By definition, the inverse cosine function, denoted as \(\cos^{-1}(x)\) or arccos(x), is defined for \(x \in [-1, 1]\).


To make the function one-to-one, its range is restricted to a specific interval.


This restricted range is known as the principal value branch.


For the function \(y = \cos^{-1}(x)\), the principal value branch (range) is \([0, \pi]\).


This interval includes angles in the first and second quadrants, where the cosine function takes on all its possible values from -1 to 1 exactly once.


Therefore, the principal branch of \(\cos^{-1} x\) is \([0, \pi]\).
Quick Tip: Memorize the principal value branches for all six inverse trigonometric functions, as they are fundamental concepts and frequently tested. For \(\cos^{-1}x\), remember the range is the upper half of the unit circle, \([0, \pi]\).

For a function \(f(x)\) to be a decreasing function for all real values of \(x\), its first derivative \(f'(x)\) must be less than or equal to zero for all \(x \in R\).


Given the function \(f(x) = \sin x - \cos x - \lambda x + C\).


First, find the derivative of \(f(x)\) with respect to \(x\):

\(f'(x) = \frac{d}{dx}(\sin x - \cos x - \lambda x + C) = \cos x - (-\sin x) - \lambda = \cos x + \sin x - \lambda\).


Now, apply the condition for a decreasing function: \(f'(x) \le 0\).

\(\cos x + \sin x - \lambda \le 0\)

\(\cos x + \sin x \le \lambda\).


This inequality must hold for all real values of \(x\). This means \(\lambda\) must be greater than or equal to the maximum value of the expression \(\cos x + \sin x\).


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


For \(\cos x + \sin x\), we have \(a = 1\) and \(b = 1\).


Maximum value = \(\sqrt{1^2 + 1^2} = \sqrt{2}\).


Therefore, for the inequality to hold for all \(x\), we must have \(\lambda \ge \sqrt{2}\).
Quick Tip: The range of the function \(a \sin(x) + b \cos(x)\) is \([-\sqrt{a^2+b^2}, \sqrt{a^2+b^2}]\). This is a very useful result for problems involving the maximum or minimum values of trigonometric expressions.

We are given two conditions for matrices A and B:

1. \(AB = A\)

2. \(BA = B\)


We need to find the value of \(A^2 + B^2\).


Let's compute \(A^2\) first.

\(A^2 = A \cdot A\)


Using the first given condition, substitute \(A\) with \(AB\):

\(A^2 = (AB)A\)


Using the associative property of matrix multiplication, \(A(BA)\):

\(A^2 = A(BA)\)


Now, using the second given condition, substitute \(BA\) with \(B\):

\(A^2 = AB\)


Finally, using the first condition again, \(AB = A\):

\(A^2 = A\).


Next, let's compute \(B^2\).

\(B^2 = B \cdot B\)


Using the second given condition, substitute \(B\) with \(BA\):

\(B^2 = (BA)B\)


Using the associative property, \(B(AB)\):

\(B^2 = B(AB)\)


Now, using the first given condition, substitute \(AB\) with \(A\):

\(B^2 = BA\)


Finally, using the second condition again, \(BA = B\):

\(B^2 = B\).


Now, we can find \(A^2 + B^2\):

\(A^2 + B^2 = A + B\).
Quick Tip: Matrices for which \(A^2 = A\) are called idempotent matrices. The problem shows that if \(AB = A\) and \(BA = B\), then both A and B must be idempotent.

For the function \(f(x)\) to be continuous at \(x = \frac{\pi}{2}\), the value of the function at that point must be equal to the limit of the function as \(x\) approaches that point.


So, \(k = f(\frac{\pi}{2}) = \lim_{x \to \frac{\pi}{2}} \frac{1-\sin^3 x}{3\cos^2 x}\).


As \(x \to \frac{\pi}{2}\), \(\sin(x) \to 1\) and \(\cos(x) \to 0\). This leads to the indeterminate form \(\frac{0}{0}\).


We can simplify the expression using algebraic identities.


The numerator is of the form \(a^3 - b^3 = (a-b)(a^2+ab+b^2)\), where \(a = 1\) and \(b = \sin x\).

\(1 - \sin^3 x = (1 - \sin x)(1 + \sin x + \sin^2 x)\).


The denominator can be written using the identity \(\cos^2 x = 1 - \sin^2 x = (1 - \sin x)(1 + \sin x)\).


So, \(3\cos^2 x = 3(1 - \sin x)(1 + \sin x)\).


Now substitute these back into the limit expression:

\(k = \lim_{x \to \frac{\pi}{2}} \frac{(1 - \sin x)(1 + \sin x + \sin^2 x)}{3(1 - \sin x)(1 + \sin x)}\).


Since \(x \to \frac{\pi}{2}\), \(x\) is not equal to \(\frac{\pi}{2}\), so \(1 - \sin x \ne 0\). We can cancel the \((1 - \sin x)\) term.

\(k = \lim_{x \to \frac{\pi}{2}} \frac{1 + \sin x + \sin^2 x}{3(1 + \sin x)}\).


Now, we can substitute \(x = \frac{\pi}{2}\):

\(k = \frac{1 + \sin(\frac{\pi}{2}) + \sin^2(\frac{\pi}{2})}{3(1 + \sin(\frac{\pi}{2}))} = \frac{1 + 1 + 1^2}{3(1 + 1)} = \frac{3}{3(2)} = \frac{3}{6} = \frac{1}{2}\).
Quick Tip: When faced with a \(\frac{0}{0}\) indeterminate form in limits involving trigonometric functions, look for ways to apply Pythagorean identities (like \(\sin^2x + \cos^2x = 1\)) and algebraic factorization formulas. L'Hôpital's Rule is also a powerful alternative.

To determine the nature of the function \(f(x) = x^3 + 5x + 1\), we will check for one-one (injective) and onto (surjective) properties.


One-one (Injective) Test:


A differentiable function is one-one if its derivative is either always positive or always negative.


Let's find the derivative of \(f(x)\):

\(f'(x) = \frac{d}{dx}(x^3 + 5x + 1) = 3x^2 + 5\).


For any real number \(x\), \(x^2\) is always non-negative (\(x^2 \ge 0\)).


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 on R.


A strictly increasing function is always one-one.


Onto (Surjective) Test:


A function \(f: R \to R\) is onto if its range is equal to its codomain (R).

\(f(x)\) is a polynomial function of odd degree (degree 3).


Let's examine the behavior of \(f(x)\) at the extremes:

\(\lim_{x \to \infty} f(x) = \lim_{x \to \infty} (x^3 + 5x + 1) = \infty\).

\(\lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} (x^3 + 5x + 1) = -\infty\).


Since \(f(x)\) is a continuous function (as all polynomial functions are) and its values range from \(-\infty\) to \(\infty\), its range is the set of all real numbers, R.


Thus, the function is onto.


Since \(f(x)\) is both one-one and onto, it is a bijective function from R to R.
Quick Tip: A quick way to analyze polynomial functions on R: If the derivative \(f'(x)\) is always positive or always negative, the function is one-one. If the polynomial is of an odd degree, it is always an onto function from R to R.

Let the direction cosines of a line be \(l, m, n\).


A fundamental property of direction cosines is that the sum of their squares is equal to 1.

\(l^2 + m^2 + n^2 = 1\).


In this problem, we are given that \(l = \lambda\), \(m = \lambda\), and \(n = \lambda\).


Substituting these values into the property:

\(\lambda^2 + \lambda^2 + \lambda^2 = 1\).

\(3\lambda^2 = 1\).

\(\lambda^2 = \frac{1}{3}\).


Taking the square root of both sides gives:

\(\lambda = \pm \sqrt{\frac{1}{3}} = \pm \frac{1}{\sqrt{3}}\).


Therefore, the value of \(\lambda\) is \(\pm \frac{1}{\sqrt{3}}\).
Quick Tip: Remember the core identity for direction cosines (\(l, m, n\)): \(l^2+m^2+n^2=1\). This is essential for solving any problem involving direction cosines. A line can have two sets of direction cosines (\((l, m, n)\) and \((-l, -m, -n)\)) representing opposite directions.

We need to solve the given determinant equation for \(x\).


Let's expand the determinant along the first row:

\((-1) \begin{vmatrix} x & 1
3 & 3x \end{vmatrix} - (2) \begin{vmatrix} 1 & 1
0 & 3x \end{vmatrix} + (4) \begin{vmatrix} 1 & x
0 & 3 \end{vmatrix} = -57\).


Now, calculate the \(2 \times 2\) determinants:

\((-1)((x)(3x) - (1)(3)) - (2)((1)(3x) - (1)(0)) + (4)((1)(3) - (x)(0)) = -57\).


Simplify the expression:

\(-1(3x^2 - 3) - 2(3x) + 4(3) = -57\).

\(-3x^2 + 3 - 6x + 12 = -57\).


Combine like terms:

\(-3x^2 - 6x + 15 = -57\).


Move all terms to one side to form a quadratic equation:

\(-3x^2 - 6x + 15 + 57 = 0\).

\(-3x^2 - 6x + 72 = 0\).


Divide the entire equation by \(-3\) to simplify:

\(x^2 + 2x - 24 = 0\).


This is a quadratic equation of the form \(ax^2 + bx + c = 0\), where \(a=1, b=2, c=-24\).


The product of the roots (possible values of \(x\)) of a quadratic equation is given by the formula \(\frac{c}{a}\).


Product of roots = \(\frac{-24}{1} = -24\).
Quick Tip: For a quadratic equation \(ax^2 + bx + c = 0\), the sum of the roots is \(-b/a\) and the product of the roots is \(c/a\). This often saves time compared to finding the individual roots and then multiplying them.

Let the given matrix be \(A = \begin{bmatrix} 0 & 1 & -2
-1 & 0 & -7
2 & 7 & 0 \end{bmatrix}\).


To classify the matrix, we need to check the properties of diagonal, symmetric, skew-symmetric, and scalar matrices.


1. A diagonal matrix has non-zero elements only on the main diagonal. This is not true for A.


2. A scalar matrix is a diagonal matrix with all diagonal elements equal. This is not true for A.


3. A symmetric matrix is a square matrix that is equal to its transpose (\(A = A^T\)).


4. A skew-symmetric matrix is a square matrix whose transpose is its negative (\(A^T = -A\)).


Let's find the transpose of A, denoted by \(A^T\). The transpose is found by interchanging rows and columns.

\(A^T = \begin{bmatrix} 0 & -1 & 2
1 & 0 & 7
-2 & -7 & 0 \end{bmatrix}\).


Now, let's find \(-A\).

\(-A = -1 \times \begin{bmatrix} 0 & 1 & -2
-1 & 0 & -7
2 & 7 & 0 \end{bmatrix} = \begin{bmatrix} 0 & -1 & 2
1 & 0 & 7
-2 & -7 & 0 \end{bmatrix}\).


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


Therefore, the matrix A is a skew-symmetric matrix.
Quick Tip: A key property of a skew-symmetric matrix is that all its main diagonal elements must be zero. Also, the elements are symmetric with respect to the main diagonal but with opposite signs, i.e., \(a_{ij} = -a_{ji}\).

The function \(f(x)\) is defined piecewise. For it to be continuous over the interval \((0, 2)\), it must be continuous at the point where the definition changes, which is \(x=1\).


The condition for continuity at a point \(x=c\) is that the left-hand limit (LHL), the right-hand limit (RHL), and the value of the function at that point are all equal.

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


For this problem, we need to ensure \(\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x)\).


First, calculate the left-hand limit (LHL) at \(x=1\). For \(x \le 1\), \(f(x) = 3x - 2\).


LHL = \(\lim_{x \to 1^-} (3x - 2) = 3(1) - 2 = 1\).


Next, calculate the right-hand limit (RHL) at \(x=1\). For \(x > 1\), \(f(x) = 2x^2 + ax\).


RHL = \(\lim_{x \to 1^+} (2x^2 + ax) = 2(1)^2 + a(1) = 2 + a\).


For continuity, we must have LHL = RHL.

\(1 = 2 + a\).


Solving for \(a\):

\(a = 1 - 2 = -1\).


Thus, the value of \(a\) is \(-1\).
Quick Tip: For piecewise functions, continuity questions almost always focus on the boundary points where the function's definition changes. Simply equate the left-hand limit and the right-hand limit at these points to find the unknown constants.

The function is defined from the set of Natural numbers \(N = \{1, 2, 3, ...\}\) to the set of Whole numbers \(W = \{0, 1, 2, 3, ...\}\).


Check for Injective (One-to-one):


A function is injective if different inputs always produce different outputs.


Let's test some odd inputs:

\(f(1) = 0\) (since 1 is odd).

\(f(3) = 0\) (since 3 is odd).


Here, we have two different inputs, \(1\) and \(3\), but they produce the same output, \(0\).


Since \(f(1) = f(3)\) but \(1 \ne 3\), the function is not injective.


Check for Surjective (Onto):


A function is surjective if every element in the codomain (W) is an output for at least one input from the domain (N).


Let \(y\) be an arbitrary element in the codomain \(W\). We need to see if there is an \(n \in N\) such that \(f(n)=y\).


Case 1: \(y = 0\).

We can choose any odd natural number, for instance \(n=1\). Then \(f(1) = 0\). So, \(0\) has a pre-image.


Case 2: \(y \in \{1, 2, 3, ...\}\).

We need to find an \(n \in N\) such that \(f(n) = y\). Since \(y > 0\), we must use the rule for even \(n\).

Let \(f(n) = \frac{n}{2} = y\).

This gives \(n = 2y\).

For any positive whole number \(y\), \(n=2y\) is an even natural number. For example, if \(y=5\), then \(n=10\) (which is in N), and \(f(10) = 10/2 = 5\).


Since every element in the codomain W has at least one pre-image in the domain N, the function is surjective.


Conclusion: The function is surjective but not injective.
Quick Tip: To test if a function is injective (one-to-one), try to find a counterexample: two different inputs that lead to the same output. To test for surjectivity (onto), take a general element 'y' from the codomain and try to solve for an input 'x' from the domain such that f(x)=y.

To determine the behavior of the function \(f(x)\), we need to analyze its first derivative, \(f'(x)\).


The given function is \(f(x) = 2x + \cos x\).


Differentiating with respect to \(x\):

\(f'(x) = \frac{d}{dx}(2x + \cos x) = 2 - \sin x\).


We know that the range of the sine function is \([-1, 1]\), which means \(-1 \le \sin x \le 1\) for all real \(x\).


Let's find the range of \(f'(x)\):


The maximum value of \(\sin x\) is 1, so the minimum value of \(f'(x)\) is \(2 - 1 = 1\).


The minimum value of \(\sin x\) is -1, so the maximum value of \(f'(x)\) is \(2 - (-1) = 3\).


Thus, \(1 \le f'(x) \le 3\) for all \(x \in R\).


Since \(f'(x) > 0\) for all real values of \(x\), the function \(f(x)\) is a strictly increasing function.
Quick Tip: To check if a function is increasing or decreasing, find its first derivative. If \(f'(x) > 0\) for all \(x\) in an interval, the function is increasing on that interval. If \(f'(x) < 0\), it is decreasing.

Let the vertex A be the origin. Then the position vectors of B and C are given by the vectors for the sides AB and AC.

\(\vec{AB} = \hat{j} + \hat{k}\)

\(\vec{AC} = 3\hat{i} - \hat{j} + 4\hat{k}\)


The median through A will connect A to the midpoint of the side BC. Let's call this midpoint D.


The position vector of the midpoint D is the average of the position vectors of B and C.

\(\vec{AD} = \frac{\vec{AB} + \vec{AC}}{2}\)

\(\vec{AD} = \frac{(\hat{j} + \hat{k}) + (3\hat{i} - \hat{j} + 4\hat{k})}{2}\)

\(\vec{AD} = \frac{3\hat{i} + (1-1)\hat{j} + (1+4)\hat{k}}{2} = \frac{3\hat{i} + 0\hat{j} + 5\hat{k}}{2} = \frac{3}{2}\hat{i} + \frac{5}{2}\hat{k}\).


The length of the median AD is the magnitude of the vector \(\vec{AD}\).

\(|\vec{AD}| = \sqrt{(\frac{3}{2})^2 + (0)^2 + (\frac{5}{2})^2}\)

\(|\vec{AD}| = \sqrt{\frac{9}{4} + \frac{25}{4}} = \sqrt{\frac{34}{4}} = \frac{\sqrt{34}}{\sqrt{4}} = \frac{\sqrt{34}}{2}\).


So, the length of the median is \(\frac{\sqrt{34}}{2}\) units.
Quick Tip: The vector representing the median from vertex A to the midpoint of side BC in a triangle ABC is given by \(\frac{\vec{AB} + \vec{AC}}{2}\). Its length is the magnitude of this resulting vector.

The function \(f(x)\) is a piecewise function. Potential points of discontinuity occur where the definition of the function changes. In this case, that point is \(x=1\).


For all \(x < 1\), \(f(x) = x\) is a polynomial, which is continuous.


For all \(x > 1\), \(f(x) = 5\) is a constant function, which is continuous.


We must check for continuity at \(x=1\). A function is continuous at \(x=c\) if the Left-Hand Limit (LHL) equals the Right-Hand Limit (RHL) and equals the function's value at that point.


Calculate the Left-Hand Limit (LHL) at \(x=1\):

LHL = \(\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x) = 1\).


Calculate the Right-Hand Limit (RHL) at \(x=1\):

RHL = \(\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (5) = 5\).


Since LHL \(\ne\) RHL (\(1 \ne 5\)), the limit of \(f(x)\) as \(x\) approaches 1 does not exist.


Therefore, the function \(f(x)\) is not continuous at \(x=1\).
Quick Tip: For piecewise functions, always check for continuity at the boundary points where the function rule changes. The function will be discontinuous if the left-hand limit and right-hand limit are not equal at that point.

This integral is in a standard form that can be solved using a specific property.


The property states that \(\int e^x [f(x) + f'(x)] dx = e^x f(x) + C\).


Let's examine the integrand \(e^x (\cos x - \sin x)\).


Let's try setting \(f(x) = \cos x\).


Now, we find the derivative of \(f(x)\):

\(f'(x) = \frac{d}{dx}(\cos x) = -\sin x\).


The integrand can be written as \(e^x [\cos x + (-\sin x)]\), which matches the form \(e^x [f(x) + f'(x)]\).


Applying the property, the integral is:

\(\int e^x (\cos x - \sin x) dx = e^x \cos x + C\).
Quick Tip: Always look for the pattern \(\int e^x [f(x) + f'(x)] dx\) when you see an integral involving \(e^x\) multiplied by a sum or difference of functions. This can save a lot of time compared to using integration by parts.

The area of the region bounded by a curve \(y = f(x)\), the x-axis, and the lines \(x=a\) and \(x=b\) is given by the definite integral \(A = \int_{a}^{b} y \,dx\).


In this case, the curve is \(y = \sqrt{x}\), and the boundaries are from \(a=0\) to \(b=4\).


So, the area is \(A = \int_{0}^{4} \sqrt{x} \,dx\).


First, rewrite the integrand with a power:

\(A = \int_{0}^{4} x^{1/2} \,dx\).


Now, apply the power rule for integration \(\int x^n \,dx = \frac{x^{n+1}}{n+1}\):

\(A = \left[ \frac{x^{1/2 + 1}}{1/2 + 1} \right]_{0}^{4} = \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{4} = \left[ \frac{2}{3}x^{3/2} \right]_{0}^{4}\).


Now, evaluate the definite integral by substituting the limits:

\(A = \frac{2}{3}(4^{3/2}) - \frac{2}{3}(0^{3/2})\).

\(A = \frac{2}{3}((\sqrt{4})^3) - 0 = \frac{2}{3}(2^3) = \frac{2}{3}(8)\).

\(A = \frac{16}{3}\).


The area is \(\frac{16}{3}\) square units.
Quick Tip: When calculating area under a curve, set up the definite integral with the correct limits. Remember that \(x^{a/b} = (\sqrt[b]{x})^a\), which is often easier to compute than \(\sqrt[b]{x^a}\).

First, we need to arrange the given differential equation into the standard linear form: \(\frac{dy}{dx} + P(x)y = Q(x)\).


The given equation is \(\frac{dy}{dx} + y \tan x - \sec x = 0\).


Rearranging it, we get:

\(\frac{dy}{dx} + (\tan x)y = \sec x\).


This is a linear differential equation where \(P(x) = \tan x\) and \(Q(x) = \sec x\).


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


Substitute \(P(x) = \tan x\) into the formula:

\(I.F. = e^{\int \tan x dx}\).


The integral of \(\tan x\) is \(\ln|\sec x|\).

\(I.F. = e^{\ln|\sec x|}\).


Using the property that \(e^{\ln a} = a\), we get:

\(I.F. = |\sec x|\).


In the context of finding an integrating factor, we typically choose the positive value, so \(I.F. = \sec x\).
Quick Tip: To solve a first-order linear differential equation, first put it in the standard form \(\frac{dy}{dx} + P(x)y = Q(x)\). Then, calculate the integrating factor \(I.F. = e^{\int P(x) dx}\). The solution is then given by \(y \cdot (I.F.) = \int Q(x) \cdot (I.F.) \,dx + C\).

The objective function is given by Z = ax + by, with a > 0 and b > 0.


It is stated that the maximum value of the objective function Z occurs at two distinct corner points: (0, 2) and (3, 0).


This implies that the value of Z at both these points is the same, and this value is the maximum value.


Let's calculate the value of Z at the point (0, 2):

\(Z_1 = a(0) + b(2) = 2b\).


Now, let's calculate the value of Z at the point (3, 0):

\(Z_2 = a(3) + b(0) = 3a\).


Since the maximum value occurs at both points, their Z values must be equal:

\(Z_1 = Z_2\).

\(2b = 3a\).


This gives the relation between a and b as \(3a = 2b\).
Quick Tip: If the optimal value (maximum or minimum) of an LPP objective function occurs at more than one corner point, then it must occur at every point on the line segment connecting these two points. For this to happen, the value of the objective function at these corner points must be equal.

Let the given integral be \(I = \int_{0}^{1} \frac{dx}{e^x + e^{-x}}\).


First, simplify the integrand by writing \(e^{-x}\) as \(\frac{1}{e^x}\):

\(I = \int_{0}^{1} \frac{dx}{e^x + \frac{1}{e^x}} = \int_{0}^{1} \frac{dx}{\frac{(e^x)^2 + 1}{e^x}}\).

\(I = \int_{0}^{1} \frac{e^x}{(e^x)^2 + 1} dx\).


Now, we use the method of substitution. Let \(u = e^x\).


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


We also need to change the limits of integration:


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


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


Substituting \(u\) and \(du\) and the new limits into the integral:

\(I = \int_{1}^{e} \frac{du}{u^2 + 1}\).


The integral of \(\frac{1}{u^2+1}\) is \(\tan^{-1}(u)\).

\(I = [\tan^{-1}(u)]_{1}^{e}\).


Now, apply the limits:

\(I = \tan^{-1}(e) - \tan^{-1}(1)\).


We know that \(\tan^{-1}(1) = \frac{\pi}{4}\).


So, \(I = \tan^{-1}(e) - \frac{\pi}{4}\).
Quick Tip: Integrals involving \(e^x\) and \(e^{-x}\) in the denominator can often be simplified by multiplying the numerator and denominator by \(e^x\). This usually leads to a form that is solvable by substitution.

Analysis of Assertion (A):


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


The condition given in the assertion is \(P(A \cap B) = 0\). This means the events A and B are mutually exclusive.


If we assume A and B are non-impossible events (i.e., \(P(A) > 0\) and \(P(B) > 0\)), then \(P(A) \cdot P(B) > 0\).


In this case, \(P(A \cap B) = 0\) while \(P(A) \cdot P(B) > 0\), so \(P(A \cap B) \ne P(A) \cdot P(B)\).


Therefore, mutually exclusive events (with non-zero probabilities) are dependent, not independent.


Thus, Assertion (A) is false.


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. This is a true statement.


Conclusion:


Since Assertion (A) is false and Reason (R) is true, the correct option is (D).
Quick Tip: Do not confuse mutually exclusive events with independent events. Mutually exclusive means they cannot happen at the same time (\(P(A \cap B) = 0\)). Independent means the outcome of one does not influence the outcome of the other (\(P(A \cap B) = P(A)P(B)\)).

Analysis of Assertion (A):


The solution to a Linear Programming Problem (LPP), if it exists, must be a point within the feasible region.


An empty feasible region means there are no points that satisfy all the given constraints simultaneously.


If there are no such points, then there can be no solution.


Therefore, Assertion (A) is true.


Analysis of Reason (R):


The definition of a feasible region in an LPP is the set of all points that satisfy all the constraints of the problem, including the non-negativity constraints (\(x \ge 0, y \ge 0\)).


This statement is the correct definition of a feasible region.


Therefore, Reason (R) is true.


Relationship between (A) and (R):


Assertion (A) is a direct consequence of the definition given in Reason (R). Because the solution must lie in the feasible region (as defined in R), it logically follows that if this region is empty, no solution exists.


Thus, (R) is the correct explanation for (A).
Quick Tip: An LPP can result in one of three outcomes: a unique optimal solution, multiple optimal solutions, an unbounded solution, or no solution. "No solution" occurs when the feasible region is empty. "Unbounded solution" occurs when the feasible region is unbounded and the objective function can be increased/decreased indefinitely.

The given pair of linear equations is:

\(5x - ky = 2\)

\(7x - 5y = 3\)


We can represent this system in the matrix form \(AX = B\), where:

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


A system of linear equations has a unique solution if and only if the determinant of the coefficient matrix A is non-zero.


So, the condition for a unique solution is \(|A| \ne 0\).


Let's calculate the determinant of A:

\(|A| = \begin{vmatrix} 5 & -k
7 & -5 \end{vmatrix}\)

\(|A| = (5)(-5) - (-k)(7)\)

\(|A| = -25 - (-7k)\)

\(|A| = -25 + 7k\).


Now, apply the condition \(|A| \ne 0\):

\(-25 + 7k \ne 0\).

\(7k \ne 25\).

\(k \ne \frac{25}{7}\).


Therefore, the pair of equations has a unique solution for all real values of \(k\) except for \(k = \frac{25}{7}\).
Quick Tip: For a system of two linear equations \(a_1x + b_1y = c_1\) and \(a_2x + b_2y = c_2\): Unique solution: \(\frac{a_1}{a_2} \ne \frac{b_1}{b_2}\) (which is equivalent to the determinant being non-zero). Infinitely many solutions: \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\). No solution: \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}\).

Let the given expression be \(y = \sin^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)\).


We can simplify the expression inside the inverse sine function using a trigonometric substitution.


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


Now substitute \(x = \tan \theta\) into the expression:

\(y = \sin^{-1}\left(\frac{\tan \theta}{\sqrt{1+\tan^2 \theta}}\right)\).


Using the trigonometric identity \(1 + \tan^2 \theta = \sec^2 \theta\):

\(y = \sin^{-1}\left(\frac{\tan \theta}{\sqrt{\sec^2 \theta}}\right) = \sin^{-1}\left(\frac{\tan \theta}{\sec \theta}\right)\).


Now, express \(\tan \theta\) and \(\sec \theta\) in terms of \(\sin \theta\) and \(\cos \theta\):

\(y = \sin^{-1}\left(\frac{\sin \theta / \cos \theta}{1 / \cos \theta}\right) = \sin^{-1}(\sin \theta)\).


Assuming \(\theta\) is in the principal value branch of \(\sin^{-1}\), we have:

\(y = \theta\).


Finally, substitute back the value of \(\theta\) in terms of \(x\):

\(y = \tan^{-1} x\).


Thus, the simplified form is \(\tan^{-1} x\).
Quick Tip: Expressions of the form \(\sqrt{a^2+x^2}\) often suggest the substitution \(x = a \tan \theta\). Similarly, for \(\sqrt{a^2-x^2}\) use \(x = a \sin \theta\), and for \(\sqrt{x^2-a^2}\) use \(x = a \sec \theta\).

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


The domain of the inverse sine function, \(\sin^{-1}(u)\), is defined for \(-1 \le u \le 1\).


In this case, the argument is \(u = \sqrt{x-1}\). So, we must satisfy the inequality:

\(-1 \le \sqrt{x-1} \le 1\).


This inequality can be split into two conditions:


1. The expression inside the square root must be non-negative for the square root to be a real number.
\(x - 1 \ge 0 \implies x \ge 1\).


2. The value of the square root must be between -1 and 1. Since the principal square root is always non-negative, the condition \(\sqrt{x-1} \ge -1\) is always true when the root is defined. We only need to consider:
\(\sqrt{x-1} \le 1\).


Squaring both sides of this inequality (which is valid as both sides are non-negative):
\(x - 1 \le 1^2 \implies x - 1 \le 1 \implies x \le 2\).


To find the domain of the function, we must satisfy both conditions simultaneously.


We need \(x \ge 1\) and \(x \le 2\).


Combining these, we get \(1 \le x \le 2\).


Therefore, the domain of the function is the interval \([1, 2]\).
Quick Tip: When finding the domain of a composite function, work from the outside in. First, find the domain requirements of the outer function (here, \(\sin^{-1}u\) requires \(-1 \le u \le 1\)), and then apply those constraints to the inner function. Also, always remember to check the domain of the inner function itself (here, \(\sqrt{...}\) requires its argument to be non-negative).

The given equation \(\frac{x^2}{9} + \frac{y^2}{4} = 1\) represents an ellipse centered at the origin.


The semi-major axis is \(a = \sqrt{9} = 3\) along the x-axis, and the semi-minor axis is \(b = \sqrt{4} = 2\) along the y-axis.


The region bounded by the ellipse and the x-axis corresponds to the area of the upper semi-ellipse, which extends from \(x = -3\) to \(x = 3\).


First, we express \(y\) in terms of \(x\) for the upper half of the ellipse (where \(y \ge 0\)).

\(\frac{y^2}{4} = 1 - \frac{x^2}{9} \implies y^2 = 4\left(1 - \frac{x^2}{9}\right) = \frac{4}{9}(9 - x^2)\).

\(y = \frac{2}{3}\sqrt{9 - x^2}\).


The area \(A\) is given by the definite integral:

\(A = \int_{-3}^{3} y \,dx = \int_{-3}^{3} \frac{2}{3}\sqrt{9 - x^2} \,dx\).


Using the standard integral formula \(\int \sqrt{a^2 - x^2} \,dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)\), with \(a=3\):

\(A = \frac{2}{3} \left[ \frac{x}{2}\sqrt{9 - x^2} + \frac{9}{2}\sin^{-1}\left(\frac{x}{3}\right) \right]_{-3}^{3}\).


Now, evaluate at the limits:


At the upper limit (\(x=3\)):
\(\frac{2}{3} \left[ \frac{3}{2}\sqrt{9 - 9} + \frac{9}{2}\sin^{-1}\left(\frac{3}{3}\right) \right] = \frac{2}{3} \left[ 0 + \frac{9}{2}\sin^{-1}(1) \right] = \frac{2}{3} \left( \frac{9}{2} \cdot \frac{\pi}{2} \right) = \frac{3\pi}{2}\).


At the lower limit (\(x=-3\)):
\(\frac{2}{3} \left[ \frac{-3}{2}\sqrt{9 - 9} + \frac{9}{2}\sin^{-1}\left(\frac{-3}{3}\right) \right] = \frac{2}{3} \left[ 0 + \frac{9}{2}\sin^{-1}(-1) \right] = \frac{2}{3} \left( \frac{9}{2} \cdot \left(-\frac{\pi}{2}\right) \right) = -\frac{3\pi}{2}\).


Total Area = (Value at upper limit) - (Value at lower limit)

\(A = \frac{3\pi}{2} - \left(-\frac{3\pi}{2}\right) = \frac{3\pi}{2} + \frac{3\pi}{2} = 3\pi\).


The area of the region is \(3\pi\) square units.
Quick Tip: The area of a full ellipse with equation \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) is given by the formula \(A = \pi ab\). In this case, the area bounded by the ellipse and the x-axis is the area of a semi-ellipse, which is \(\frac{1}{2}\pi ab = \frac{1}{2}\pi(3)(2) = 3\pi\). This can be used to quickly verify the result from integration.

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


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


First, we find the derivative:
\(f'(x) = \frac{d}{dx}(2x^2 - ax + 3) = 4x - a\).


The condition is that \(f'(x) \ge 0\) for all \(x \in [2, 4]\).
\(4x - a \ge 0 \implies a \le 4x\).


This inequality must hold for every value of \(x\) in the interval \([2, 4]\). Therefore, 'a' must be less than or equal to the minimum value of \(4x\) on this interval.


The expression \(4x\) is an increasing function of \(x\). Its minimum value on the closed interval \([2, 4]\) occurs at the leftmost point, \(x=2\).


Minimum value of \(4x\) on \([2, 4]\) is \(4(2) = 8\).


So, the condition becomes \(a \le 8\).


The set of all possible values for 'a' is \((-\infty, 8]\). This set has a greatest value, which is 8, but no least value. The question asks for the "least value", which is likely a misstatement for the "greatest value" or the "boundary value". The greatest value of 'a' for which the condition holds is 8.


Therefore, the required value of 'a' is 8.
Quick Tip: To ensure a function is increasing over an interval, find the derivative and set the condition \(f'(x) \ge 0\). If \(f'(x)\) is itself monotonic (like the linear function \(4x-a\)), you only need to check the condition at the endpoint where \(f'(x)\) has its minimum value in the interval.

To show that the function \(f(x)\) is increasing on the interval \([1, \infty)\), we need to show that its derivative, \(f'(x)\), is greater than or equal to zero for all \(x \ge 1\).


The function is \(f(x) = x + \frac{1}{x} = x + x^{-1}\).


Differentiating with respect to \(x\):
\(f'(x) = \frac{d}{dx}(x + x^{-1}) = 1 - 1 \cdot x^{-2} = 1 - \frac{1}{x^2}\).


Now, we need to analyze the sign of \(f'(x)\) for the given domain \(x \ge 1\).


Case 1: \(x = 1\).
\(f'(1) = 1 - \frac{1}{1^2} = 1 - 1 = 0\).


Case 2: \(x > 1\).

If \(x > 1\), then \(x^2 > 1\).

Dividing by \(x^2\) (a positive number), we get \(1 > \frac{1}{x^2}\).

Also, since \(x^2\) is positive, \(\frac{1}{x^2}\) is positive. So, \(0 < \frac{1}{x^2} < 1\).

This means that \(f'(x) = 1 - \frac{1}{x^2}\) will be \(1\) minus a positive number less than 1, which results in a positive value.

Thus, \(f'(x) > 0\) for all \(x > 1\).


Combining both cases, we have \(f'(x) \ge 0\) for all \(x \ge 1\).

Since the derivative is non-negative on the interval \([1, \infty)\), the function \(f(x)\) is an increasing function on this interval.
Quick Tip: To prove a function is increasing or decreasing on an interval, the most direct method is to analyze the sign of its first derivative over that interval. If \(f'(x) \ge 0\), it is increasing; if \(f'(x) \le 0\), it is decreasing.

To find local maxima and minima, we use the first and second derivative tests.


Step 1: Find the first derivative and the critical points.
\(f(x) = \frac{8}{3}x^3 - 12x^2 + 18x + 5\).
\(f'(x) = \frac{d}{dx}f(x) = \frac{8}{3}(3x^2) - 12(2x) + 18 = 8x^2 - 24x + 18\).


Set \(f'(x) = 0\) to find the critical points:
\(8x^2 - 24x + 18 = 0\).

Dividing the equation by 2 gives:
\(4x^2 - 12x + 9 = 0\).

This is a perfect square trinomial: \((2x - 3)^2 = 0\).

This gives a single critical point at \(x = \frac{3}{2}\).


Step 2: Use the second derivative test to classify the critical point.

Find the second derivative, \(f''(x)\):
\(f''(x) = \frac{d}{dx}(8x^2 - 24x + 18) = 16x - 24\).


Evaluate \(f''(x)\) at the critical point \(x = \frac{3}{2}\):
\(f''\left(\frac{3}{2}\right) = 16\left(\frac{3}{2}\right) - 24 = 24 - 24 = 0\).


Since \(f''(x) = 0\), the second derivative test is inconclusive. We must use the first derivative test.


Step 3: Use the first derivative test.

We check the sign of \(f'(x) = 2(2x-3)^2\) around the critical point \(x = \frac{3}{2}\).

Since \((2x-3)^2\) is a square, its value is always non-negative.

For any value of \(x\) less than \(\frac{3}{2}\), \(f'(x)\) is positive.

For any value of \(x\) greater than \(\frac{3}{2}\), \(f'(x)\) is also positive.


Since the sign of the first derivative does not change as it passes through the critical point \(x = \frac{3}{2}\), this point is neither a local maximum nor a local minimum. It is a point of inflection.


Therefore, the function has no local maxima and no local minima.
Quick Tip: If the second derivative test results in \(f''(c)=0\) at a critical point \(c\), the test is inconclusive. You must then revert to the first derivative test. If the sign of \(f'(x)\) does not change across \(c\), it's a point of inflection. If it changes from + to -, it's a local maximum. If it changes from - to +, it's a local minimum.

Let 'B' denote a boy and 'G' denote a girl. The probability of having a boy is \(P(B) = \frac{1}{2}\) and a girl is \(P(G) = \frac{1}{2}\).


For a family with three children, the sample space of possible outcomes is:

S = \{BBB, BBG, BGB, GBB, GGG, GGB, GBG, BGG\.


The total number of outcomes is \(2^3 = 8\). Each outcome has a probability of \((\frac{1}{2})^3 = \frac{1}{8}\).


Let X be the random variable representing the number of boys. The possible values for X are 0, 1, 2, 3.


We calculate the probability for each value of X:

\(P(X=0)\) (no boys, i.e., GGG):
\(P(X=0) = P(GGG) = \frac{1}{8}\).

\(P(X=1)\) (one boy, i.e., GGB, GBG, BGG):

There are 3 such outcomes. \(P(X=1) = 3 \times \frac{1}{8} = \frac{3}{8}\).

\(P(X=2)\) (two boys, i.e., BBG, BGB, GBB):

There are 3 such outcomes. \(P(X=2) = 3 \times \frac{1}{8} = \frac{3}{8}\).

\(P(X=3)\) (three boys, i.e., BBB):
\(P(X=3) = P(BBB) = \frac{1}{8}\).


The probability distribution of X is summarized in the table below:

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

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

\hline
\end{tabular
Quick Tip: This is a classic example of a binomial distribution with \(n=3\) (number of trials) and \(p=1/2\) (probability of success, i.e., having a boy). The probability of k successes is given by \(P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}\).

When a coin is tossed twice, the sample space is S = \{HH, HT, TH, TT\.


Let X be the random variable defined as (number of heads) - (number of tails).


We determine the value of X for each outcome:

For HH: Number of heads = 2, Number of tails = 0. So, \(X = 2 - 0 = 2\).

For HT: Number of heads = 1, Number of tails = 1. So, \(X = 1 - 1 = 0\).

For TH: Number of heads = 1, Number of tails = 1. So, \(X = 1 - 1 = 0\).

For TT: Number of heads = 0, Number of tails = 2. So, \(X = 0 - 2 = -2\).


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


Now we find the probabilities for each value of X:
\(P(X = -2) = P(TT) = \frac{1}{4}\).
\(P(X = 0) = P(HT or TH) = \frac{2}{4} = \frac{1}{2}\).
\(P(X = 2) = P(HH) = \frac{1}{4}\).


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


The mean (or expected value) of X, denoted E(X), is calculated as \(\sum x_i P(x_i)\).
\(E(X) = (-2)\left(\frac{1}{4}\right) + (0)\left(\frac{1}{2}\right) + (2)\left(\frac{1}{4}\right)\).
\(E(X) = -\frac{2}{4} + 0 + \frac{2}{4} = 0\).


Thus, the mean of X is 0.
Quick Tip: The mean of a random variable, also known as its expected value, represents the long-term average value of the variable. It's calculated by summing the product of each possible value and its corresponding probability.

To prove that the function \(f(x) = \log_a x\) is a bijection, we must show that it is both injective (one-to-one) and surjective (onto).


1. Injective (One-to-one):

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

Assume that \(f(x_1) = f(x_2)\).

This means \(\log_a x_1 = \log_a x_2\).

By the property of logarithms, if \(\log_a m = \log_a n\), then \(m=n\).

Alternatively, raising 'a' to the power of both sides: \(a^{\log_a x_1} = a^{\log_a x_2}\).

This simplifies to \(x_1 = x_2\).

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


2. Surjective (Onto):

Let \(y\) be an arbitrary element in the codomain \(R\). We need to show that there exists an element \(x\) in the domain \(R^+\) such that \(f(x) = y\).

Let \(f(x) = y\).
\(\log_a x = y\).

Using the definition of a logarithm, we can rewrite this in exponential form as \(x = a^y\).

Since \(a > 0\), the value of \(a^y\) is always a positive real number for any real number \(y\).

Therefore, \(x = a^y \in R^+\).

This means that for every \(y\) in the codomain \(R\), there exists a pre-image \(x = a^y\) in the domain \(R^+\).

Hence, the function \(f\) is surjective.


Since the function \(f(x) = \log_a x\) is both injective and surjective, it is a bijection.
Quick Tip: A bijection is a function that is both one-to-one and onto. To prove a function is one-to-one, show that \(f(x_1) = f(x_2) \implies x_1 = x_2\). To prove it is onto, show that for any \(y\) in the codomain, there exists an \(x\) in the domain such that \(f(x) = y\).

Given sets A = \{1, 2, 3\, B = \{4, 5, 6\ and the relation R = \{(x, y) : x + y = 6, x \(\in\) A, y \(\in\) B\.


(i) Write all elements of R.

We test each element of A to find a corresponding element in B that satisfies the condition \(x+y=6\).

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

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

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

Therefore, the elements of R are: R = \{(1, 5), (2, 4)\.


(ii) Is R a function ? Justify.

A relation R from set A to set B is a function if every element in the domain A has exactly one image in the codomain B.

In this case, the element 3, which is in set A, does not have any image in set B under the relation R.

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


(iii) Determine domain and range of R.

The domain of a relation is the set of all first components of the ordered pairs in the relation.

Domain(R) = \{1, 2\.

The range of a relation is the set of all second components of the ordered pairs in the relation.

Range(R) = \{4, 5\.
Quick Tip: For a relation from A to B to be a function, two conditions must be met: 1) Every element in A must have an image in B. 2) No element in A can have more than one image in B. The domain of a relation is a subset of A, and the range is a subset of B.

Let the integral be \(I = \int \frac{\cos x}{1+\cos x+\sin x} dx\).


We can express the numerator in terms of the denominator and its derivative.

Let Numerator = \(A(Denominator) + B(Derivative of Denominator) + C\).
\(\cos x = A(1+\cos x+\sin x) + B(-\sin x+\cos x)\).

The constant C is not needed here if we manipulate the integral differently. A more direct method is:

Let's write \(2\cos x = (\cos x - \sin x) + (\cos x + \sin x)\).

Adding and subtracting 1, we get \(2\cos x = (\cos x - \sin x) + (1 + \cos x + \sin x) - 1\).

So, \(\cos x = \frac{1}{2}(\cos x - \sin x) + \frac{1}{2}(1 + \cos x + \sin x) - \frac{1}{2}\).


Now, substitute this into the integral:
\(I = \int \frac{\frac{1}{2}(\cos x - \sin x) + \frac{1}{2}(1 + \cos x + \sin x) - \frac{1}{2}}{1+\cos x+\sin x} dx\).


Split the integral into three parts:
\(I = \frac{1}{2} \int \frac{\cos x - \sin x}{1+\cos x+\sin x} dx + \frac{1}{2} \int \frac{1+\cos x+\sin x}{1+\cos x+\sin x} dx - \frac{1}{2} \int \frac{1}{1+\cos x+\sin x} dx\).

\(I = \frac{1}{2} \int \frac{\cos x - \sin x}{1+\cos x+\sin x} dx + \frac{1}{2} \int 1 \,dx - \frac{1}{2} \int \frac{1}{1+\cos x+\sin x} dx\).


The first integral is of the form \(\int \frac{f'(t)}{f(t)} dt = \ln|f(t)|\). Here, the derivative of the denominator is \(\cos x - \sin x\).

So, the first part is \(\frac{1}{2} \ln|1+\cos x+\sin x|\).


The second part is \(\frac{1}{2}x\).


For the third part, we use half-angle formulas: \(1+\cos x = 2\cos^2(x/2)\) and \(\sin x = 2\sin(x/2)\cos(x/2)\).
\(\int \frac{1}{2\cos^2(x/2) + 2\sin(x/2)\cos(x/2)} dx = \int \frac{1}{2\cos(x/2)(\cos(x/2)+\sin(x/2))} dx\).

Divide numerator and denominator by \(\cos^2(x/2)\):
\(= \int \frac{\frac{1}{2}\sec^2(x/2)}{1+\tan(x/2)} dx\).

Let \(u = 1+\tan(x/2)\), then \(du = \frac{1}{2}\sec^2(x/2) dx\). The integral becomes \(\int \frac{du}{u} = \ln|u| = \ln|1+\tan(x/2)|\).


Combining all parts:
\(I = \frac{1}{2}\ln|1+\cos x+\sin x| + \frac{x}{2} - \frac{1}{2}\ln|1+\tan(x/2)| + C\).
Quick Tip: For integrals of the form \(\int \frac{a\cos x + b\sin x}{c\cos x + d\sin x + e} dx\), a standard method is to express the numerator as \(N = A(D) + B(D') + C\), where D is the denominator and D' is its derivative.

The sample space S for this experiment consists of outcomes from two stages.

Stage 1: Toss a coin. Outcomes are H (Head) or T (Tail), with \(P(H)=P(T)=1/2\).

Stage 2: If H, toss again. Outcomes are HH, HT. So \(P(HH)=P(H)\times P(H)=1/4\), \(P(HT)=1/4\).

If T, throw a die. Outcomes are T1, T2, T3, T4, T5, T6. \(P(Ti) = P(T) \times P(i) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}\).


The complete sample space is S = \{HH, HT, T1, T2, T3, T4, T5, T6\.


Event A: 'the die shows a number greater than 3'.

A = \{T4, T5, T6\.
\(P(A) = P(T4) + P(T5) + P(T6) = 3 \times \frac{1}{12} = \frac{3}{12} = \frac{1}{4}\).


Event B: 'there is at least one tail'.

B = \{HT, T1, T2, T3, T4, T5, T6\.
\(P(B) = P(HT) + P(\{T1, ..., T6\}) = \frac{1}{4} + 6 \times \frac{1}{12} = \frac{1}{4} + \frac{1}{2} = \frac{3}{4}\).


We need to find the conditional probability \(P(A|B) = \frac{P(A \cap B)}{P(B)}\).


First, find the intersection \(A \cap B\):
\(A \cap B\) is the event where the die shows a number greater than 3 AND there is at least one tail.

This is exactly the event A itself, since A can only occur if a tail appeared first.

So, \(A \cap B = A = \{T4, T5, T6\}\).
\(P(A \cap B) = P(A) = \frac{1}{4}\).


Now, calculate \(P(A|B)\):
\(P(A|B) = \frac{1/4}{3/4} = \frac{1}{3}\).
Quick Tip: The formula for conditional probability is \(P(A|B) = \frac{P(A \cap B)}{P(B)}\). Always clearly define the events A, B, and their intersection based on the sample space before calculating probabilities.

(i) Calculate \(\lambda\), if E(X) = 3.2.

The expected value (mean) of a random variable is given by \(E(X) = \sum x_i P(x_i)\).

Given \(E(X) = 3.2 = \frac{32}{10}\).
\(E(X) = 1(\frac{11}{30}) + 2(\frac{1}{15}) + 3(\frac{1}{10}) + 2\lambda(\frac{3}{10}) + 3\lambda(\frac{1}{15}) + 4\lambda(\frac{1}{10})\).


Let's simplify the expression:
\(\frac{32}{10} = \frac{11}{30} + \frac{2}{15} + \frac{3}{10} + \frac{6\lambda}{10} + \frac{3\lambda}{15} + \frac{4\lambda}{10}\).

To work with a common denominator of 30:
\(\frac{96}{30} = \frac{11}{30} + \frac{4}{30} + \frac{9}{30} + \frac{18\lambda}{30} + \frac{6\lambda}{30} + \frac{12\lambda}{30}\).

Multiply the entire equation by 30:
\(96 = (11 + 4 + 9) + (18\lambda + 6\lambda + 12\lambda)\).
\(96 = 24 + 36\lambda\).
\(96 - 24 = 36\lambda\).
\(72 = 36\lambda\).
\(\lambda = \frac{72}{36} = 2\).


(ii) Find P(X > 1).

The event \(X > 1\) includes all outcomes where X is not equal to 1.

We can calculate this using the complement rule: \(P(X > 1) = 1 - P(X=1)\).

From the table, \(P(X=1) = \frac{11}{30}\).
\(P(X > 1) = 1 - \frac{11}{30} = \frac{30 - 11}{30} = \frac{19}{30}\).


Alternatively, sum the probabilities for \(X > 1\):
\(P(X>1) = P(2) + P(3) + P(2\lambda) + P(3\lambda) + P(4\lambda)\)
\(P(X>1) = \frac{1}{15} + \frac{1}{10} + \frac{3}{10} + \frac{1}{15} + \frac{1}{10} = \frac{2}{30} + \frac{3}{30} + \frac{9}{30} + \frac{2}{30} + \frac{3}{30} = \frac{19}{30}\).
Quick Tip: For a valid probability distribution, the sum of all probabilities must equal 1. This can be used as a check or sometimes to find an unknown. The mean is the weighted average of the values, where the weights are the probabilities.

Let the two lines be \(L_1\) and \(L_2\).
\(L_1: \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4} = s\) (say).

A general point on \(L_1\) is \(P_1 = (2s+1, 3s+2, 4s+3)\).

\(L_2: \frac{x-4}{5} = \frac{y-1}{2} = \frac{z-0}{1} = t\) (say).

A general point on \(L_2\) is \(P_2 = (5t+4, 2t+1, t)\).


At the point of intersection, \(P_1 = P_2\). We equate the corresponding coordinates:

(i) \(2s + 1 = 5t + 4 \implies 2s - 5t = 3\).

(ii) \(3s + 2 = 2t + 1 \implies 3s - 2t = -1\).

(iii) \(4s + 3 = t\).


We can solve this system of equations. Substitute equation (iii) into (ii):
\(3s - 2(4s + 3) = -1\).
\(3s - 8s - 6 = -1\).
\(-5s = 5 \implies s = -1\).


Now substitute \(s = -1\) back into equation (iii) to find \(t\):
\(t = 4(-1) + 3 = -1\).


We must verify these values in the first equation (i):
\(2(-1) - 5(-1) = -2 + 5 = 3\). This is correct.

Since the values of \(s\) and \(t\) satisfy all three equations, the lines intersect.


To find the point of intersection, substitute \(s = -1\) into the coordinates for \(P_1\):
\(x = 2(-1) + 1 = -1\).
\(y = 3(-1) + 2 = -1\).
\(z = 4(-1) + 3 = -1\).

The point of intersection is Q = (-1, -1, -1).


Now, we find the distance between the given point P = (-1, -5, -10) and the point of intersection Q = (-1, -1, -1) using the distance formula:

Distance \(PQ = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}\).
\(PQ = \sqrt{(-1 - (-1))^2 + (-1 - (-5))^2 + (-1 - (-10))^2}\).
\(PQ = \sqrt{(0)^2 + (4)^2 + (9)^2}\).
\(PQ = \sqrt{0 + 16 + 81} = \sqrt{97}\).


The required distance is \(\sqrt{97}\) units.
Quick Tip: To find the intersection of two lines in 3D, write the general coordinates for a point on each line using different parameters (e.g., s and t). Equate the coordinates to get a system of three linear equations. If a consistent solution for the parameters exists, the lines intersect.

The problem is to minimise \(Z = 3x + 5y\) subject to the given constraints.

First, we plot the lines corresponding to the constraints:
\(L_1: x + 2y = 10\) (passes through (10, 0) and (0, 5)).
\(L_2: x + y = 6\) (passes through (6, 0) and (0, 6)).
\(L_3: 3x + y = 8\) (passes through (8/3, 0) and (0, 8)).


Since the inequalities are all '\(\ge\)', the feasible region is the area on or above all these lines in the first quadrant (due to \(x, y \ge 0\)). This region is unbounded.


Next, we find the corner points of the feasible region by finding the intersection of these lines.

Intersection of \(L_1\) and \(L_2\):
\(x + 2y = 10\) and \(x + y = 6\). Subtracting the second from the first gives \(y = 4\). Substituting back, \(x + 4 = 6 \implies x = 2\). Point is A(2, 4).


Intersection of \(L_2\) and \(L_3\):
\(x + y = 6 \implies y = 6-x\). Substitute into \(L_3\): \(3x + (6-x) = 8 \implies 2x = 2 \implies x = 1\). Then \(y = 6-1=5\). Point is B(1, 5).


The corner points of the feasible region are the points of intersection and the intercepts on the axes. The vertices are A(2, 4), B(1, 5), and the points on the axes C(10, 0) and D(0, 8).


Now, we evaluate the objective function Z at these corner points:

At A(2, 4): \(Z = 3(2) + 5(4) = 6 + 20 = 26\).

At B(1, 5): \(Z = 3(1) + 5(5) = 3 + 25 = 28\).

At C(10, 0): \(Z = 3(10) + 5(0) = 30\).

At D(0, 8): \(Z = 3(0) + 5(8) = 40\).


The minimum value of Z from the corner points is 26, which occurs at A(2, 4).


Since the feasible region is unbounded, we must check if a smaller value of Z is possible. We graph the inequality \(3x + 5y < 26\). This is an open half-plane. The line \(3x+5y=26\) passes through the corner point A(2, 4).


We check if this half-plane has any points in common with the feasible region. All other corner points (B, C, D) result in Z values greater than 26, meaning they lie outside and "above" the line \(3x+5y=26\). The entire feasible region lies on one side of this line.


Therefore, the open half-plane \(3x + 5y < 26\) has no points in common with the feasible region.

The minimum value of Z is 26, which occurs at the point (2, 4).
Quick Tip: For an LPP with an unbounded feasible region, find the minimum (or maximum) value at the corner points. Let this value be M. Then, graph the inequality \(Z < M\) (for minimization) or \(Z > M\) (for maximization). If this open half-plane has no points in common with the feasible region, then M is the optimal solution. Otherwise, no optimal solution exists.

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


(i) Rate of growth


The rate of growth of the plant is the derivative of its height with respect to time (days), which is \(\frac{dy}{dx}\).

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


Differentiating with respect to \(x\):
\(\frac{dy}{dx} = 4 - \frac{1}{2}(2x) = 4 - x\).


So, the rate of growth of the plant is \((4-x)\) cm/day.


(ii) Maximum Height


To find the maximum height, we first need to find the value of \(x\) for which the rate of growth is zero.

Set \(\frac{dy}{dx} = 0\):
\(4 - x = 0 \implies x = 4\).


To confirm that this value of \(x\) corresponds to a maximum, we use the second derivative test.

Find the second derivative:
\(\frac{d^2y}{dx^2} = \frac{d}{dx}(4-x) = -1\).


Since \(\frac{d^2y}{dx^2} = -1 < 0\), the function has a maximum at \(x=4\).


This means the plant will attain its maximum height in 4 days.


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


The maximum height is 8 cm.
Quick Tip: To find the maximum or minimum value of a function, find its first derivative, set it to zero, and solve for the critical points. Then, use the second derivative test to classify them: if \(f''(x) < 0\), it's a maximum; if \(f''(x) > 0\), it's a minimum.

Part 1: Proof


By the definition of an inverse matrix, for any invertible matrix M, we have \(M M^{-1} = I\).


Let \(M = kA\). Then, \((kA)(kA)^{-1} = I\).


Since A is invertible, its inverse \(A^{-1}\) exists. Pre-multiply both sides by \(A^{-1}\):
\(A^{-1}(kA)(kA)^{-1} = A^{-1}I\).


Using the associative property and the property that scalar multiplication is commutative:
\((A^{-1}k A)(kA)^{-1} = A^{-1}\).
\((k A^{-1} A)(kA)^{-1} = A^{-1}\).


Since \(A^{-1}A = I\), the equation becomes:
\((kI)(kA)^{-1} = A^{-1}\).


Pre-multiply both sides by the scalar \(\frac{1}{k}\) (this is possible since \(k \ne 0\)):
\(\frac{1}{k}(kI)(kA)^{-1} = \frac{1}{k}A^{-1}\).

\((1 \cdot I)(kA)^{-1} = \frac{1}{k}A^{-1}\).
\(I(kA)^{-1} = \frac{1}{k}A^{-1}\).
\((kA)^{-1} = \frac{1}{k}A^{-1}\). Hence proved.


Part 2: Calculation of \((3A)^{-1}\)


Using the property we just proved, \((3A)^{-1} = \frac{1}{3}A^{-1}\).

We need to find the inverse of A. The formula is \(A^{-1} = \frac{1}{|A|}adj(A)\).


First, calculate the determinant of A:
\(|A| = 2(2 \cdot 2 - (-1)(-1)) - (-1)(-1 \cdot 2 - (-1) \cdot 1) + 1(-1 \cdot -1 - 2 \cdot 1)\)
\(|A| = 2(4-1) + 1(-2+1) + 1(1-2) = 2(3) + 1(-1) + 1(-1) = 6 - 1 - 1 = 4\).


Next, find the matrix of cofactors:
\(C_{11} = (4-1)=3\), \(C_{12} = -(-2+1)=1\), \(C_{13} = (1-2)=-1\).
\(C_{21} = -(-2+1)=1\), \(C_{22} = (4-1)=3\), \(C_{23} = -(-2+1)=1\).
\(C_{31} = (1-2)=-1\), \(C_{32} = -(-2+1)=1\), \(C_{33} = (4-1)=3\).


Cofactor Matrix \(C = \begin{bmatrix} 3 & 1 & -1
1 & 3 & 1
-1 & 1 & 3 \end{bmatrix}\).


The adjugate of A is the transpose of the cofactor matrix:
\(adj(A) = C^T = \begin{bmatrix} 3 & 1 & -1
1 & 3 & 1
-1 & 1 & 3 \end{bmatrix}\).


Now, find the inverse of A:
\(A^{-1} = \frac{1}{4} \begin{bmatrix} 3 & 1 & -1
1 & 3 & 1
-1 & 1 & 3 \end{bmatrix}\).


Finally, calculate \((3A)^{-1}\):
\((3A)^{-1} = \frac{1}{3}A^{-1} = \frac{1}{3} \cdot \frac{1}{4} \begin{bmatrix} 3 & 1 & -1
1 & 3 & 1
-1 & 1 & 3 \end{bmatrix} = \frac{1}{12} \begin{bmatrix} 3 & 1 & -1
1 & 3 & 1
-1 & 1 & 3 \end{bmatrix}\).
Quick Tip: To calculate the inverse of a 3x3 matrix, follow these steps: 1. Calculate the determinant. If it's zero, the inverse doesn't exist. 2. Calculate the matrix of cofactors. Remember the sign pattern! 3. Transpose the cofactor matrix to get the adjugate. 4. Multiply the adjugate by 1/determinant.

Let the integral be \(I = \int_{0}^{\pi/4} \frac{\sin x \cos x}{\cos^4 x + \sin^4 x} dx\).


To simplify the integrand, divide both the numerator and the denominator by \(\cos^4 x\):
\(I = \int_{0}^{\pi/4} \frac{\frac{\sin x \cos x}{\cos^4 x}}{\frac{\cos^4 x}{\cos^4 x} + \frac{\sin^4 x}{\cos^4 x}} dx = \int_{0}^{\pi/4} \frac{\frac{\sin x}{\cos x} \cdot \frac{1}{\cos^2 x}}{1 + (\frac{\sin x}{\cos x})^4} dx\).
\(I = \int_{0}^{\pi/4} \frac{\tan x \sec^2 x}{1 + \tan^4 x} dx\).


Now, we use substitution. Let \(t = \tan^2 x\).

Then, the differential is \(dt = 2 \tan x \cdot \frac{d}{dx}(\tan x) dx = 2 \tan x \sec^2 x dx\).

This gives \(\tan x \sec^2 x dx = \frac{1}{2} dt\).


Next, we change the limits of integration:

When \(x=0\), \(t = \tan^2(0) = 0\).

When \(x=\pi/4\), \(t = \tan^2(\pi/4) = 1^2 = 1\).


Substitute \(t\) and the new limits into the integral:
\(I = \int_{0}^{1} \frac{1}{1 + t^2} \left(\frac{1}{2} dt\right) = \frac{1}{2} \int_{0}^{1} \frac{1}{1 + t^2} dt\).


This is a standard integral:
\(I = \frac{1}{2} [\tan^{-1}(t)]_{0}^{1}\).


Evaluate the definite integral using the limits:
\(I = \frac{1}{2} (\tan^{-1}(1) - \tan^{-1}(0))\).
\(I = \frac{1}{2} (\frac{\pi}{4} - 0) = \frac{\pi}{8}\).
Quick Tip: When dealing with integrals involving powers of \(\sin x\) and \(\cos x\), a useful strategy is often to divide the numerator and denominator by a suitable power of \(\cos x\) to convert the expression into terms of \(\tan x\) and \(\sec^2 x\), setting up a simple substitution.

Let the integral be \(I = \int \frac{\sqrt{x^2+1}}{x^2} (\log(x^2+1) - 2\log x) dx\).


First, simplify the logarithmic term using the property \(\log m - \log n = \log(m/n)\) and \(k \log m = \log(m^k)\):
\(\log(x^2+1) - 2\log x = \log(x^2+1) - \log(x^2) = \log\left(\frac{x^2+1}{x^2}\right) = \log\left(1+\frac{1}{x^2}\right)\).


The integral becomes:
\(I = \int \frac{\sqrt{x^2+1}}{x^2} \log\left(1+\frac{1}{x^2}\right) dx = \int \sqrt{\frac{x^2+1}{x^4}} \log\left(1+\frac{1}{x^2}\right) dx\). \(I = \int \sqrt{\frac{1}{x^2}+\frac{1}{x^4}} \log\left(1+\frac{1}{x^2}\right) dx\). This is not simpler.

Let's use integration by parts with the form \(I = \int u \, dv\).

Let \(u = \log\left(1+\frac{1}{x^2}\right)\) and \(dv = \frac{\sqrt{x^2+1}}{x^2} dx\).


First, find \(v = \int \frac{\sqrt{x^2+1}}{x^2} dx\). We use integration by parts for this sub-problem.
Let \(u_1 = \sqrt{x^2+1}\) and \(dv_1 = x^{-2}dx\).
Then \(du_1 = \frac{x}{\sqrt{x^2+1}}dx\) and \(v_1 = -x^{-1} = -\frac{1}{x}\). \(v = u_1 v_1 - \int v_1 du_1 = -\frac{\sqrt{x^2+1}}{x} - \int \left(-\frac{1}{x}\right) \frac{x}{\sqrt{x^2+1}} dx\). \(v = -\frac{\sqrt{x^2+1}}{x} + \int \frac{1}{\sqrt{x^2+1}} dx\). \(v = -\frac{\sqrt{x^2+1}}{x} + \log|x+\sqrt{x^2+1}|\).


Now, find \(du\):
\(u = \log\left(1+\frac{1}{x^2}\right) \implies du = \frac{1}{1+1/x^2} \cdot \left(-\frac{2}{x^3}\right) dx = \frac{x^2}{x^2+1} \cdot \left(-\frac{2}{x^3}\right) dx = \frac{-2}{x(x^2+1)} dx\).


Applying the integration by parts formula \(I = uv - \int v du\):

The expression becomes extremely complex and suggests this might not be the intended method for an exam.


Let's try a substitution approach. Let \(t = \sqrt{1+\frac{1}{x^2}}\). \(t^2 = 1+\frac{1}{x^2} \implies \log(t^2) = \log(1+\frac{1}{x^2})\). \(2t\,dt = -\frac{2}{x^3} dx \implies t\,dt = -\frac{1}{x^3} dx\).
The integral is \(\int x \sqrt{1+\frac{1}{x^2}} \log(1+\frac{1}{x^2}) \frac{1}{x^3} dx = \int x \cdot t \cdot \log(t^2) (-t\,dt)\). This is also problematic.


Given the complexity, we recognize this might fit a very specific pattern. Let's consider the function \(F(x) = \frac{2}{3}\left(1+\frac{1}{x^2}\right)^{3/2}\). Its derivative is \(F'(x) = -\frac{2}{x^3}\sqrt{1+\frac{1}{x^2}}\). This suggests the problem is likely beyond standard high-school curriculum or contains a misprint. A plausible intended answer based on similar forms is \(\frac{2}{3}\left(1+\frac{1}{x^2}\right)^{3/2}\left[\log\left(1+\frac{1}{x^2}\right)-\frac{2}{3}\right]+C\).
Quick Tip: When an integral appears overwhelmingly complex, re-examine the expression for potential simplifications or substitutions. Trigonometric substitutions (\(x=\tan\theta, x=\sin\theta\)) are powerful tools when terms like \(\sqrt{x^2+a^2}\) or \(\sqrt{a^2-x^2}\) appear. If all standard methods fail, there might be a misprint in the question.

Part 1: Proof of the Area Formula

Let the adjacent sides of a parallelogram be represented by the vectors \(\vec{p}\) and \(\vec{q}\).

The area of this parallelogram is given by Area = \(|\vec{p} \times \vec{q}|\).


The diagonals of the parallelogram can be represented in terms of its sides:

First diagonal, \(\vec{a} = \vec{p} + \vec{q}\).

Second diagonal, \(\vec{b} = \vec{p} - \vec{q}\).


Now, let's compute the cross product of the diagonals:
\(\vec{a} \times \vec{b} = (\vec{p} + \vec{q}) \times (\vec{p} - \vec{q})\).

Using the distributive property of the cross product:
\(\vec{a} \times \vec{b} = (\vec{p} \times \vec{p}) - (\vec{p} \times \vec{q}) + (\vec{q} \times \vec{p}) - (\vec{q} \times \vec{q})\).


We know that the cross product of any vector with itself is the zero vector (\(\vec{p} \times \vec{p} = \vec{0}\)), and the cross product is anti-commutative (\(\vec{q} \times \vec{p} = -\vec{p} \times \vec{q}\)).
\(\vec{a} \times \vec{b} = \vec{0} - (\vec{p} \times \vec{q}) - (\vec{p} \times \vec{q}) - \vec{0} = -2(\vec{p} \times \vec{q})\).


Taking the magnitude of both sides:
\(|\vec{a} \times \vec{b}| = |-2(\vec{p} \times \vec{q})| = 2|\vec{p} \times \vec{q}|\).


Since Area = \(|\vec{p} \times \vec{q}|\), we have \(|\vec{a} \times \vec{b}| = 2 \times Area\).

Therefore, Area = \(\frac{1}{2}|\vec{a} \times \vec{b}|\). Hence proved.


Part 2: Calculation of Area

The given diagonals are \(\vec{a} = 2\hat{i} - \hat{j} + \hat{k}\) and \(\vec{b} = \hat{i} + 3\hat{j} - \hat{k}\).

First, calculate the cross product \(\vec{a} \times \vec{b}\):
\(\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
2 & -1 & 1
1 & 3 & -1 \end{vmatrix}\).
\(= \hat{i}((-1)(-1) - (1)(3)) - \hat{j}((2)(-1) - (1)(1)) + \hat{k}((2)(3) - (-1)(1))\).
\(= \hat{i}(1 - 3) - \hat{j}(-2 - 1) + \hat{k}(6 + 1) = -2\hat{i} + 3\hat{j} + 7\hat{k}\).


Next, find the magnitude of this vector:
\(|\vec{a} \times \vec{b}| = \sqrt{(-2)^2 + 3^2 + 7^2} = \sqrt{4 + 9 + 49} = \sqrt{62}\).


The area of the parallelogram is:

Area = \(\frac{1}{2}|\vec{a} \times \vec{b}| = \frac{1}{2}\sqrt{62}\) square units.
Quick Tip: Be careful to distinguish between the two area formulas for a parallelogram. If the adjacent sides are given as \(\vec{p}\) and \(\vec{q}\), the area is \(|\vec{p} \times \vec{q}|\). If the diagonals are given as \(\vec{a}\) and \(\vec{b}\), the area is \(\frac{1}{2}|\vec{a} \times \vec{b}|\).

Let the two given lines be \(L_1\) and \(L_2\).


The first line is \(L_1: \frac{x-8}{3} = \frac{y+19}{-16} = \frac{z-10}{7}\).

The direction vector of \(L_1\) is \(\vec{d_1} = 3\hat{i} - 16\hat{j} + 7\hat{k}\).


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

The direction vector of \(L_2\) is \(\vec{d_2} = 3\hat{i} + 8\hat{j} - 5\hat{k}\).


The required line is perpendicular to both \(L_1\) and \(L_2\). Therefore, its direction vector, let's call it \(\vec{d}\), will be parallel to the cross product of \(\vec{d_1}\) and \(\vec{d_2}\).
\(\vec{d} = \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
3 & -16 & 7
3 & 8 & -5 \end{vmatrix}\).

\(\vec{d} = \hat{i}((-16)(-5) - (7)(8)) - \hat{j}((3)(-5) - (7)(3)) + \hat{k}((3)(8) - (-16)(3))\).
\(\vec{d} = \hat{i}(80 - 56) - \hat{j}(-15 - 21) + \hat{k}(24 + 48)\).
\(\vec{d} = 24\hat{i} - \hat{j}(-36) + \hat{k}(72) = 24\hat{i} + 36\hat{j} + 72\hat{k}\).


We can use a simpler direction vector parallel to \(\vec{d}\) by dividing by the greatest common divisor, 12.

Let the simplified direction vector be \(\vec{b} = \frac{1}{12}\vec{d} = 2\hat{i} + 3\hat{j} + 6\hat{k}\).


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


Vector Equation of the Line

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


Cartesian Equation of the Line

The cartesian equation is of the form \(\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}\).

Here, \((x_1, y_1, z_1) = (1, 2, -4)\) and \((a, b, c) = (2, 3, 6)\).
\(\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-(-4)}{6}\).
\(\frac{x-1}{2} = \frac{y-2}{3} = \frac{z+4}{6}\).
Quick Tip: A line perpendicular to two other lines will have a direction vector that is parallel to the cross product of the direction vectors of the two given lines. Always simplify the resulting direction vector by factoring out common scalars to make the equations cleaner.

Let's define the following events:
\(E_1\): The student has the misconception.
\(E_2\): The student does not have the misconception.

A: The student answers 'Bijoy'.


From the problem statement, we have the following probabilities:
\(P(E_1) = 40% = 0.4\).
\(P(E_2) = 1 - P(E_1) = 1 - 0.4 = 0.6\).
\(P(A|E_1)\) (Probability of answering Bijoy given misconception) = \(80% = 0.8\).
\(P(A'|E_2)\) (Probability of NOT answering Bijoy given no misconception) = \(90% = 0.9\).

From this, we can find \(P(A|E_2) = 1 - P(A'|E_2) = 1 - 0.9 = 0.1\).


(i) Probability of not having misconception and answering Bijoy

This corresponds to finding \(P(E_2 \cap A)\).

Using the multiplication rule of probability: \(P(E_2 \cap A) = P(E_2) \times P(A|E_2)\).
\(P(E_2 \cap A) = 0.6 \times 0.1 = 0.06\).


(ii) Probability that a student answers Bijoy

This corresponds to finding the total probability of event A, \(P(A)\).

Using the Law of Total Probability: \(P(A) = P(E_1)P(A|E_1) + P(E_2)P(A|E_2)\).
\(P(A) = (0.4 \times 0.8) + (0.6 \times 0.1)\).
\(P(A) = 0.32 + 0.06 = 0.38\).


(iii) (a) Probability of having misconception given the answer is Bijoy

This is a conditional probability, \(P(E_1|A)\).

Using Bayes' Theorem: \(P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(A)}\).
\(P(E_1|A) = \frac{0.4 \times 0.8}{0.38} = \frac{0.32}{0.38} = \frac{32}{38} = \frac{16}{19}\).


OR (b) Probability of not having misconception given the answer is Bijoy

This is a conditional probability, \(P(E_2|A)\).

Using Bayes' Theorem: \(P(E_2|A) = \frac{P(E_2)P(A|E_2)}{P(A)}\).
\(P(E_2|A) = \frac{0.6 \times 0.1}{0.38} = \frac{0.06}{0.38} = \frac{6}{38} = \frac{3}{19}\).
Quick Tip: This is a classic application of Bayes' Theorem. It's helpful to first list all the known probabilities and conditional probabilities from the text. Then use the Law of Total Probability to find the probability of the "evidence" event before applying Bayes' formula.

From the given equations:

Line A (\(l_1\)): Passes through point \(\vec{a_1} = 2\hat{i} - \hat{j} + 3\hat{k}\) with direction vector \(\vec{d_1} = 3\hat{i} - 2\hat{j} + 4\hat{k}\).

Line B (\(l_2\)): Passes through point \(\vec{a_2} = \hat{i} + 3\hat{j} - 2\hat{k}\) with direction vector \(\vec{d_2} = 2\hat{i} + \hat{j} - 3\hat{k}\).


(i) Check for Parallel Lines

Two lines are parallel if their direction vectors are proportional, i.e., \(\vec{d_1} = \lambda \vec{d_2}\) for some scalar \(\lambda\).

Comparing the ratios of the direction vector components:
\(\frac{3}{2} \ne \frac{-2}{1} \ne \frac{4}{-3}\).

Since the ratios are not equal, the direction vectors are not proportional. Thus, the lines are not parallel.


(ii) Equation of Line Parallel to Line A

The new line is parallel to Line A, so it has the same direction vector \(\vec{d_1} = 3\hat{i} - 2\hat{j} + 4\hat{k}\).

It passes through the point \((1, -2, -3)\), so its position vector is \(\vec{a} = \hat{i} - 2\hat{j} - 3\hat{k}\).

The equation of the line is \(\vec{r} = \vec{a} + \lambda \vec{d_1}\).
\(\vec{r} = (\hat{i} - 2\hat{j} - 3\hat{k}) + \lambda(3\hat{i} - 2\hat{j} + 4\hat{k})\).


(iii) (a) Equation of Perpendicular Walkway

The walkway is perpendicular to both lines, so its direction vector \(\vec{d_p}\) is the cross product of their direction vectors.
\(\vec{d_p} = \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
3 & -2 & 4
2 & 1 & -3 \end{vmatrix} = \hat{i}(6-4) - \hat{j}(-9-8) + \hat{k}(3 - (-4))\).
\(\vec{d_p} = 2\hat{i} + 17\hat{j} + 7\hat{k}\).

The walkway passes through \((3, 2, 1)\), so its position vector is \(\vec{a_p} = 3\hat{i} + 2\hat{j} + \hat{k}\).

The equation is \(\vec{r} = (3\hat{i} + 2\hat{j} + \hat{k}) + \mu(2\hat{i} + 17\hat{j} + 7\hat{k})\).


OR (b) Shortest Distance between Line A and Line B

The formula for the shortest distance between two skew lines is \(SD = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{d_1} \times \vec{d_2})|}{|\vec{d_1} \times \vec{d_2}|}\).

From part (iii)(a), we have \(\vec{d_1} \times \vec{d_2} = 2\hat{i} + 17\hat{j} + 7\hat{k}\).
\(|\vec{d_1} \times \vec{d_2}| = \sqrt{2^2 + 17^2 + 7^2} = \sqrt{4 + 289 + 49} = \sqrt{342}\).

Calculate the vector connecting the points on the lines:
\(\vec{a_2} - \vec{a_1} = (1-2)\hat{i} + (3-(-1))\hat{j} + (-2-3)\hat{k} = -\hat{i} + 4\hat{j} - 5\hat{k}\).

Now, the dot product for the numerator:
\((\vec{a_2} - \vec{a_1}) \cdot (\vec{d_1} \times \vec{d_2}) = (-1)(2) + (4)(17) + (-5)(7) = -2 + 68 - 35 = 31\).
\(SD = \frac{|31|}{\sqrt{342}} = \frac{31}{\sqrt{342}}\) units.
Quick Tip: For 3D line problems, first identify the point on the line (\(\vec{a}\)) and the direction vector (\(\vec{d}\)) for each line. Parallel lines have proportional direction vectors. The shortest distance between skew lines involves both a cross product and a dot product.

The given differential equation is Newton's Law of Cooling: \(\frac{dT}{dt} = -k(T - 25)\).


(i) Expression for Temperature T(t)

This is a separable differential equation. We can rearrange it as:
\(\frac{dT}{T - 25} = -k \, dt\).


Integrate both sides:
\(\int \frac{1}{T - 25} dT = \int -k \, dt\).
\(\ln|T - 25| = -kt + C_1\).


Since the processor is cooling towards 25\(^{\circ}\)C, \(T \ge 25\), so we can drop the absolute value.
\(\ln(T - 25) = -kt + C_1\).


Exponentiate both sides to solve for T:
\(T - 25 = e^{-kt + C_1} = e^{C_1}e^{-kt}\).

Let \(A = e^{C_1}\). The general solution is \(T(t) = 25 + Ae^{-kt}\).


Now, use the initial condition \(T(0) = 85^{\circ}\)C to find the constant A.
\(85 = 25 + Ae^{-k(0)} = 25 + A(1)\).
\(A = 85 - 25 = 60\).


So, the specific expression for the temperature is \(T(t) = 25 + 60e^{-kt}\).


(ii) Time to reach 40\(^{\circ}\)C

We need to find the time \(t\) when \(T(t) = 40^{\circ}\)C, given \(k = 0.03\).

Using the expression from part (i):
\(40 = 25 + 60e^{-0.03t}\).


Solve for \(t\):
\(40 - 25 = 60e^{-0.03t}\).
\(15 = 60e^{-0.03t}\).
\(\frac{15}{60} = e^{-0.03t} \implies \frac{1}{4} = e^{-0.03t}\).


Take the natural logarithm of both sides:
\(\ln\left(\frac{1}{4}\right) = -0.03t\).
\(-\ln(4) = -0.03t\).
\(t = \frac{\ln(4)}{0.03}\).


Now, substitute the given value \(\ln(4) = 1.3863\):
\(t = \frac{1.3863}{0.03} = \frac{138.63}{3} = 46.21\).


It will take 46.21 minutes for the processor's temperature to reach 40\(^{\circ}\)C.
Quick Tip: Differential equations of the form \(\frac{dy}{dt} = k(y-a)\) are common in growth and decay problems. The general solution is always of the form \(y(t) = a + Ce^{kt}\). Solving involves separating variables, integrating, and then using the initial condition to find the constant C.