CBSE Class 12 2025 Mathematics Set-2 (Available): Download Answer Key and Solution PDF

The angle \( \theta \) between two vectors \( \vec{a} \) and \( \vec{b} \) is obtuse if \( \cos \theta < 0 \). This occurs when the dot product \( \vec{a} \cdot \vec{b} < 0 \).

The dot product of \( \vec{a} = 2x^2 \hat{i} + 4x \hat{j} + \hat{k} \) and \( \vec{b} = 7 \hat{i} - 2 \hat{j} + x \hat{k} \) is: \(\)\vec{a \cdot \vec{b = (2x^2)(7) + (4x)(-2) + (1)(x)\(\) \(\)\vec{a \cdot \vec{b = 14x^2 - 8x + x\(\) \(\)\vec{a \cdot \vec{b = 14x^2 - 7x\(\)

For the angle to be obtuse, we need \( \vec{a} \cdot \vec{b} < 0 \): \(\)14x^2 - 7x < 0\(\)
Factor out \( 7x \): \(\)7x(2x - 1) < 0\(\)

The critical points are found by setting each factor to zero: \(\)7x = 0 \implies x = 0\(\) \(\)2x - 1 = 0 \implies x = \frac{1{2\(\)

We analyze the sign of \( 7x(2x - 1) \) in the intervals \( (-\infty, 0) \), \( (0, \frac{1}{2}) \), and \( (\frac{1}{2}, \infty) \).


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


The inequality \( 14x^2 - 7x < 0 \) is satisfied when \( 0 < x < \frac{1}{2} \). Therefore, the values of \( x \) for which the angle between the vectors is obtuse are in the interval \( \left( 0, \frac{1}{2} \right) \). Quick Tip: The angle between two non-zero vectors \( \vec{a} \) and \( \vec{b} \) is obtuse if and only if their dot product \( \vec{a} \cdot \vec{b} \) is negative. Remember the formula for the dot product in terms of components: if \( \vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k} \) and \( \vec{b} = b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k} \), then \( \vec{a} \cdot \vec{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 \). Solving the resulting inequality gives the range of the required variable.

We need to evaluate the integral \( \int \frac{dx}{\sin^2 x \cos^2 x} \).
We can rewrite the integrand using the identity \( \sin^2 x + \cos^2 x = 1 \): \(\)\frac{1{\sin^2 x \cos^2 x = \frac{\sin^2 x + \cos^2 x{\sin^2 x \cos^2 x\(\) \(\)= \frac{\sin^2 x{\sin^2 x \cos^2 x + \frac{\cos^2 x{\sin^2 x \cos^2 x\(\) \(\)= \frac{1{\cos^2 x + \frac{1{\sin^2 x\(\) \(\)= \sec^2 x + \csc^2 x\(\)

Now, we can integrate term by term: \(\)\int (\sec^2 x + \csc^2 x) dx = \int \sec^2 x dx + \int \csc^2 x dx\(\)
We know that the integral of \( \sec^2 x \) is \( \tan x \) and the integral of \( \csc^2 x \) is \( -\cot x \). Therefore, \(\)\int \sec^2 x dx + \int \csc^2 x dx = \tan x - \cot x + C\(\)

Thus, \( \int \frac{dx}{\sin^2 x \cos^2 x} = \tan x - \cot x + C \). Quick Tip: When dealing with integrals involving powers of \( \sin x \) and \( \cos x \) in the denominator, it is often useful to use the identity \( 1 = \sin^2 x + \cos^2 x \) to split the fraction. Alternatively, you can use the identity \( \sin 2x = 2 \sin x \cos x \) to rewrite the denominator as \( \left( \frac{\sin 2x}{2} \right)^2 = \frac{\sin^2 2x}{4} \). Then the integral becomes \( \int \frac{4}{\sin^2 2x} dx = 4 \int \csc^2 2x dx \). Integrating \( \csc^2 ax \) gives \( -\frac{1}{a} \cot ax \). So, \( 4 \int \csc^2 2x dx = 4 \left( -\frac{1}{2} \cot 2x \right) + C = -2 \cot 2x + C \). You can show that \( -2 \cot 2x \) is equivalent to \( \tan x - \cot x \) using the double angle formula for cotangent: \( \cot 2x = \frac{\cot^2 x - 1}{2 \cot x} \). \(\)-2 \cot 2x = -2 \left( \frac{\cot^2 x - 1}{2 \cot x} \right) = -\frac{\cot^2 x - 1}{\cot x} = -\cot x + \frac{1}{\cot x} = \tan x - \cot x\(\)

A skew-symmetric matrix \( P \) satisfies the condition \( P^T = -P \).
Taking the determinant of both sides, we get: \(\)\det(P^T) = \det(-P)\(\)

We know that \( \det(P^T) = \det(P) \). Also, for a matrix of order \( n \), \( \det(kP) = k^n \det(P) \). Here, the order of matrix \( P \) is \( n = 3 \) and \( k = -1 \). Therefore, \(\)\det(-P) = (-1)^3 \det(P) = -\det(P)\(\)

So, we have: \(\)\det(P) = -\det(P)\(\) \(\)2 \det(P) = 0\(\) \(\)\det(P) = 0\(\)

Given that \( \det(P) = \alpha \), we have \( \alpha = 0 \).

Now we need to find the value of \( (2025)^\alpha \): \(\)(2025)^\alpha = (2025)^0\(\)

Since any non-zero number raised to the power of 0 is 1, we have: \(\)(2025)^0 = 1\(\)

Thus, \( (2025)^\alpha = 1 \). Quick Tip: A crucial property to remember is that the determinant of any skew-symmetric matrix of odd order is always zero. This can be proven using the properties of determinants: \( \det(A^T) = \det(A) \) and \( \det(cA) = c^n \det(A) \). For a skew-symmetric matrix \( P \) of odd order \( n \), \( \det(P) = \det(P^T) = \det(-P) = (-1)^n \det(P) = -\det(P) \), which implies \( 2 \det(P) = 0 \), so \( \det(P) = 0 \).

The area of a triangle formed by three points P, Q, and R can be found using the magnitude of the cross product of the vectors representing two of its sides originating from a common vertex. For example, the area of \( \triangle PQR \) is given by half the magnitude of the cross product of \( \vec{PQ} \) and \( \vec{PR} \): \(\) Area(\triangle PQR) = \frac{1{2 |\vec{PQ \times \vec{PR| \(\)

We are given that \( \vec{PQ \times \vec{PR} = 4 \hat{i} + 8 \hat{j} - 8 \hat{k} \). We need to find the magnitude of this vector: \(\) |\vec{PQ \times \vec{PR| = \sqrt{(4)^2 + (8)^2 + (-8)^2 \(\) \(\) = \sqrt{16 + 64 + 64 \(\) \(\) = \sqrt{144 \(\) \(\) = 12 \(\)

Now, we can find the area of \( \triangle PQR \): \(\) Area(\triangle PQR) = \frac{1{2 (12) = 6 \(\)

The area of \( \triangle PQR \) is 6 square units. Quick Tip: The magnitude of the cross product of two vectors \( \vec{a \) and \( \vec{b} \), \( |\vec{a} \times \vec{b}| \), represents the area of the parallelogram formed by these two vectors as adjacent sides. The area of the triangle formed by these two vectors as two of its sides is half the area of the parallelogram, i.e., \( \frac{1}{2} |\vec{a} \times \vec{b}| \).

We need to evaluate the integral \( I_p = \int_{0}^{\pi} \frac{\sin 2px}{\sin x} dx \), where \( p \in N \).

We can use the identity \( \sin(A+B) - \sin(A-B) = 2 \cos A \sin B \).
Let \( A = (2p-1)x \) and \( B = x \), then \( A+B = 2px \) and \( A-B = (2p-2)x \).
So, \( \sin 2px - \sin (2p-2)x = 2 \cos (2p-1)x \sin x \).
Therefore, \( \frac{\sin 2px}{\sin x} = \frac{\sin (2p-2)x}{\sin x} + 2 \cos (2p-1)x \).

Integrating from \( 0 \) to \( \pi \): \(\)I_p = \int_{0^{\pi \frac{\sin (2p-2)x{\sin x dx + \int_{0^{\pi 2 \cos (2p-1)x dx\(\) \(\)I_p = I_{p-1 + \left[ \frac{2 \sin (2p-1)x{2p-1 \right]_{0^{\pi\(\) \(\)I_p = I_{p-1 + \frac{2 \sin (2p-1)\pi{2p-1 - \frac{2 \sin 0{2p-1\(\)
Since \( p \in N \), \( 2p-1 \) is an integer, so \( \sin (2p-1)\pi = 0 \). Also, \( \sin 0 = 0 \).
Thus, \( I_p = I_{p-1} \).

This shows that the value of the integral is independent of \( p \).
Let's evaluate for \( p = 1 \): \(\)I_1 = \int_{0^{\pi \frac{\sin 2x{\sin x dx = \int_{0^{\pi \frac{2 \sin x \cos x{\sin x dx = \int_{0^{\pi 2 \cos x dx\(\) \(\)= [2 \sin x]_{0^{\pi = 2 \sin \pi - 2 \sin 0 = 2(0) - 2(0) = 0\(\)

Since \( I_p = I_{p-1} \) and \( I_1 = 0 \), it follows that \( I_p = 0 \) for all \( p \in N \). Quick Tip: Using reduction formulas or recurrence relations can be helpful for integrals involving parameters. In this case, by relating \( I_p \) to \( I_{p-1} \), we found a pattern. Remember the values of trigonometric functions at multiples of \( \pi \): \( \sin(n\pi) = 0 \) for any integer \( n \).

We are given the vectors: \(\)\vec{p = 2\hat{i - 3\hat{j - \hat{k\(\) \(\)\vec{q = -3\hat{i + 4\hat{j + \hat{k\(\) \(\)\vec{r = \hat{i + \hat{j + 2\hat{k\(\)
We need to find scalars \( \lambda \) and \( \mu \) such that \( \vec{r} = \lambda \vec{p} + \mu \vec{q} \).
Substituting the vectors: \(\)\hat{i + \hat{j + 2\hat{k = \lambda (2\hat{i - 3\hat{j - \hat{k) + \mu (-3\hat{i + 4\hat{j + \hat{k)\(\) \(\)\hat{i + \hat{j + 2\hat{k = (2\lambda - 3\mu)\hat{i + (-3\lambda + 4\mu)\hat{j + (-\lambda + \mu)\hat{k\(\)
Equating the coefficients of \( \hat{i}, \hat{j}, \) and \( \hat{k} \), we get the following system of linear equations: \(\)2\lambda - 3\mu = 1 \quad (1)\(\) \(\)-3\lambda + 4\mu = 1 \quad (2)\(\) \(\)-\lambda + \mu = 2 \quad (3)\(\)
From equation (3), we can express \( \mu \) in terms of \( \lambda \): \(\)\mu = \lambda + 2\(\)
Substitute this into equation (1): \(\)2\lambda - 3(\lambda + 2) = 1\(\) \(\)2\lambda - 3\lambda - 6 = 1\(\) \(\)-\lambda = 7\(\) \(\)\lambda = -7\(\)
Now, substitute the value of \( \lambda \) back into the expression for \( \mu \): \(\)\mu = -7 + 2\(\) \(\)\mu = -5\(\)
To verify, substitute \( \lambda = -7 \) and \( \mu = -5 \) into equation (2): \(\)-3(-7) + 4(-5) = 21 - 20 = 1\(\)
The values satisfy all three equations.
Thus, \( \vec{r} = -7\vec{p} - 5\vec{q} \), with \( \lambda = -7 \) and \( \mu = -5 \). Quick Tip: When expressing a vector as a linear combination of other vectors, the problem reduces to solving a system of linear equations for the scalar coefficients. If the vectors are in three dimensions, you will typically get a system of three equations with three unknowns. Ensure your algebraic manipulations are accurate to find the correct values of the scalars.

We are given a function \( f : N \rightarrow N \) defined by \( f(x) = ax + b \), where \( a \) and \( b \) are natural numbers (\( N = \{1, 2, 3, \dots\} \)). We need to prove that this function is one-one but not onto.

Proof for One-One (Injective):
A function \( f \) is one-one if for any \( x_1, x_2 \) in the domain, \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \).
Let \( x_1, x_2 \in N \) such that \( f(x_1) = f(x_2) \). \(\)ax_1 + b = ax_2 + b\(\)
Subtract \( b \) from both sides: \(\)ax_1 = ax_2\(\)
Since \( a \in N \), \( a \neq 0 \). We can divide both sides by \( a \): \(\)x_1 = x_2\(\)
Thus, \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \), which means the function \( f \) is one-one.

Proof for Not Onto (Not Surjective):
A function \( f : A \rightarrow B \) is onto if for every \( y \in B \), there exists an \( x \in A \) such that \( f(x) = y \). In our case, \( A = N \) and \( B = N \).
Consider an arbitrary \( y \in N \). We want to find if there exists an \( x \in N \) such that \( f(x) = y \). \(\)ax + b = y\(\) \(\)ax = y - b\(\) \(\)x = \frac{y - b{a\(\)
For \( f \) to be onto, for every \( y \in N \), the value of \( x = \frac{y - b}{a} \) must also be a natural number.

Let's take a specific example. Let \( a = 2 \) and \( b = 1 \). Then \( f(x) = 2x + 1 \). The range of this function for \( x \in N \) is \( \{3, 5, 7, \dots\} \), which is the set of odd natural numbers greater than or equal to 3.
Consider \( y = 1 \in N \). If \( f(x) = 1 \), then \( 2x + 1 = 1 \implies 2x = 0 \implies x = 0 \), which is not a natural number.
Consider \( y = 2 \in N \). If \( f(x) = 2 \), then \( 2x + 1 = 2 \implies 2x = 1 \implies x = \frac{1}{2} \), which is not a natural number.
In general, for \( f(x) = ax + b \), if we choose \( y \) such that \( y - b \) is not a positive multiple of \( a \), or if \( y - b \le 0 \), then \( x \) will not be a natural number.

Since \( a \ge 1 \) and \( b \ge 1 \), let's consider \( y = 1 \in N \). If there exists \( x \in N \) such that \( ax + b = 1 \), then \( ax = 1 - b \). Since \( b \ge 1 \), \( 1 - b \le 0 \). If \( 1 - b = 0 \), then \( b = 1 \) and \( ax = 0 \), which implies \( x = 0 \) (not in \( N \)). If \( 1 - b < 0 \), then \( ax \) is negative, and since \( a > 0 \), \( x \) must be negative (not in \( N \)).
Therefore, there exists \( y \in N \) (for example, \( y = 1 \)) for which there is no \( x \in N \) such that \( f(x) = y \). Hence, the function \( f \) is not onto. Quick Tip: To prove a function is one-one, show that \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \). To prove a function is not onto, find an element in the codomain for which there is no pre-image in the domain. For functions involving natural numbers, consider the properties of arithmetic progressions formed by \( ax + b \).

The feasible region is defined by the intersection of several linear inequalities. We need to determine the equations of the boundary lines and the corresponding inequalities that define the shaded region.

Constraint from the line passing through (35, 0) and (30, 10):
The equation of the line passing through \( (x_1, y_1) \) and \( (x_2, y_2) \) is \( \frac{y - y_1}{x - x_1} = \frac{y_2 - y_1}{x_2 - x_1} \).
Using points (35, 0) and (30, 10): \(\)\frac{y - 0{x - 35 = \frac{10 - 0{30 - 35\(\) \(\)\frac{y{x - 35 = \frac{10{-5\(\) \(\)\frac{y{x - 35 = -2\(\) \(\)y = -2(x - 35)\(\) \(\)y = -2x + 70\(\) \(\)2x + y = 70\(\)
The feasible region is below this line, so the inequality is \( 2x + y \le 70 \).

Constraint from the line passing through (0, 30) and (15, 25):
Using points (0, 30) and (15, 25): \(\)\frac{y - 30{x - 0 = \frac{25 - 30{15 - 0\(\) \(\)\frac{y - 30{x = \frac{-5{15\(\) \(\)\frac{y - 30{x = -\frac{1{3\(\) \(\)3(y - 30) = -x\(\) \(\)3y - 90 = -x\(\) \(\)x + 3y = 90\(\)
The feasible region is below this line, so the inequality is \( x + 3y \le 90 \).

Constraint from the horizontal line passing through (30, 10) and extending leftwards:
This line is \( y = 10 \).
The feasible region is above this line, so the inequality is \( y \ge 10 \).

Since the feasible region is in the first quadrant (including the axes), we also have the non-negativity constraints: \(\)x \ge 0\(\) \(\)y \ge 0\(\)

Therefore, the constraints for the given linear programming problem are: \(\)2x + y \le 70\(\) \(\)x + 3y \le 90\(\) \(\)y \ge 10\(\) \(\)x \ge 0\(\) \(\)y \ge 0\(\) Quick Tip: To determine the inequality from a line on a graph, pick a test point in the feasible region (not on the line) and substitute its coordinates into the equation of the line. If the inequality holds true for the test point, then that inequality defines the feasible region relative to that line. For example, for \( 2x + y = 70 \), the origin (0, 0) is in the feasible region, and \( 2(0) + 0 = 0 \le 70 \), confirming \( 2x + y \le 70 \).

Let the given integral be \( I \): \(\)I = \int_{0^{a f(x) g(x) dx \quad \cdots (1)\(\)
Using the property of definite integrals \( \int_{0}^{a} h(x) dx = \int_{0}^{a} h(a - x) dx \), we can write: \(\)I = \int_{0^{a f(a - x) g(a - x) dx\(\)
We are given that \( f(a - x) = f(x) \), so substituting this into the integral: \(\)I = \int_{0^{a f(x) g(a - x) dx \quad \cdots (2)\(\)
We are also given that \( g(x) + g(a - x) = a \), which implies \( g(a - x) = a - g(x) \). Substituting this into equation (2): \(\)I = \int_{0^{a f(x) (a - g(x)) dx\(\) \(\)I = \int_{0^{a (a f(x) - f(x) g(x)) dx\(\)
Using the linearity of integrals: \(\)I = \int_{0^{a a f(x) dx - \int_{0^{a f(x) g(x) dx\(\) \(\)I = a \int_{0^{a f(x) dx - I \quad (from equation (1))\(\)
Now, we solve for \( I \): \(\)I + I = a \int_{0^{a f(x) dx\(\) \(\)2I = a \int_{0^{a f(x) dx\(\) \(\)I = \frac{a{2 \int_{0^{a f(x) dx\(\)
Thus, we have shown that \( \int_{0^{a} f(x) g(x) dx = \frac{a}{2} \int_{0}^{a} f(x) dx \). Quick Tip: The property \( \int_{0}^{a} h(x) dx = \int_{0}^{a} h(a - x) dx \) is very useful when dealing with integrals where the integrand has symmetry about the line \( x = a/2 \). Combining this property with the given conditions on the functions often leads to the solution.

Let A, B, and C denote the events that the car is manufactured by Amber, Bonzi, and Comet respectively.
Let E denote the event that the car is electric.
We are given the following probabilities: \( P(A) = 0.60 \) (Production share of Amber) \( P(B) = 0.30 \) (Production share of Bonzi) \( P(C) = 0.10 \) (Production share of Comet)
Note that \( P(A) + P(B) + P(C) = 0.60 + 0.30 + 0.10 = 1.00 \).

We are also given the conditional probabilities of a car being electric given the manufacturer: \( P(E|A) = 0.20 \) (Probability that Amber manufactures an electric car) \( P(E|B) = 0.10 \) (Probability that Bonzi manufactures an electric car) \( P(E|C) = 0.05 \) (Probability that Comet manufactures an electric car)

(i) (a) Probability that a randomly selected car is an electric car:
We can use the law of total probability to find \( P(E) \): \(\) P(E) = P(E|A)P(A) + P(E|B)P(B) + P(E|C)P(C) \(\) \(\) P(E) = (0.20)(0.60) + (0.10)(0.30) + (0.05)(0.10) \(\) \(\) P(E) = 0.12 + 0.03 + 0.005 \(\) \(\) P(E) = 0.155 \(\)
The probability that a randomly selected car is an electric car is \( 0.155 \).

(i) (b) Probability that a randomly selected car is a petrol car:
Let \( E' \) denote the event that the car is a petrol car. Then \( P(E') = 1 - P(E) \). \(\) P(E') = 1 - 0.155 = 0.845 \(\)
The probability that a randomly selected car is a petrol car is \( 0.845 \).

(ii) Probability that a car was manufactured by Comet given that it is electric:
We need to find \( P(C|E) \). Using Bayes' theorem: \(\) P(C|E) = \frac{P(E|C)P(C){P(E) \(\) \(\) P(C|E) = \frac{(0.05)(0.10){0.155 \(\) \(\) P(C|E) = \frac{0.005{0.155 = \frac{5{155 = \frac{1{31 \(\)
The probability that the electric car was manufactured by Comet is \( \frac{1}{31} \).

(iii) Probability that a car was manufactured by Amber or Bonzi given that it is electric:
We need to find \( P(A \cup B | E) \). \(\) P(A \cup B | E) = P(A|E) + P(B|E) \(\)
Using Bayes' theorem: \(\) P(A|E) = \frac{P(E|A)P(A){P(E) = \frac{(0.20)(0.60){0.155 = \frac{0.12{0.155 = \frac{120{155 = \frac{24{31 \(\) \(\) P(B|E) = \frac{P(E|B)P(B){P(E) = \frac{(0.10)(0.30){0.155 = \frac{0.03{0.155 = \frac{30{155 = \frac{6{31 \(\) \(\) P(A \cup B | E) = \frac{24{31 + \frac{6{31 = \frac{30{31 \(\)
Alternatively, \( P(A \cup B | E) = 1 - P(C|E) = 1 - \frac{1}{31} = \frac{30}{31} \).
The probability that the electric car was manufactured by Amber or Bonzi is \( \frac{30}{31} \). Quick Tip: This problem involves conditional probability and the law of total probability. Bayes' theorem is crucial for finding the probability of the cause given the effect. Remember to carefully identify the events and their probabilities as given in the problem statement.

Given the function \( f(x) = e^x \sin x \) for \( x \in [0, \pi] \).

(i) Intervals of increasing or decreasing function:
First, we find the first derivative \( f'(x) \): \(\) f'(x) = \frac{d{dx (e^x \sin x) = e^x \sin x + e^x \cos x = e^x (\sin x + \cos x) \(\)
To find the intervals where \( f(x) \) is increasing or decreasing, we need to determine the sign of \( f'(x) \). Since \( e^x \) is always positive, the sign of \( f'(x) \) depends on the sign of \( \sin x + \cos x \).
We consider \( \sin x + \cos x > 0 \) and \( \sin x + \cos x < 0 \) for \( x \in [0, \pi] \). \(\) \sin x + \cos x > 0 \implies \sqrt{2 \sin \left( x + \frac{\pi{4 \right) > 0 \implies \sin \left( x + \frac{\pi{4 \right) > 0 \(\)
For \( x \in [0, \pi] \), \( x + \frac{\pi}{4} \in \left[ \frac{\pi}{4}, \frac{5\pi}{4} \right] \).
In this interval, \( \sin \left( x + \frac{\pi}{4} \right) > 0 \) when \( \frac{\pi}{4} < x + \frac{\pi}{4} < \pi \), which means \( 0 < x < \frac{3\pi}{4} \).
So, \( f'(x) > 0 \) for \( x \in \left( 0, \frac{3\pi}{4} \right) \), and \( f(x) \) is increasing on \( \left[ 0, \frac{3\pi}{4} \right] \).
\(\) \sin x + \cos x < 0 \implies \sqrt{2 \sin \left( x + \frac{\pi{4 \right) < 0 \implies \sin \left( x + \frac{\pi{4 \right) < 0 \(\)
For \( x \in [0, \pi] \), \( x + \frac{\pi}{4} \in \left[ \frac{\pi}{4}, \frac{5\pi}{4} \right] \).
In this interval, \( \sin \left( x + \frac{\pi}{4} \right) < 0 \) when \( \pi < x + \frac{\pi}{4} < \frac{5\pi}{4} \), which means \( \frac{3\pi}{4} < x < \pi \).
So, \( f'(x) < 0 \) for \( x \in \left( \frac{3\pi}{4}, \pi \right) \), and \( f(x) \) is decreasing on \( \left[ \frac{3\pi}{4}, \pi \right] \).

(ii) Critical points and their nature:
Critical points occur where \( f'(x) = 0 \) or \( f'(x) \) is undefined. \(\) f'(x) = e^x (\sin x + \cos x) = 0 \(\)
Since \( e^x \neq 0 \), we have \( \sin x + \cos x = 0 \implies \tan x = -1 \).
For \( x \in [0, \pi] \), the solution is \( x = \frac{3\pi}{4} \).
So, the critical point is \( x = \frac{3\pi}{4} \).

We use the second derivative test to determine the nature of this critical point. \(\) f''(x) = \frac{d{dx (e^x (\sin x + \cos x)) = e^x (\sin x + \cos x) + e^x (\cos x - \sin x) = e^x (2 \cos x) \(\)
At \( x = \frac{3\pi}{4} \), \(\) f''\left( \frac{3\pi{4 \right) = e^{3\pi/4 \left( 2 \cos \left( \frac{3\pi{4 \right) \right) = e^{3\pi/4 \left( 2 \left( -\frac{1{\sqrt{2 \right) \right) = -\sqrt{2 e^{3\pi/4 \(\)
Since \( f''\left( \frac{3\pi}{4} \right) < 0 \), the critical point \( x = \frac{3\pi}{4} \) is a point of local maximum.

Points of inflexion occur where \( f''(x) = 0 \) and the concavity changes. \(\) f''(x) = 2 e^x \cos x = 0 \(\)
Since \( e^x \neq 0 \), we have \( \cos x = 0 \).
For \( x \in [0, \pi] \), the solution is \( x = \frac{\pi}{2} \).
We need to check if the concavity changes at \( x = \frac{\pi}{2} \).
For \( x \in \left( 0, \frac{\pi}{2} \right) \), \( \cos x > 0 \), so \( f''(x) > 0 \) (concave up).
For \( x \in \left( \frac{\pi}{2}, \pi \right) \), \( \cos x < 0 \), so \( f''(x) < 0 \) (concave down).
Since the concavity changes at \( x = \frac{\pi}{2} \), it is a point of inflexion.

The critical point is \( x = \frac{3\pi}{4} \), which is a point of local maximum.
The point \( x = \frac{\pi}{2} \) is a point of inflexion. Quick Tip: To find intervals of increasing/decreasing functions, analyze the sign of the first derivative. Critical points occur where the first derivative is zero or undefined. The second derivative test helps determine if a critical point is a local maximum, local minimum, or neither. Points of inflexion occur where the second derivative is zero and the concavity changes.

Given sets \( S = \{S_1, S_2, S_3, S_4\} \) with \( |S| = 4 \) and \( J = \{J_1, J_2, J_3\} \) with \( |J| = 3 \).

(i) Number of relations from set S to set J:
A relation from set S to set J is a subset of the Cartesian product \( S \times J \). The number of elements in \( S \times J \) is \( |S| \times |J| = 4 \times 3 = 12 \). The number of subsets of a set with \( n \) elements is \( 2^n \). Therefore, the number of relations from S to J is \( 2^{12} = 4096 \).

(ii) Check if the function \( f = \{(S_1, J_1), (S_2, J_2), (S_3, J_2), (S_4, J_3)\} \) is bijective:
For \( f \) to be a function from S to J, each element of S must be mapped to a unique element of J. In the given \( f \), each element of S is mapped to exactly one element of J, so it is a function.
For \( f \) to be one-one (injective), different elements of S must be mapped to different elements of J. Here, \( f(S_2) = J_2 \) and \( f(S_3) = J_2 \), so two different elements of S are mapped to the same element in J. Thus, \( f \) is not one-one.
For \( f \) to be onto (surjective), every element of J must be the image of some element in S. The images of elements in S are \( \{J_1, J_2, J_3\} \), which is equal to J. Thus, \( f \) is onto.
Since \( f \) is not one-one, it is not bijective.

(iii) (a) Number of one-one functions from set S to set J:
A one-one function from S to J requires that each of the 4 distinct elements in S is mapped to a distinct element in J. However, the number of elements in J (\( |J| = 3 \)) is less than the number of elements in S (\( |S| = 4 \)). By the Pigeonhole Principle, it is not possible to have a one-one function from S to J. Therefore, the number of one-one functions from S to J is 0.

(iii) (b) Minimum ordered pairs to be included in \( R_1 = \{(S_1, S_2), (S_2, S_4)\} \) in set S so that \( R_1 \) is reflexive but not symmetric:
For \( R_1 \) to be reflexive on S, for every \( a \in S \), \( (a, a) \) must be in \( R_1 \). The elements in S are \( S_1, S_2, S_3, S_4 \). So, we need to include \( (S_1, S_1), (S_2, S_2), (S_3, S_3), (S_4, S_4) \) in \( R_1 \).
After including these, \( R_1 = \{(S_1, S_2), (S_2, S_4), (S_1, S_1), (S_2, S_2), (S_3, S_3), (S_4, S_4)\} \).
Now, we need to ensure that \( R_1 \) is not symmetric. A relation is symmetric if whenever \( (a, b) \in R_1 \), then \( (b, a) \in R_1 \).
We have \( (S_1, S_2) \in R_1 \), but \( (S_2, S_1) \notin R_1 \).
We have \( (S_2, S_4) \in R_1 \), but \( (S_4, S_2) \notin R_1 \).
The pairs \( (S_1, S_1), (S_2, S_2), (S_3, S_3), (S_4, S_4) \) satisfy the symmetric property.
To make \( R_1 \) not symmetric, we need at least one pair \( (a, b) \in R_1 \) such that \( (b, a) \notin R_1 \). The given relation \( R_1 \) already has this property with \( (S_1, S_2) \) and \( (S_2, S_4) \).
The minimum ordered pairs to be included to make \( R_1 \) reflexive are \( (S_1, S_1), (S_2, S_2), (S_3, S_3), (S_4, S_4) \). The resulting relation \( \{(S_1, S_2), (S_2, S_4), (S_1, S_1), (S_2, S_2), (S_3, S_3), (S_4, S_4)\} \) is reflexive and not symmetric. The minimum number of ordered pairs to be included is 4.

Final Answer: The final answer is \(\boxed{4}\) (for part (iii)(b)) Quick Tip: Remember the definitions of relations, functions, one-one functions, onto functions, bijective functions, reflexive relations, and symmetric relations. The number of relations from a set A to a set B is \( 2^{|A| \times |B|} \). A function requires each element of the domain to have a unique image in the codomain. For a relation to be reflexive on a set A, every element of A must be related to itself. For a relation to be symmetric, if a is related to b, then b must be related to a.