular to the direction vectors of both given lines. We can find such a vector by taking the cross product of the direction vectors of the given lines.


Step 2: Identifying Given Information:

The required line passes through the point P(1, 2, -4). The position vector of this point is \( \vec{a} = \hat{i} + 2\hat{j} - 4\hat{k} \).

The first given line is \( L_1: \frac{x-8}{3} = \frac{y+19}{-16} = \frac{z-10}{7} \). Its direction vector is \( \vec{b_1} = 3\hat{i} - 16\hat{j} + 7\hat{k} \).

The second given line is \( L_2: \vec{r} = (15\hat{i} + 29\hat{j} + 5\hat{k}) + \mu(3\hat{i} + 8\hat{j} - 5\hat{k}) \). Its direction vector is \( \vec{b_2} = 3\hat{i} + 8\hat{j} - 5\hat{k} \).


Step 3: Finding the Direction Vector of the Required Line:

Let the direction vector of the required line be \( \vec{b} \). Since the line is perpendicular to \( L_1 \) and \( L_2 \), \( \vec{b} \) is perpendicular to both \( \vec{b_1} \) and \( \vec{b_2} \). We can find \( \vec{b} \) by computing their cross product. \[ \vec{b} = \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
3 & -16 & 7
3 & 8 & -5 \end{vmatrix} \] \[ = \hat{i}((-16)(-5) - (7)(8)) - \hat{j}((3)(-5) - (7)(3)) + \hat{k}((3)(8) - (-16)(3)) \] \[ = \hat{i}(80 - 56) - \hat{j}(-15 - 21) + \hat{k}(24 + 48) \] \[ = 24\hat{i} - (-36)\hat{j} + 72\hat{k} = 24\hat{i} + 36\hat{j} + 72\hat{k} \]
The direction ratios are (24, 36, 72). We can use a simpler vector parallel to this one by dividing by their greatest common divisor, which is 12. \[ \vec{b}_{simplified} = \frac{1}{12}(24\hat{i} + 36\hat{j} + 72\hat{k}) = 2\hat{i} + 3\hat{j} + 6\hat{k} \]

Step 4: Writing the Equations of the Line:

We have the point \( \vec{a} = \hat{i} + 2\hat{j} - 4\hat{k} \) and the direction vector \( \vec{b} = 2\hat{i} + 3\hat{j} + 6\hat{k} \).

Vector Equation: The equation is \( \vec{r} = \vec{a} + \lambda\vec{b} \). \[ \vec{r} = (\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k}) \]
where \( \lambda \) is a scalar parameter.

Cartesian Equation: The equation is \( \frac{x-x_1}{l} = \frac{y-y_1}{m} = \frac{z-z_1}{n} \).
The point is (1, 2, -4) and the direction ratios are (2, 3, 6). \[ \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-(-4)}{6} \] \[ \frac{x-1}{2} = \frac{y-2}{3} = \frac{z+4}{6} \]

Step 5: Final Answer:

The vector equation is \( \vec{r} = (\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k}) \).

The Cartesian equation is \( \frac{x-1}{2} = \frac{y-2}{3} = \frac{z+4}{6} \).
Quick Tip: A vector perpendicular to two given non-parallel vectors \( \vec{u} \) and \( \vec{v} \) can always be found using their cross product, \( \vec{u} \times \vec{v} \). This is a fundamental concept for finding the direction of a line perpendicular to two other lines, or for finding the normal vector to a plane containing two vectors.

Step 1: Understanding the Concept:

This question asks for the probability of the intersection of two events: the event that a student does not have the misconception (M') AND the event that the student answers Bijoy (B). We need to find \( P(M' \cap B) \).


Step 2: Key Formula or Approach:

The multiplication rule of probability states that for two events A and E:
\[ P(A \cap E) = P(E) \times P(A|E) \]
In our case, we need to find \( P(M' \cap B) \), which can be calculated as:
\[ P(M' \cap B) = P(M') \times P(B|M') \]

Step 3: Detailed Explanation:

From the setup derived from the problem description, we have:

- The probability that a student does not have the misconception, \( P(M') = 0.60 \).

- The probability that a student answers Bijoy given they do not have the misconception, \( P(B|M') = 0.10 \).

Now, we can substitute these values into the formula:
\[ P(M' \cap B) = 0.60 \times 0.10 \] \[ P(M' \cap B) = 0.06 \]

Step 4: Final Answer:

The probability of a student not having a misconception but still answering Bijoy is 0.06 or 6%.
Quick Tip: The word "but" or "and" in a probability question typically signifies the intersection of events. It is important to distinguish this from conditional probability, which is usually indicated by phrases like "given that" or "if".

Step 1: Understanding the Concept:

This question asks for the total probability of a student answering Bijoy, regardless of whether they have the misconception or not. We need to find \( P(B) \). A student can answer Bijoy in two mutually exclusive ways: (1) having the misconception AND answering Bijoy, or (2) not having the misconception AND answering Bijoy.


Step 2: Key Formula or Approach:

We will use the Law of Total Probability. For an event B and a partition of the sample space (in this case, M and M'), the formula is:
\[ P(B) = P(M \cap B) + P(M' \cap B) \]
Which can also be written as:
\[ P(B) = P(M) \times P(B|M) + P(M') \times P(B|M') \]

Step 3: Detailed Explanation:

We need to calculate the probability of each of the two cases:

Case 1: The student has the misconception and answers Bijoy, \( P(M \cap B) \).
\[ P(M \cap B) = P(M) \times P(B|M) = 0.40 \times 0.80 = 0.32 \]
Case 2: The student does not have the misconception and answers Bijoy, \( P(M' \cap B) \).

From part (i), we already calculated this: \( P(M' \cap B) = 0.06 \).

Now, we add the probabilities of these two disjoint events to find the total probability of answering Bijoy:
\[ P(B) = P(M \cap B) + P(M' \cap B) = 0.32 + 0.06 = 0.38 \]

Step 4: Final Answer:

The probability that a randomly selected student answers Bijoy is 0.38 or 38%.
Quick Tip: The Law of Total Probability is fundamental for solving problems that lead into Bayes' Theorem. It helps you find the probability of a subsequent event by summing the probabilities of all paths that lead to that event.

Step 1: Understanding the Concept:

This is a conditional probability problem. We are given that a student has answered "Bijoy" (event B has occurred), and we need to find the probability that this student is one who has the misconception (event M). We need to calculate \( P(M | B) \).


Step 2: Key Formula or Approach:

We use Bayes' Theorem, which is derived from the definition of conditional probability:
\[ P(M|B) = \frac{P(M \cap B)}{P(B)} \]
This can be expanded as:
\[ P(M|B) = \frac{P(M) \times P(B|M)}{P(B)} \]

Step 3: Detailed Explanation:

We have already calculated the necessary components in the previous parts:

- The probability that a student has the misconception AND answers Bijoy is \( P(M \cap B) = P(M) \times P(B|M) = 0.40 \times 0.80 = 0.32 \).

- The total probability that a student answers Bijoy is \( P(B) = 0.38 \) (from part ii).

Now, we can apply Bayes' Theorem:
\[ P(M|B) = \frac{0.32}{0.38} \]
To simplify the fraction, we can write it as:
\[ P(M|B) = \frac{32}{38} = \frac{16}{19} \]

Step 4: Final Answer:

The probability that a student who answered Bijoy has the misconception is \( \frac{16}{19} \).
Quick Tip: Bayes' Theorem helps us update our beliefs about an event based on new evidence. Here, the "new evidence" is that the student answered Bijoy. We use this to find the "updated" probability that the student has a misconception. The formula essentially says: \( Posterior = \frac{Likelihood \times Prior}{Evidence} \).

Step 1: Understanding the Concept:

This is another conditional probability problem, similar to the previous one. We are given that a student has answered "Bijoy" (event B), and we want to find the probability that this student does NOT have the misconception (event M'). We need to calculate \( P(M' | B) \).


Step 2: Key Formula or Approach:

We can use Bayes' Theorem again:
\[ P(M'|B) = \frac{P(M' \cap B)}{P(B)} \]
Alternatively, since M and M' are complementary events, we can use the property of conditional probabilities:
\[ P(M'|B) = 1 - P(M|B) \]

Step 3: Detailed Explanation:

Method 1: Using Bayes' Theorem directly

We have the required values from previous parts:

- The probability that a student does not have the misconception AND answers Bijoy is \( P(M' \cap B) = 0.06 \) (from part i).

- The total probability that a student answers Bijoy is \( P(B) = 0.38 \) (from part ii).

Substituting these into the formula:
\[ P(M'|B) = \frac{0.06}{0.38} \]
Simplifying the fraction:
\[ P(M'|B) = \frac{6}{38} = \frac{3}{19} \]

Method 2: Using the complement rule

From part (iii)(a), we found that \( P(M|B) = \frac{16}{19} \).

Since a student who answered Bijoy either has the misconception or does not have it, these two outcomes are complementary.
\[ P(M'|B) = 1 - P(M|B) \] \[ P(M'|B) = 1 - \frac{16}{19} = \frac{19}{19} - \frac{16}{19} = \frac{3}{19} \]
Both methods yield the same result.


Step 4: Final Answer:

The probability that a student who answered Bijoy does not have the misconception is \( \frac{3}{19} \).
Quick Tip: When an "OR" question in an exam asks for a probability that is the complement of a previously calculated probability, using the complement rule \( P(A'|B) = 1 - P(A|B) \) is often the fastest and easiest method, and it also serves as a good check for your previous answer.

Step 1: Understanding the Concept:

Two lines in 3D space are parallel if and only if their direction vectors are parallel. Two vectors are parallel if one is a scalar multiple of the other. This means their corresponding direction ratios must be proportional.


Step 2: Key Formula or Approach:

Let the direction ratios of the first line be \( (a_1, b_1, c_1) \) and the second line be \( (a_2, b_2, c_2) \). The lines are parallel if \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \).


Step 3: Detailed Explanation:

The equation for Line A is \( l_1 : \frac{x-2}{3} = \frac{y+1}{-2} = \frac{z-3}{4} \).
The direction ratios of Line A are \( (a_1, b_1, c_1) = (3, -2, 4) \).

The direction vector for Line A is \( \vec{d_1} = 3\hat{i} - 2\hat{j} + 4\hat{k} \).


The equation for Line B is \( l_2 : \frac{x-1}{2} = \frac{y-3}{1} = \frac{z+2}{-3} \).
The direction ratios of Line B are \( (a_2, b_2, c_2) = (2, 1, -3) \).

The direction vector for Line B is \( \vec{d_2} = 2\hat{i} + \hat{j} - 3\hat{k} \).


Now, we check if the direction ratios are proportional: \[ \frac{a_1}{a_2} = \frac{3}{2} \] \[ \frac{b_1}{b_2} = \frac{-2}{1} = -2 \] \[ \frac{c_1}{c_2} = \frac{4}{-3} \]
Since \( \frac{3}{2} \ne -2 \ne \frac{4}{-3} \), the direction ratios are not proportional.


Step 4: Final Answer:

Because the direction ratios of the two lines are not proportional, the two metro tracks are not parallel.
Quick Tip: To quickly check if two lines are parallel, compare the ratios of their direction ratios. If all the ratios are equal, the lines are parallel. If not, they are either intersecting or skew.

Step 1: Understanding the Concept:

We need to find the equation of a line. The equation of a line in 3D space is determined by a point on the line and a direction vector that is parallel to the line.


Step 2: Key Formula or Approach:

The Cartesian equation of a line passing through the point \( (x_1, y_1, z_1) \) with direction ratios \( (a, b, c) \) is given by: \[ \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} \]

Step 3: Detailed Explanation:

We are given that the line for the solar panels passes through the point \( (x_1, y_1, z_1) = (1, -2, -3) \).

We are also given that this line is parallel to Line A's track, \( l_1 \). Parallel lines have the same direction ratios.
The direction ratios of Line A (\( l_1 \)) are \( (a, b, c) = (3, -2, 4) \).
Therefore, the direction ratios for the line of solar panels are also (3, -2, 4).

Now, we can substitute the point and the direction ratios into the Cartesian equation formula: \[ \frac{x-1}{3} = \frac{y-(-2)}{-2} = \frac{z-(-3)}{4} \]

Step 4: Final Answer:

The equation of the line representing the placement of the solar panels is: \[ \frac{x-1}{3} = \frac{y+2}{-2} = \frac{z+3}{4} \] Quick Tip: When a problem asks for the equation of a line parallel to another given line, you can directly use the direction ratios from the given line for your new line's equation.

Step 1: Understanding the Concept:

The pedestrian walkway is a line that is perpendicular to both metro line A and metro line B. The direction vector of a line that is perpendicular to two other lines can be found by taking the cross product of their direction vectors. Once we have the direction vector and the given point, we can write the equation of the line.


Step 2: Key Formula or Approach:

If a line is perpendicular to two vectors \( \vec{d_1} \) and \( \vec{d_2} \), its direction vector \( \vec{d} \) can be found by \( \vec{d} = \vec{d_1} \times \vec{d_2} \). The equation of the line passing through \( (x_1, y_1, z_1) \) with direction vector \( a\hat{i} + b\hat{j} + c\hat{k} \) is \( \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} \).


Step 3: Detailed Explanation:

The direction vector for Line A is \( \vec{d_1} = 3\hat{i} - 2\hat{j} + 4\hat{k} \).

The direction vector for Line B is \( \vec{d_2} = 2\hat{i} + \hat{j} - 3\hat{k} \).

Let the direction vector of the pedestrian walkway be \( \vec{d} \). Then \( \vec{d} = \vec{d_1} \times \vec{d_2} \). \[ \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
3 & -2 & 4
2 & 1 & -3 \end{vmatrix} \] \[ = \hat{i}((-2)(-3) - (4)(1)) - \hat{j}((3)(-3) - (4)(2)) + \hat{k}((3)(1) - (-2)(2)) \] \[ = \hat{i}(6 - 4) - \hat{j}(-9 - 8) + \hat{k}(3 + 4) \] \[ = 2\hat{i} + 17\hat{j} + 7\hat{k} \]
The direction ratios of the walkway are (2, 17, 7).

The walkway passes through the point \( (x_1, y_1, z_1) = (3, 2, 1) \).

Using the Cartesian form for the equation of a line: \[ \frac{x-3}{2} = \frac{y-2}{17} = \frac{z-1}{7} \]

Step 4: Final Answer:

The equation of the pedestrian walkway is \( \frac{x-3}{2} = \frac{y-2}{17} = \frac{z-1}{7} \).
Quick Tip: The cross product of two vectors gives a third vector that is perpendicular to both of the original vectors. This is a fundamental tool for solving problems involving perpendicularity in 3D geometry.

Step 1: Understanding the Concept:

Since the lines are not parallel (as determined in part i), they are either intersecting or skew. The shortest distance between two skew lines is the length of the line segment that is perpendicular to both lines. We can calculate this using a standard formula.


Step 2: Key Formula or Approach:

The shortest distance (SD) between two skew lines \( \vec{r} = \vec{a_1} + \lambda\vec{b_1} \) and \( \vec{r} = \vec{a_2} + \mu\vec{b_2} \) is given by: \[ SD = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})}{|\vec{b_1} \times \vec{b_2}|} \right| \]

Step 3: Detailed Explanation:

From the equations of the lines:
For Line A (\( l_1 \)): It passes through the point \( P_1(2, -1, 3) \). So, \( \vec{a_1} = 2\hat{i} - \hat{j} + 3\hat{k} \).
Its direction vector is \( \vec{b_1} = 3\hat{i} - 2\hat{j} + 4\hat{k} \).


For Line B (\( l_2 \)): It passes through the point \( P_2(1, 3, -2) \). So, \( \vec{a_2} = \hat{i} + 3\hat{j} - 2\hat{k} \).
Its direction vector is \( \vec{b_2} = 2\hat{i} + \hat{j} - 3\hat{k} \).


Now we compute the components needed for the formula:
1. \( \vec{a_2} - \vec{a_1} = (1-2)\hat{i} + (3-(-1))\hat{j} + (-2-3)\hat{k} = -\hat{i} + 4\hat{j} - 5\hat{k} \).
2. \( \vec{b_1} \times \vec{b_2} \): We calculated this in the previous part (iii)(a).
\( \vec{b_1} \times \vec{b_2} = 2\hat{i} + 17\hat{j} + 7\hat{k} \).
3. \( |\vec{b_1} \times \vec{b_2}| = \sqrt{2^2 + 17^2 + 7^2} = \sqrt{4 + 289 + 49} = \sqrt{342} \).
4. \( (\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) \): This is the scalar triple product.
\( = (-\hat{i} + 4\hat{j} - 5\hat{k}) \cdot (2\hat{i} + 17\hat{j} + 7\hat{k}) \)
\( = (-1)(2) + (4)(17) + (-5)(7) \)
\( = -2 + 68 - 35 = 31 \).

Now substitute these values into the shortest distance formula: \[ SD = \left| \frac{31}{\sqrt{342}} \right| = \frac{31}{\sqrt{342}} \]

Step 4: Final Answer:

The shortest distance between Line A and Line B is \( \frac{31}{\sqrt{342}} \) units.
Quick Tip: The formula for the shortest distance between skew lines involves the scalar triple product in the numerator. If this scalar triple product is zero, it means the vectors \( (\vec{a_2} - \vec{a_1}) \), \( \vec{b_1} \), and \( \vec{b_2} \) are coplanar, which implies the lines intersect, and the shortest distance is zero.

Step 1: Understanding the Concept:

The problem provides a differential equation that models Newton's Law of Cooling. We need to solve this differential equation to find the function \(T(t)\) which gives the temperature at any time \(t\). This is an initial value problem, as we are given an initial condition \(T(0) = 85\).


Step 2: Key Formula or Approach:

The given differential equation is a first-order linear differential equation which is also separable. We will use the method of separation of variables to solve it.

1. Separate the variables involving T and t.

2. Integrate both sides.

3. Use the initial condition to find the value of the constant of integration.

4. Write the final expression for T(t).


Step 3: Detailed Explanation:

The differential equation is:
\[ \frac{dT}{dt} = -k(T - 25) \]
Separate the variables by moving all terms with T to one side and terms with t to the other.
\[ \frac{dT}{T - 25} = -k \, dt \]
Now, integrate both sides of the equation:
\[ \int \frac{1}{T - 25} \, dT = \int -k \, dt \]
The integration yields:
\[ \ln|T - 25| = -kt + C \]
where C is the constant of integration. To solve for T, we exponentiate both sides:
\[ |T - 25| = e^{-kt + C} = e^C \cdot e^{-kt} \]
Let \( A = \pm e^C \) be a new constant. Since the processor is cooling, \(T \geq 25\), so \(|T-25| = T-25\).
\[ T - 25 = A e^{-kt} \] \[ T(t) = 25 + A e^{-kt} \]
This is the general solution. Now we use the initial condition \(T(0) = 85\) to find A.
\[ 85 = 25 + A e^{-k(0)} \] \[ 85 = 25 + A e^0 \] \[ 85 = 25 + A \cdot 1 \] \[ A = 85 - 25 = 60 \]
Substitute A = 60 back into the general solution to get the particular solution:
\[ T(t) = 25 + 60e^{-kt} \]

Step 4: Final Answer:

The expression for the temperature of the processor at time t is \( T(t) = 25 + 60e^{-kt} \).
Quick Tip: Differential equations modeling physical phenomena like cooling, heating, or radioactive decay often follow a similar pattern. Recognizing this pattern (\(dy/dt = k(y-a)\)) allows you to quickly write down the general solution form (\(y(t) = a + Ce^{kt}\)) and then solve for the constants using the given conditions.

Step 1: Understanding the Concept:

Using the specific temperature function \(T(t)\) derived in the previous part, we need to find the time \(t\) at which the temperature \(T\) becomes 40\(^{\circ}\)C. We are given the value of the cooling constant \(k\).


Step 2: Key Formula or Approach:

We will use the equation found in part (i):
\[ T(t) = 25 + 60e^{-kt} \]
We will substitute the given values \(T(t) = 40\) and \(k = 0.03\) and solve the resulting exponential equation for \(t\).


Step 3: Detailed Explanation:

From part (i), the expression for temperature is:
\[ T(t) = 25 + 60e^{-kt} \]
We are given that we want to find the time \(t\) when \(T(t) = 40\). We are also given \(k = 0.03\). Substitute these values into the equation:
\[ 40 = 25 + 60e^{-0.03t} \]
Now, we solve for \(t\). First, isolate the exponential term.
\[ 40 - 25 = 60e^{-0.03t} \] \[ 15 = 60e^{-0.03t} \] \[ \frac{15}{60} = e^{-0.03t} \] \[ \frac{1}{4} = e^{-0.03t} \]
To solve for \(t\), take the natural logarithm (\(\log_e\) or \(\ln\)) of both sides:
\[ \ln\left(\frac{1}{4}\right) = \ln(e^{-0.03t}) \]
Using the properties of logarithms, \( \ln(a^b) = b\ln(a) \) and \( \ln(1/a) = -\ln(a) \):
\[ -\ln(4) = -0.03t \cdot \ln(e) \]
Since \( \ln(e) = 1 \):
\[ -\ln(4) = -0.03t \] \[ \ln(4) = 0.03t \]
Now, solve for \(t\):
\[ t = \frac{\ln(4)}{0.03} \]
We are given that \( \log_e 4 = 1.3863 \). Substitute this value:
\[ t = \frac{1.3863}{0.03} \] \[ t = \frac{138.63}{3} \] \[ t = 46.21 \]
The time is in minutes.


Step 4: Final Answer:

It will take 46.21 minutes for the processor's temperature to reach 40\(^{\circ}\)C.
Quick Tip: When solving exponential equations of the form \(a = be^{ct}\), the standard procedure is to isolate the exponential term first (\(e^{ct} = a/b\)) and then take the natural logarithm of both sides to bring the exponent down. Be careful with the signs and logarithm properties.