CBSE Class 12 Mathematics Sample Paper 2026 27 with Solutions PDF Download

SECTION - A (20 x 1 = 20)

This section comprises of 18 multiple choice questions and two assertion and reason type questions of 1 mark each.

Q1. Identify the function whose graph is given below: [1 Mark]
(A) \( \cos^{-1}\left(\dfrac{x}{2}\right) \)
(B) \( \cos^{-1}(2x) \)
(C) \( \pi - \cos^{-1}x \)
(D) \( \pi + \cos^{-1}x \)

[Figure: A graph with axes X, X' and Y, Y'; y-axis marked at \( -\pi, \pi, 2\pi, 3\pi \); a decreasing curve is plotted for x between -1 and 1, passing approximately through (-1, \(2\pi\)) and (1, \(\pi\))]

Answer: (D) \( \pi + \cos^{-1}x \)

Teacher's Note:
a) Check the graph values at \( x = -1 \) and \( x = 1 \): they give \( 2\pi \) and \( \pi \), matching \( \pi + \cos^{-1}x \).
b) Remember the range of \( \cos^{-1}x \) is \( [0, \pi] \), so adding \( \pi \) shifts the graph to lie between \( \pi \) and \( 2\pi \).

For Visually Impaired Candidates (in lieu of Q. 1)

Find the range of \( \dfrac{\pi}{2} + \sec^{-1}\left(\dfrac{x}{2}\right) \) [1 Mark]
(A) \( [0, \pi] \)
(B) \( [0, \pi] - \left\{\dfrac{\pi}{2}\right\} \)
(C) \( \left[\dfrac{\pi}{2}, \dfrac{3\pi}{2}\right] \)
(D) \( \left[\dfrac{\pi}{2}, \dfrac{3\pi}{2}\right] - \{\pi\} \)

Answer: (D) \( \left[\dfrac{\pi}{2}, \dfrac{3\pi}{2}\right] - \{\pi\} \)

Teacher's Note:
a) The range of \( \sec^{-1}x \) is \( [0, \pi] - \left\{\dfrac{\pi}{2}\right\} \), and replacing x by x/2 does not change the range.
b) Adding \( \dfrac{\pi}{2} \) to every value shifts the range to \( \left[\dfrac{\pi}{2}, \dfrac{3\pi}{2}\right] \) with \( \pi \) excluded.

Q2. Domain of the function: \( \sec^{-1}(2x) \) is [1 Mark]
(A) \( (-\infty, \infty) \)
(B) \( \left(-\dfrac{1}{2}, \dfrac{1}{2}\right) \)
(C) \( \left(-\infty, -\dfrac{1}{2}\right) \cup \left(\dfrac{1}{2}, \infty\right) \)
(D) \( \left(-\infty, -\dfrac{1}{2}\right] \cup \left[\dfrac{1}{2}, \infty\right) \)

Answer: (D) \( \left(-\infty, -\dfrac{1}{2}\right] \cup \left[\dfrac{1}{2}, \infty\right) \)

Teacher's Note:
a) \( \sec^{-1}y \) is defined for \( |y| \geq 1 \).
b) Here \( y = 2x \), so \( |2x| \geq 1 \) gives the required domain.

Q3. Value of \( \cot^{-1}\left(2\cos\left(\dfrac{1}{2}\sin^{-1}\left(\dfrac{\sqrt{3}}{2}\right)\right)\right) \) [1 Mark]
(A) \( \dfrac{\pi}{6} \)
(B) \( \dfrac{\pi}{4} \)
(C) \( \dfrac{\pi}{3} \)
(D) \( \dfrac{\pi}{2} \)

Answer: (A) \( \dfrac{\pi}{6} \)

Teacher's Note:
a) First evaluate \( \sin^{-1}\left(\dfrac{\sqrt{3}}{2}\right) = \dfrac{\pi}{3} \).
b) Then \( \cot^{-1}(2\cos(\pi/6)) = \cot^{-1}(\sqrt{3}) = \dfrac{\pi}{6} \).

Q4. If \( A = \begin{pmatrix} 2 & 1 & -2 \\ 3a & 0 & 4 \\ b & 2c & 1 \end{pmatrix} \) is a symmetric matrix, then value of \( \sqrt{6a + 2b + 3c} \) is: [1 Mark]
(A) \( -2 \)
(B) \( 2 \)
(C) \( \pm 2 \)
(D) \( 4 \)

Answer: (B) \( 2 \)

Teacher's Note:
a) For a symmetric matrix, corresponding off-diagonal elements are equal: \( 3a = 1, b = -2, 2c = 4 \).
b) Substituting gives \( \sqrt{2 - 4 + 6} = \sqrt{4} = 2 \).

Q5. If \( P \) and \( Q \) are matrices of same order such that \( P = \begin{pmatrix} 2 & 1 \\ 0 & 1 \end{pmatrix} \), \( |Q| = 3 \), then find \( |adj(PQ)| \) [1 Mark]
(A) \( -3 \)
(B) \( 1 \)
(C) \( 6 \)
(D) \( 36 \)

Answer: (C) \( 6 \)

Teacher's Note:
a) For a \( 2 \times 2 \) matrix M, \( |adj\,M| = |M| \).
b) So \( |adj(PQ)| = |PQ| = |P||Q| = 2 \times 3 = 6 \).

Q6. If \( [1 \ x]\begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}\begin{pmatrix} 4 \\ -1 \end{pmatrix} = O \), then value of \( x \) is [1 Mark]
(A) \( -2 \)
(B) \( -1 \)
(C) \( 1 \)
(D) \( 2 \)

Answer: (D) \( 2 \)

Teacher's Note:
a) Multiply out to get \( [1 \ 2x]\begin{pmatrix} 4 \\ -1 \end{pmatrix} = 4 - 2x = 0 \).
b) Solving gives \( x = 2 \).

Q7. The derivative of \( \cos^{-1}(2x^2 - 1) \) with respect to \( \cos^{-1}(x) \) is [1 Mark]
(A) \( 2 \)
(B) \( \pm 1 \)
(C) \( \dfrac{1}{2} \)
(D) \( -2 \)

Answer: (A) \( 2 \)

Teacher's Note:
a) Put \( x = \cos u \), so \( \cos^{-1}(2x^2-1) = \cos^{-1}(\cos 2u) = 2u \).
b) So \( \dfrac{dy}{du} = 2 \), a common substitution trick for such derivatives.

Q8. If \( a, b, c \) are arbitrary constants such that the function defined by
\( f(x) = \begin{cases} ax^2 + b, & b \neq 0, x \leq 1 \\ bx^2 + ax + c, & x \gt 1 \end{cases} \) is differentiable at \( x = 1 \), then which of the following could be a triplet for \( (a,b,c) \)? [1 Mark]
(A) \( (3,2,0) \)
(B) \( (0,0,3) \)
(C) \( \left(\dfrac{3}{2}, 2, 0\right) \)
(D) \( \left(3, \dfrac{3}{2}, 0\right) \)

Answer: (D) \( \left(3, \dfrac{3}{2}, 0\right) \)

Teacher's Note:
a) Differentiability implies continuity, which forces \( c = 0 \) and \( a = 2b \).
b) Only option (D) satisfies \( a = 2b \) with \( c = 0 \).

Q9. If \( y = a\left\{x + 2a\sin\left(\dfrac{x}{a}\right)\right\}, a \gt 0, x \in [0, \pi a] \); is a decreasing function, then the interval in which \( x \) lies is [1 Mark]
(A) \( \left(\dfrac{5\pi}{6}, \pi\right] \)
(B) \( \left(\dfrac{2\pi}{3}, \pi\right] \)
(C) \( \left(\dfrac{5\pi a}{6}, \pi a\right] \)
(D) \( \left(\dfrac{2\pi a}{3}, \pi a\right] \)

Answer: (D) \( \left(\dfrac{2\pi a}{3}, \pi a\right] \)

Teacher's Note:
a) For a decreasing function \( \dfrac{dy}{dx} \lt 0 \), which gives \( \cos\left(\dfrac{x}{a}\right) \lt -\dfrac{1}{2} \).
b) This gives \( \dfrac{x}{a} \in \left(\dfrac{2\pi}{3}, \pi\right] \), so \( x \in \left(\dfrac{2\pi a}{3}, \pi a\right] \).

Q10. The values of \( x \) for which the derivative of the function \( y = x^4 \) is greater than the value of the function are [1 Mark]
(A) \( 0 \lt x \lt 4 \)
(B) \( 0 \lt x \leq 4 \)
(C) \( x \lt 0 \text{ or } x \gt 4 \)
(D) \( x \lt 0 \text{ or } x \geq 4 \)

Answer: (A) \( 0 \lt x \lt 4 \)

Teacher's Note:
a) Set up \( 4x^3 \gt x^4 \), which gives \( x^3(4-x) \gt 0 \).
b) Solving this inequality gives \( 0 \lt x \lt 4 \).

Q11. The degree and order of the differential equation \( y - x\left(\dfrac{dy}{dx}\right)^3 = a\left(y + \dfrac{d^2y}{dx^2}\right)^2 \) are respectively [1 Mark]
(A) \( 3, 1 \)
(B) \( 2, 2 \)
(C) \( 3, 2 \)
(D) \( 4, 2 \)

Answer: (B) \( 2, 2 \)

Teacher's Note:
a) The order is decided by the highest derivative present, which is \( \dfrac{d^2y}{dx^2} \), giving order 2.
b) The power of this highest derivative term gives the degree, which is 2.

Q12. If \( \int \dfrac{1}{1+e^{2x}}dx = x - P\log(1+e^{2x}) + C \), then 'P' is equal to [1 Mark]
(A) \( -\dfrac{1}{2} \)
(B) \( -\dfrac{1}{3} \)
(C) \( \dfrac{1}{3} \)
(D) \( \dfrac{1}{2} \)

Answer: (D) \( \dfrac{1}{2} \)

Teacher's Note:
a) Write \( \dfrac{1}{1+e^{2x}} = 1 - \dfrac{e^{2x}}{1+e^{2x}} \) and integrate term by term.
b) This gives \( x - \dfrac{1}{2}\log(1+e^{2x}) + C \), so \( P = \dfrac{1}{2} \).

Q13. If vector \( 2\hat{i} - u\hat{j} + 3\hat{k} \) has magnitude 4, then value of 'u' is [1 Mark]
(A) \( 1 \)
(B) \( \sqrt{2} \)
(C) \( \pm\sqrt{3} \)
(D) \( \pm 2 \)

Answer: (C) \( \pm\sqrt{3} \)

Teacher's Note:
a) Use \( \sqrt{4 + u^2 + 9} = 4 \), giving \( u^2 = 3 \).
b) Remember to take both positive and negative square roots.

Q14. Angle made by the vector \( \hat{i} + \hat{j} + \sqrt{2}\hat{k} \) with positive direction of z-axis is [1 Mark]
(A) \( \dfrac{\pi}{6} \)
(B) \( \dfrac{\pi}{4} \)
(C) \( \dfrac{\pi}{3} \)
(D) \( \dfrac{\pi}{2} \)

Answer: (B) \( \dfrac{\pi}{4} \)

Teacher's Note:
a) The direction cosine along z-axis is \( \dfrac{\sqrt{2}}{2} = \dfrac{1}{\sqrt{2}} \).
b) So \( \cos\gamma = \dfrac{1}{\sqrt{2}} \) gives \( \gamma = \dfrac{\pi}{4} \).

Q15. A vector perpendicular to the vectors \( \hat{i} + \hat{j} \) and \( \hat{i} - \hat{j} \) is [1 Mark]
(A) \( -2\hat{j} \)
(B) \( 3\hat{k} \)
(C) \( 2\hat{i} \)
(D) \( 2\hat{i} + \hat{k} \)

Answer: (B) \( 3\hat{k} \)

Teacher's Note:
a) Both given vectors lie in the XY-plane.
b) So any vector purely along the z-axis is perpendicular to both, matching option (B).

Q16. The probability that it rains on a particular day is \( \dfrac{1}{5} \). Assuming weather conditions of any two days are independent, find the probability that it rains on three consecutive days. [1 Mark]
(A) \( \dfrac{4}{5} \)
(B) \( \dfrac{1}{5} \)
(C) \( \dfrac{1}{25} \)
(D) \( \dfrac{1}{125} \)

Answer: (D) \( \dfrac{1}{125} \)

Teacher's Note:
a) For independent events, multiply the probabilities: \( \dfrac{1}{5} \times \dfrac{1}{5} \times \dfrac{1}{5} \).
b) This gives \( \dfrac{1}{125} \).

Q17. A die is thrown twice. What is the probability that at least one number is 4, given the sum of the numbers appearing is a multiple of 5? [1 Mark]
(A) \( \dfrac{3}{5} \)
(B) \( \dfrac{4}{5} \)
(C) \( \dfrac{3}{7} \)
(D) \( \dfrac{4}{7} \)

Answer: (D) \( \dfrac{4}{7} \)

Teacher's Note:
a) List the outcomes where the sum is a multiple of 5: there are 7 such outcomes.
b) Of these, 4 outcomes contain at least one 4, giving \( P(B|A) = \dfrac{4}{7} \).

Q18. \( A \) and \( B \) are two independent events such that \( P(A) = \dfrac{1}{6}, P(B) = k \). If the probability of occurrence of exactly one of them is \( \dfrac{1}{3} \), then find the value of \( k \). [1 Mark]
(A) \( \dfrac{1}{4} \)
(B) \( \dfrac{2}{3} \)
(C) \( \dfrac{3}{4} \)
(D) \( \dfrac{5}{6} \)

Answer: (A) \( \dfrac{1}{4} \)

Teacher's Note:
a) "Exactly one occurs" means \( P(A')P(B) + P(A)P(B') = \dfrac{1}{3} \).
b) Substituting \( P(A) = \dfrac{1}{6} \) and solving gives \( k = \dfrac{1}{4} \).

Assertion-Reason Based Questions

Directions: Question numbers 19 and 20 are Assertion and Reason based questions carrying 1 mark each. Two statements are given, one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the options (A), (B), (C) and (D) as given below:

Q19. Assertion: The value of the integral \( \int_{-1}^{1} |x-2|\,dx \) is 4. Reason: If \( f(-x) = f(x) \) then \( \int_{-a}^{a} f(x)dx = 2\int_{0}^{a} f(x)dx \) [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(B) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
(C) Assertion (A) is true but Reason (R) is false.
(D) Assertion (A) is false but Reason (R) is true.

Answer: (B) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).

Teacher's Note:
a) Evaluating \( \int_{-1}^{1}|x-2|dx = \int_{-1}^{1}(2-x)dx = 4 \), so the assertion is true.
b) The reason is a true property of even functions, but it is not the reason the assertion holds since \( |x-2| \) is not an even function.

Q20. Assertion: The general solution of a differential equation represents a family of curves. Reason: The derivative of an arbitrary constant is zero. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(B) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
(C) Assertion (A) is true but Reason (R) is false.
(D) Assertion (A) is false but Reason (R) is true.

Answer: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

Teacher's Note:
a) The general solution contains an arbitrary constant, and different values of the constant give different curves, forming a family.
b) Since the derivative of the constant is zero, it does not affect the differential equation, which explains why all these curves satisfy the same equation.

21. (A). Evaluate \( \int \frac{1}{x\sqrt{x^6-1}}dx \) [2 Marks]

Answer:
1. Write \( I=\int \frac{1}{x\sqrt{x^6-1}}dx=\frac{1}{6}\int \frac{6x^5}{x^6\sqrt{x^6-1}}dx \).
2. Put \( x^6-1=t^2 \), so \( 6x^5dx=2t\,dt \), giving \( I=\frac{1}{3}\int \frac{1}{t^2+1}dt \).
3. Integrating, \( I=\frac{1}{3}\tan^{-1}(t)+C=\frac{1}{3}\tan^{-1}\left(\sqrt{x^6-1}\right)+C \).

Teacher's Note:
a) The key step is spotting the substitution \( x^6-1=t^2 \) after adjusting the numerator to \( x^5 \).
b) Remember the standard result \( \int \frac{dt}{t^2+1}=\tan^{-1}t+C \).

OR

(B). Evaluate \( \int \frac{\cos x-\sin x}{\sqrt{\sin(2x)}}dx \) [2 Marks]

Answer:
1. Write \( \sin(2x)=(\sin x+\cos x)^2-1 \), so \( I=\int \frac{\cos x-\sin x}{\sqrt{(\sin x+\cos x)^2-1}}dx \).
2. Put \( \sin x+\cos x=t \), so \( (\cos x-\sin x)dx=dt \), giving \( I=\int \frac{dt}{\sqrt{t^2-1}} \).
3. Integrating, \( I=\log\left|t+\sqrt{t^2-1}\right|+C=\log\left|\sin x+\cos x+\sqrt{\sin(2x)}\right|+C \).

Teacher's Note:
a) Converting \( \sin 2x \) using \( (\sin x+\cos x)^2-1 \) is the trick that allows substitution.
b) Recall \( \int \frac{dt}{\sqrt{t^2-1}}=\log|t+\sqrt{t^2-1}|+C \).

SECTION - B (5 x 2 = 10)

This section comprises of 5 very short answer (VSA) type questions of 2 marks each.

Q21 (A). Evaluate \( \int \dfrac{1}{x\sqrt{x^6-1}}dx \) [2 Marks]

Answer:
Let \( x^6 - 1 = t^2 \), so \( 6x^5 dx = 2t\,dt \).
\( I = \dfrac{1}{6}\int \dfrac{6x^5}{x^6\sqrt{x^6-1}}dx = \dfrac{1}{3}\int \dfrac{dt}{t^2+1} \)
\( I = \dfrac{1}{3}\tan^{-1}(\sqrt{x^6-1}) + C \)

Teacher's Note:
a) Multiply and divide by \( x^5 \) to create a substitutable form.
b) Do not forget the constant of integration C.

OR

Q21 (B). Evaluate \( \int \dfrac{\cos x - \sin x}{\sqrt{\sin(2x)}}dx \) [2 Marks]

Answer:
Write \( \sin(2x) = (\sin x + \cos x)^2 - 1 \).
Let \( \sin x + \cos x = t \Rightarrow (\cos x - \sin x)dx = dt \)
\( I = \int \dfrac{dt}{\sqrt{t^2-1}} = \log|t + \sqrt{t^2-1}| + C \)
\( I = \log|\sin x + \cos x + \sqrt{\sin(2x)}| + C \)

Teacher's Note:
a) Recognise \( \sin(2x)+1 = (\sin x + \cos x)^2 \) as the key trick.
b) Use the standard formula for \( \int \dfrac{dt}{\sqrt{t^2-1}} \).

Q22. If \( y = \sin(x+y) \) and \( \dfrac{dy}{dx} = 0 \), then prove that \( y = \pm 1 \). [2 Marks]

Answer:
Differentiating, \( \dfrac{dy}{dx} = \cos(x+y)\left(1+\dfrac{dy}{dx}\right) \)
\( \Rightarrow \dfrac{dy}{dx}(1-\cos(x+y)) = \cos(x+y) \)
Since \( \dfrac{dy}{dx} = 0 \), we get \( \cos(x+y) = 0 \), so \( \sin(x+y) = \pm 1 \)
As \( y = \sin(x+y) \), we get \( y = \pm 1 \)

Teacher's Note:
a) Differentiate implicitly, treating y as a function of x.
b) Substitute \( \dfrac{dy}{dx}=0 \) only after simplifying the equation.

23. (A). A line passing through the points (2,6,5), and (3, p, 7) is parallel to the line joining the points (-4, 3,4) and (-1,0, q). Find the value of \( \frac{q}{p} \). [2 Marks]

Answer:
1. Direction ratios of \( AB \) are \( \langle 1, p-6, 2\rangle \) and of \( CD \) are \( \langle 3,-3,q-4\rangle \).
2. Since the lines are parallel, \( \frac{1}{3}=\frac{p-6}{-3}=\frac{2}{q-4} \), giving \( p=5 \) and \( q=10 \).
3. Thus \( \frac{q}{p}=\frac{10}{5}=2 \).

Teacher's Note:
a) Parallel lines have proportional direction ratios; set up the equal ratios carefully.
b) Solve each ratio pair separately for \( p \) and \( q \) before computing the required value.

OR

(B). A rectangle ABCD is such that \( \vec{AC}=\vec{u} \), and \( \vec{BD}=\vec{v} \), then find the vectors representing four sides of the rectangle (in a cyclic order starting from \( \vec{AB} \)). [2 Marks]

Answer:
1. Let \( \vec{AB}=\vec{a}, \vec{BC}=\vec{b}, \vec{CD}=\vec{c}, \vec{DA}=\vec{d} \). Then \( \vec{u}=\vec{a}+\vec{b} \) and \( \vec{v}=\vec{b}-\vec{a} \).
2. Solving these, \( \vec{a}=\frac{1}{2}(\vec{u}-\vec{v}) \) and \( \vec{b}=\frac{1}{2}(\vec{u}+\vec{v}) \).
3. Also \( \vec{c}=-\vec{a}=\frac{1}{2}(\vec{v}-\vec{u}) \) and \( \vec{d}=-\vec{b}=-\frac{1}{2}(\vec{u}+\vec{v}) \).

Teacher's Note:
a) Use the triangle law along the diagonals to express \( \vec{u} \) and \( \vec{v} \) in terms of the sides.
b) Opposite sides of a rectangle are equal and opposite, so \( \vec{c}=-\vec{a} \) and \( \vec{d}=-\vec{b} \).

Q23 (A). A line passing through the points \( (2,6,5) \) and \( (3,p,7) \) is parallel to the line joining the points \( (-4,3,4) \) and \( (-1,0,q) \). Find the value of \( \dfrac{q}{p} \). [2 Marks]

Answer:
Direction ratios of AB: \( \langle 1, p-6, 2 \rangle \); of CD: \( \langle 3, -3, q-4 \rangle \)
As lines are parallel: \( \dfrac{1}{3} = \dfrac{p-6}{-3} = \dfrac{2}{q-4} \)
Solving: \( p = 5 \), \( q = 10 \)
\( \dfrac{q}{p} = \dfrac{10}{5} = 2 \)

Teacher's Note:
a) Set up direction ratios carefully from the given points.
b) Equate the ratios of direction ratios since the lines are parallel.

OR

Q23 (B). A rectangle ABCD is such that \( \vec{AC} = \vec{u} \), and \( \vec{BD} = \vec{v} \), then find the vectors representing four sides of the rectangle (in a cyclic order starting from \( \vec{AB} \)). [2 Marks]

Answer:
Let \( \vec{AB} = \vec{a}, \vec{BC} = \vec{b}, \vec{CD} = \vec{c}, \vec{DA} = \vec{d} \)
\( \vec{u} = \vec{a}+\vec{b} \) and \( \vec{v} = \vec{b}-\vec{a} \)
Solving: \( \vec{a} = \dfrac{1}{2}(\vec{u}-\vec{v}), \vec{b} = \dfrac{1}{2}(\vec{u}+\vec{v}) \)
\( \vec{c} = -\vec{a} = \dfrac{1}{2}(\vec{v}-\vec{u}), \vec{d} = -\vec{b} = -\dfrac{1}{2}(\vec{u}+\vec{v}) \)

Teacher's Note:
a) Draw the rectangle and mark the diagonals to set up the vector equations correctly.
b) Use the property that opposite sides of a rectangle give equal and opposite vectors.

Q24. For two non-zero vectors \( \vec{a} \) and \( \vec{b} \), \( |\vec{a}| = \sqrt{6} \), \( \vec{a} \cdot \vec{b} = 1 \) and \( \vec{b} \times \vec{a} = 3\hat{i}+4\hat{j}-2\hat{k} \). Find \( |\vec{b}| \). [2 Marks]

Answer:
\( |\vec{a} \times \vec{b}| = |\vec{b} \times \vec{a}| = \sqrt{9+16+4} = \sqrt{29} \)
Using \( |\vec{a}\times\vec{b}|^2 + |\vec{a}\cdot\vec{b}|^2 = |\vec{a}|^2|\vec{b}|^2 \)
\( 29+1 = 6|\vec{b}|^2 \Rightarrow |\vec{b}|^2 = 5 \Rightarrow |\vec{b}| = \sqrt{5} \)

Teacher's Note:
a) Note that \( |\vec{b}\times\vec{a}| = |\vec{a}\times\vec{b}| \) since magnitude doesn't depend on the order.
b) Use the identity connecting cross product, dot product and magnitudes.

Q25. The feasible region of a linear programming problem is shown in the given figure:
Write all the constraints on the decision variables involved [2 Marks]

[Figure: A graph showing lines \( 3x+2y=10 \) and \( x+2y=6 \) intersecting at point C, with the shaded feasible region bounded between points A, B and C in the first quadrant]

Answer:
\( 3x + 2y \geq 10 \)
\( x + 2y \leq 6 \)
\( y \geq 0 \)

Teacher's Note:
a) Check whether the shaded region lies above or below each line to decide the inequality sign.
b) Non-negativity constraints are usually needed unless the region shown allows negative values.

For Visually Impaired Candidates (in lieu of Q. 25)

For a linear programming problem, the maximum value of an objective function \( Z = 3x+4y \) exists at the points \( (p,q) \) and \( \left(2p, \dfrac{q}{2}\right) \). Find the relation between \( p \) and \( q \). [2 Marks]

Answer:
Since maximum exists at both points, \( Z \) at \( (p,q) \) = \( Z \) at \( \left(2p, \dfrac{q}{2}\right) \)
\( 3p+4q = 6p+2q \Rightarrow 3p = 2q \)

Teacher's Note:
a) When Z has the same maximum value at two different points, both points must give equal Z values.
b) Equate the two expressions for Z and simplify to get the relation.

SECTION - C (6 x 3 = 18)

This section comprises of 6 short answer (SA) type questions of 3 marks each.

Q26. For two matrices, \( P = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 1 & 2 \end{pmatrix} \) and \( Q = \begin{pmatrix} 1 & 0 \\ 0 & 2 \\ 1 & 0 \end{pmatrix} \), find \( PQ \) and \( QP \), whichever is defined. Also, find the inverse of the product matrix, if it exists. [3 Marks]

Answer:
1. \( PQ = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 1 & 2 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & 2 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 2 & 0 \\ 2 & 2 \end{pmatrix} \)
2. \( QP = \begin{pmatrix} 1 & 0 \\ 0 & 2 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 2 & 0 & 0 \\ 0 & 1 & 2 \end{pmatrix} = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 4 \\ 2 & 0 & 0 \end{pmatrix} \)
3. \( |PQ| = 4 \neq 0 \), so \( (PQ)^{-1} \) exists: \( (PQ)^{-1} = \begin{pmatrix} \dfrac{1}{2} & 0 \\ -\dfrac{1}{2} & \dfrac{1}{2} \end{pmatrix} \). Since \( |QP| = 0 \), \( QP \) is singular and its inverse does not exist.

Teacher's Note:
a) Check matrix compatibility (columns of first = rows of second) before multiplying.
b) A matrix has an inverse only if its determinant is non-zero.
c) Use \( M^{-1} = \dfrac{1}{|M|}adj(M) \) for a \( 2\times2 \) matrix.

Q27. The circular area of vision on the ground by a surveillance drone fitted with camera increases at the rate of 44 \( m^2 \) per metre as it gains vertical height. The radius of the circle on the ground increases at the rate of 6 m/s. Find the rate at which the height of the drone increases with time, at the instant when the radius of the circular area on the ground is 14 m. (Use \( \pi = \dfrac{22}{7} \)) [3 Marks]

[Figure: A drone with a camera pointed downward, hovering above a dashed circular area marked on the ground]

Answer:
1. Given \( \dfrac{dA}{dh} = 44 \), \( \dfrac{dr}{dt} = 6 \) m/s, and \( A = \pi r^2 \).
2. \( \dfrac{dA}{dt} = 2\pi r \dfrac{dr}{dt} \), and also \( \dfrac{dA}{dt} = \dfrac{dA}{dh}\cdot\dfrac{dh}{dt} \).
3. So \( 44\dfrac{dh}{dt} = 2\left(\dfrac{22}{7}\right)(14)(6) \Rightarrow \dfrac{dh}{dt} = 12 \) m/s.

Teacher's Note:
a) Identify A as a function of both r and h, and use the chain rule.
b) Substitute r = 14 only after setting up the equation in terms of derivatives.
c) Answer with correct units: 12 m/s.

28. (A). If \( x=\sin^3 t \), \( y=\cos^3 t \), Evaluate \( \frac{d^2y}{dx^2} \) at \( \left(\frac{1}{2\sqrt2},\frac{1}{2\sqrt2}\right) \). [3 Marks]

Answer:
1. \( \frac{dx}{dt}=3\sin^2t\cos t \) and \( \frac{dy}{dt}=-3\cos^2t\sin t \), so \( \frac{dy}{dx}=-\cot t \).
2. \( \frac{d^2y}{dx^2}=\frac{d}{dt}\left(\frac{dy}{dx}\right)\cdot\frac{dt}{dx}=\csc^2t\left(\frac{1}{3\sin^2t\cos t}\right)=\frac{1}{3}\sec t\csc^4t \).
3. At the given point, \( \sin^3t=\frac{1}{2\sqrt2}\Rightarrow t=\frac{\pi}{4} \).
4. So \( \frac{d^2y}{dx^2}\Big|_{t=\pi/4}=\frac{4\sqrt2}{3} \).

Teacher's Note:
a) First find \( \frac{dy}{dx} \) in terms of the parameter \( t \), then differentiate again with the chain rule.
b) Identify the correct value of \( t \) from the given point before substituting.

OR

(B). If \( y=x^{\log x}+(\log x)^x \); \( x\gt 1 \), then evaluate \( \frac{dy}{dx} \) [3 Marks]

Answer:
1. Let \( y=u+v \) where \( u=x^{\log x} \) and \( v=(\log x)^x \).
2. Taking logs, \( \log u=(\log x)^2 \Rightarrow \frac{1}{u}\frac{du}{dx}=\frac{2\log x}{x}\Rightarrow \frac{du}{dx}=x^{\log x}\left(\frac{2\log x}{x}\right) \).
3. Also \( \log v=x\log(\log x)\Rightarrow \frac{1}{v}\frac{dv}{dx}=\log(\log x)+\frac{1}{\log x}\Rightarrow \frac{dv}{dx}=(\log x)^x\left(\log(\log x)+\frac{1}{\log x}\right) \).
4. So \( \frac{dy}{dx}=\frac{2}{x}x^{\log x}\log x+(\log x)^x\left(\log(\log x)+\frac{1}{\log x}\right) \).

Teacher's Note:
a) Use logarithmic differentiation separately for each term since both base and exponent are variable.
b) Keep the two derivatives separate and simply add them at the end since \( y=u+v \).

29. (A). Find the particular solution of the differential equation \( xdy+(y-x^2e^x)dx=0 \), given that y (1) =0 [3 Marks]

Answer:
1. Rewrite as \( \frac{dy}{dx}+\frac{1}{x}y=xe^x \), a linear differential equation.
2. \( I.F.=e^{\int \frac{1}{x}dx}=e^{\log x}=x \).
3. Solution: \( yx=\int x^2e^xdx=x^2e^x-2(xe^x-e^x)+C \).
4. Using \( y(1)=0 \), \( C=-e \), so the particular solution is \( yx=e^x(x^2-2x+2)-e \).

Teacher's Note:
a) Recognise the equation as linear in \( y \) and find the integrating factor first.
b) Use integration by parts twice on \( \int x^2e^xdx \) and apply the initial condition to find \( C \).

OR

(B). Find the particular solution of the differential equation \( (x^2+y^2)dy=xydx \), given that \( y(0)=1 \) [3 Marks]

Answer:
1. Rewrite as \( \frac{dy}{dx}=\frac{xy}{x^2+y^2} \), a homogeneous equation.
2. Put \( y=vx \), so \( \frac{dy}{dx}=v+x\frac{dv}{dx} \); the equation reduces to \( \int\left(\frac{1}{v^3}+\frac{1}{v}\right)dv=-\int \frac{dx}{x} \).
3. Integrating, \( -\frac{1}{2v^2}+\log|v|=-\log|x|+\log C \), giving \( \log\left|\frac{y}{C}\right|=\frac{x^2}{2y^2} \) or \( y=Ce^{x^2/2y^2} \).
4. Using \( y(0)=1 \), \( C=1 \), so the particular solution is \( y=e^{x^2/2y^2} \).

Teacher's Note:
a) Check the equation is homogeneous of degree zero before substituting \( y=vx \).
b) Apply the initial condition carefully at the end to evaluate the constant \( C \).

30. (A). Find the foot of the perpendicular drawn from the point (2,1,4) on the line \( x-2=\frac{2y+3}{4}=2-z \). Also find the equation of the line passing through this point and the point (2,1,4). [3 Marks]

Answer:
1. General point on the line: \( P=\left(k+2,\frac{4k-3}{2},2-k\right) \); direction ratios of the line are \( \langle1,2,-1\rangle \).
2. Direction ratios of \( AP \) (where \( A=(2,1,4) \)) are \( \langle k,\frac{4k-5}{2},-k-2\rangle \).
3. Since \( AP \) is perpendicular to the line, \( k+2\left(\frac{4k-5}{2}\right)+(-1)(-k-2)=0 \Rightarrow k=\frac{1}{2} \).
4. Foot of perpendicular: \( P=\left(\frac{5}{2},-\frac{1}{2},\frac{3}{2}\right) \).
5. Equation of line \( AP \): \( \frac{x-2}{1}=\frac{y-1}{-3}=\frac{z-4}{-5} \).

Teacher's Note:
a) Take a general point on the line using the parameter, then use the perpendicularity condition (dot product = 0).
b) Substitute the value of the parameter back to get the exact foot of the perpendicular.

OR

(B). Find the angle between the lines: \( \frac{x-1}{2}=\frac{y-1}{2}=\frac{z-2}{1} \) and \( \frac{x-5}{2}=y-\frac{1}{2}=\frac{3-z}{2} \). Also find the cartesian equation of a line passing through the point (1,1,1) and perpendicular to both lines. [3 Marks]

Answer:
1. Rewrite the second line as \( \frac{x-5}{2}=\frac{y-\frac{1}{2}}{1}=\frac{z-3}{-2} \).
2. \( \cos\theta=\frac{2(2)+2(1)+1(-2)}{\sqrt9\times\sqrt9}=\frac{4}{9}\Rightarrow \theta=\cos^{-1}\left(\frac{4}{9}\right) \).
3. Direction of the required line: \( \vec{b_1}\times\vec{b_2}=\begin{vmatrix}\hat i&\hat j&\hat k\\2&2&1\\2&1&-2\end{vmatrix}=-5\hat i+6\hat j-2\hat k \).
4. Required line: \( \frac{x-1}{-5}=\frac{y-1}{6}=\frac{z-1}{-2} \).

Teacher's Note:
a) Use the direction ratios of both lines in the standard cosine formula to find the angle.
b) The line perpendicular to both given lines must be along their cross product vector.

Q28 (A). If \( x = \sin^3 t, y = \cos^3 t \), Evaluate \( \dfrac{d^2y}{dx^2} \) at \( \left(\dfrac{1}{2\sqrt{2}}, \dfrac{1}{2\sqrt{2}}\right) \). [3 Marks]

Answer:
1. \( \dfrac{dx}{dt} = 3\sin^2 t \cos t \), \( \dfrac{dy}{dt} = -3\cos^2 t \sin t \)
2. \( \dfrac{dy}{dx} = -\cot t \), and \( \dfrac{d^2y}{dx^2} = \dfrac{1}{3}(\sec t \cdot \csc^4 t) \)
3. At the given point, \( \sin^3 t = \dfrac{1}{2\sqrt{2}} \Rightarrow t = \dfrac{\pi}{4} \), so \( \dfrac{d^2y}{dx^2} = \dfrac{4\sqrt{2}}{3} \)

Teacher's Note:
a) Use the parametric formula \( \dfrac{d^2y}{dx^2} = \dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)\div\dfrac{dx}{dt} \).
b) Find the parameter t from the given point before substituting.

OR

Q28 (B). If \( y = x^{\log x} + (\log x)^x; x \gt 1 \), then evaluate \( \dfrac{dy}{dx} \) [3 Marks]

Answer:
1. Let \( u = x^{\log x} \): \( \log u = (\log x)^2 \Rightarrow \dfrac{du}{dx} = x^{\log x}\left(\dfrac{2\log x}{x}\right) \)
2. Let \( v = (\log x)^x \): \( \log v = x\log(\log x) \Rightarrow \dfrac{dv}{dx} = (\log x)^x\left(\log(\log x) + \dfrac{1}{\log x}\right) \)
3. \( \dfrac{dy}{dx} = \dfrac{2}{x}x^{\log x}\log x + (\log x)^x\left(\log(\log x) + \dfrac{1}{\log x}\right) \)

Teacher's Note:
a) Use logarithmic differentiation since both base and power involve x.
b) Differentiate each term (u and v) separately before adding.

Q29 (A). Find the particular solution of the differential equation \( xdy + (y - x^2e^x)dx = 0 \), given that \( y(1) = 0 \) [3 Marks]

Answer:
1. Rewrite as \( \dfrac{dy}{dx} + \dfrac{1}{x}y = xe^x \), a linear equation with I.F. \( = e^{\int \frac{1}{x}dx} = x \)
2. Solution: \( yx = \int x^2 e^x dx = x^2e^x - 2(xe^x - e^x) + C \)
3. Using \( y(1) = 0 \), \( C = -e \), so the particular solution is \( yx = e^x(x^2-2x+2) - e \)

Teacher's Note:
a) Identify the equation as linear in y and find the integrating factor first.
b) Apply integration by parts twice to evaluate \( \int x^2e^x dx \).
c) Use the initial condition only at the end to find C.

OR

Q29 (B). Find the particular solution of the differential equation \( (x^2+y^2)dy = xy\,dx \), given that \( y(0) = 1 \) [3 Marks]

Answer:
1. This is homogeneous; let \( y = vx \Rightarrow \dfrac{dy}{dx} = v + x\dfrac{dv}{dx} \)
2. Substituting and simplifying: \( \int\left(\dfrac{1}{v^3}+\dfrac{1}{v}\right)dv = -\int \dfrac{dx}{x} \), giving \( \log\left|\dfrac{y}{C}\right| = \dfrac{x^2}{2y^2} \)
3. Using \( y(0)=1 \), \( C=1 \), so the particular solution is \( y = e^{\frac{x^2}{2y^2}} \)

Teacher's Note:
a) Check that the equation is homogeneous (same degree in x, y throughout) before substituting y = vx.
b) Apply the initial condition after obtaining the general solution.

Q30 (A). Find the foot of the perpendicular drawn from the point \( (2,1,4) \) on the line \( x - 2 = \dfrac{2y+3}{4} = 2-z \). Also find the equation of the line passing through this point and the point \( (2,1,4) \). [3 Marks]

Answer:
1. General point on the line: \( P\left(k+2, \dfrac{4k-3}{2}, 2-k\right) \), direction ratios of line: \( \langle 1,2,-1 \rangle \)
2. As AP is perpendicular to the line: \( k + 4k-5+k+2 = 0 \Rightarrow k = \dfrac{1}{2} \)
3. Foot of perpendicular: \( \left(\dfrac{5}{2}, -\dfrac{1}{2}, \dfrac{3}{2}\right) \); equation of AP: \( \dfrac{x-2}{1} = \dfrac{y-1}{-3} = \dfrac{z-4}{-5} \)

Teacher's Note:
a) Express a general point on the line using a parameter k.
b) Use the perpendicularity condition (dot product = 0) to find k.
c) Substitute k back to get the foot and then form the line equation.

OR

Q30 (B). Find the angle between the lines: \( \dfrac{x-1}{2} = \dfrac{y-1}{2} = \dfrac{z-2}{1} \) and \( \dfrac{x-5}{2} = y - \dfrac{1}{2} = \dfrac{3-z}{2} \). Also find the cartesian equation of a line passing through the point \( (1,1,1) \) and perpendicular to both lines. [3 Marks]

Answer:
1. Direction ratios: \( \langle 2,2,1 \rangle \) and \( \langle 2,1,-2 \rangle \)
2. \( \cos\theta = \dfrac{4+2-2}{\sqrt{9}\sqrt{9}} = \dfrac{4}{9} \Rightarrow \theta = \cos^{-1}\left(\dfrac{4}{9}\right) \)
3. Required line direction \( = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 2 & 1 \\ 2 & 1 & -2 \end{vmatrix} = -5\hat{i}+6\hat{j}-2\hat{k} \); equation: \( \dfrac{x-1}{-5} = \dfrac{y-1}{6} = \dfrac{z-1}{-2} \)

Teacher's Note:
a) Use the formula \( \cos\theta = \dfrac{\vec{b_1}\cdot\vec{b_2}}{|\vec{b_1}||\vec{b_2}|} \) for the angle between lines.
b) The cross product of the two direction vectors gives a vector perpendicular to both.

Q31. Solve the linear programming problem graphically: Maximize \( Z = 40x+30y \) Subject to constraints: \( x+y \leq 6, 2x \geq y, y \geq 2 \) [3 Marks]

Answer:
1. The feasible region is bounded with vertices \( A(1,2), B(4,2), C(2,4) \).
2. Evaluating Z: \( Z_A = 100, Z_B = 220, Z_C = 200 \).
3. Maximum value of Z is 220 at the point \( B(4,2) \).

Teacher's Note:
a) Plot all three constraint lines and shade the common feasible region.
b) Evaluate the objective function only at the corner points of the feasible region.
c) The maximum occurs at the vertex giving the largest Z value.

32. (A). Let \( A=N\times N \), where \( N \) is the set of natural numbers and a relation \( R \) is defined on \( A \) such that \( (a,b)R(c,d) \) if and only if \( a^2+d^2=b^2+c^2 \). Check whether \( R \) is an equivalence relation or not? [5 Marks]

Answer:
1. Reflexive: For \( (p,q)\in N\times N \), \( p^2+q^2=q^2+p^2 \), so \( (p,q)R(p,q) \). Thus \( R \) is reflexive.
2. Symmetric: If \( (p,q)R(r,s) \), then \( p^2+s^2=q^2+r^2\Rightarrow r^2+q^2=s^2+p^2\Rightarrow (r,s)R(p,q) \). Thus \( R \) is symmetric.
3. Transitive: If \( (p,q)R(r,s) \) and \( (r,s)R(u,v) \), then \( p^2+s^2=q^2+r^2 \) and \( r^2+v^2=s^2+u^2 \). Adding gives \( p^2+v^2=q^2+u^2\Rightarrow (p,q)R(u,v) \). Thus \( R \) is transitive.
4. Since \( R \) is reflexive, symmetric and transitive, \( R \) is an equivalence relation.

Teacher's Note:
a) Check all three properties separately using the definition of the relation directly with algebra.
b) Adding the two given equations is the key trick for proving transitivity.
c) State the final conclusion clearly: equivalence relation needs all three properties.

OR

(B). A relation \( R:(-\infty,\infty)\to[0,\infty) \) is defined as \( R=\{(x,y):y=x^2+5\} \) Draw its graph and check whether it is a function or not. If the given relation is a function, then examine whether the function is injective, and whether it is surjective or not. [5 Marks]

[Figure: Graph of an upward parabola \( y=x^2+5 \), symmetric about the y-axis, with vertex at (0,5), opening upwards.]

Answer:
1. Since every element of \( (-\infty,\infty) \) has a unique image in \( [0,\infty) \) as shown by the graph, \( R \) is a function \( f(x)=x^2+5 \).
2. Injective: \( f(-1)=6=f(1) \), so two different inputs give the same output. Hence \( f \) is not injective (many-one).
3. Surjective: Range of \( f \) is \( [5,\infty) \), which is not equal to the co-domain \( [0,\infty) \). Hence \( f \) is not surjective.
4. So \( f \) is a function, but it is neither injective nor surjective.

Teacher's Note:
a) A vertical line test on the graph confirms it is a function.
b) Comparing the range with the co-domain is the quickest way to check surjectivity.
c) Finding two distinct inputs with the same output is enough to disprove injectivity.

For Visually Impaired Candidates (in lieu of Q. 31)

For a linear programming problem, vertices of the feasible region are \( (20,0), (40,0), (30,10), (0,20) \) and \( (0,10) \). Minimize and maximize the objective function \( Z = 50x-20y \) [3 Marks]

Answer:
1. Evaluate Z at each vertex: \( Z(20,0)=1000, Z(40,0)=2000, Z(30,10)=1300, Z(0,20)=-400, Z(0,10)=-200 \).
2. Minimum value of Z is \( -400 \) at \( (0,20) \).
3. Maximum value of Z is \( 2000 \) at \( (40,0) \).

Teacher's Note:
a) Since vertices are already given, simply substitute each into the objective function.
b) Compare all values carefully to correctly identify minimum and maximum.

SECTION - D (4 x 5 = 20)

This section comprises of 4 long answer (LA) type questions of 5 marks each.

Q32 (A). Let \( A = N \times N \), where \( N \) is the set of natural numbers and a relation \( R \) is defined on \( A \) such that \( (a,b)R(c,d) \) if and only if \( a^2+d^2 = b^2+c^2 \). Check whether \( R \) is an equivalence relation or not? [5 Marks]

Answer:
1. Reflexive: For \( (p,q) \in A \), \( p^2+q^2 = q^2+p^2 \), so \( (p,q)R(p,q) \) always holds; R is reflexive.
2. Symmetric: If \( (p,q)R(r,s) \), then \( p^2+s^2=q^2+r^2 \), which gives \( r^2+q^2=s^2+p^2 \), so \( (r,s)R(p,q) \); R is symmetric.
3. Transitive: If \( (p,q)R(r,s) \) and \( (r,s)R(u,v) \), adding the two equations gives \( p^2+v^2=q^2+u^2 \), so \( (p,q)R(u,v) \); R is transitive.
4. Since R is reflexive, symmetric and transitive, R is an equivalence relation.

Teacher's Note:
a) Always check all three properties in the order reflexive, symmetric, transitive.
b) For transitivity, adding the two given equations is the key trick.
c) State the final conclusion clearly to earn the concluding mark.

OR

Q32 (B). A relation \( R: (-\infty, \infty) \to [0, \infty) \) is defined as \( R = \{(x,y): y = x^2+5\} \) Draw its graph and check whether it is a function or not. If the given relation is a function, then examine whether the function is injective, and whether it is surjective or not. [5 Marks]

[Figure: An upward-opening parabola with vertex at (0,5), symmetric about the y-axis]

Answer:
1. The graph shows that every element of \( (-\infty,\infty) \) has a unique image in \( [0,\infty) \), so R is a function \( f(x) = x^2+5 \).
2. Injective: \( f(-1) = f(1) = 6 \), so two different inputs give the same output; f is not injective.
3. Surjective: Range of f is \( [5,\infty) \), which is not equal to the co-domain \( [0,\infty) \); f is not surjective.
4. So f is neither injective nor surjective.

Teacher's Note:
a) A vertical line test on the graph confirms it is a function.
b) Finding two x-values giving the same y-value is the quickest way to disprove injectivity.
c) Compare the range with the co-domain to check surjectivity.

For Visually Impaired Candidates (in lieu of Q. 32(B))

A relation \( R: (-\infty, \infty) \to [0, \infty) \) is defined as \( R = \{(x,y): y = x^2+7\} \). Check whether it is a function or not. If the given relation is a function, then examine whether the function is injective, surjective, or both, and conclude whether it is bijective or not. [5 Marks]

Answer:
1. For every \( a \in (-\infty,\infty) \), \( a^2 \in [0,\infty) \), so \( a^2+7 \in [7,\infty) \); every element has a unique image, so \( f(x)=x^2+7 \) is a function.
2. Injective: \( f(-1) = f(1) = 8 \), so f is many-one, hence not injective.
3. Surjective: Range of f is \( [7,\infty) \), which does not equal the co-domain \( [0,\infty) \); f is not surjective.
4. Since f is neither injective nor surjective, f is not bijective.

Teacher's Note:
a) Confirm uniqueness of image for every input before calling it a function.
b) A function must be both injective and surjective to be called bijective.

Q33. Evaluate \( \int_{-a}^{a} \sqrt{\dfrac{a-x}{a+x}}\,dx \) [5 Marks]

Answer:
1. Multiply numerator and denominator inside the root by \( (a-x) \): \( I = \int_{-a}^{a}\dfrac{a-x}{\sqrt{a^2-x^2}}dx = I_1 - I_2 \)
2. \( I_1 = a\int_{-a}^{a}\dfrac{1}{\sqrt{a^2-x^2}}dx = 2a\int_0^a \dfrac{1}{\sqrt{a^2-x^2}}dx \) (even function) \( = 2a\left[\sin^{-1}\left(\dfrac{x}{a}\right)\right]_0^a = \pi a \)
3. \( I_2 = \int_{-a}^{a}\dfrac{x}{\sqrt{a^2-x^2}}dx = 0 \) (odd function)
4. Therefore \( I = I_1 - I_2 = \pi a \)
5. So \( \int_{-a}^{a} \sqrt{\dfrac{a-x}{a+x}}\,dx = \pi a \)

Teacher's Note:
a) Rationalising the integrand is the key step to simplify the square root.
b) Split the integral and use the properties of even and odd functions over symmetric limits.
c) Remember the standard result \( \int \dfrac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}\left(\dfrac{x}{a}\right)+C \).

For Visually Impaired Candidates (in lieu of Q.32(B))

A relation \( R:(-\infty,\infty)\to[0,\infty) \) is defined as \( R=\{(x,y):y=x^2+7\} \). Check whether it is a function or not. If the given relation is a function, then examine whether the function is injective, surjective, or both, and conclude whether it is bijective or not. [5 Marks]

Answer:
1. For every \( a\in(-\infty,\infty) \), \( a^2\in[0,\infty) \), so \( a^2+7\in[7,\infty) \); every element has a unique image, so \( f(x)=x^2+7 \) is a function.
2. Injective: \( f(-1)=8=f(1) \), so \( f \) is many-one, hence not injective.
3. Surjective: Range of \( f \) is \( [7,\infty) \), which is not equal to the co-domain \( [0,\infty) \). Hence \( f \) is not surjective.
4. Since \( f \) is neither injective nor surjective, it is not bijective.

Teacher's Note:
a) Showing that some values in the co-domain (like values less than 7) have no pre-image is enough to disprove surjectivity.
b) A function must be both injective and surjective to be called bijective.

34. (A). Using Integration, find the area of the region \( \{(x,y):0\le y\le \sqrt{4-x^2}, x\ge 2-y\} \). [5 Marks]

[Figure: A semi-circle \( y=\sqrt{4-x^2} \) of radius 2 centred at origin, and the line \( x+y=2 \), intersecting at (0,2) and (2,0); the required shaded region lies between the semicircle and the line.]

Answer:
1. \( y=\sqrt{4-x^2} \) is a semicircle of radius 2 centred at the origin, and \( x=2-y \) i.e. \( x+y=2 \) is a line meeting the axes at (2,0) and (0,2).
2. The curves meet at the points (0,2) and (2,0).
3. Area \( =\int_0^2\left(\sqrt{4-x^2}-(2-x)\right)dx \).
4. \( =\left[\frac{x}{2}\sqrt{4-x^2}+2\sin^{-1}\left(\frac{x}{2}\right)-2x+\frac{x^2}{2}\right]_0^2 \).
5. \( =\pi-4+2=(\pi-2) \) sq. units.

Teacher's Note:
a) Sketching the region and finding the points of intersection is essential before setting up the integral.
b) Use the standard formula \( \int\sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)+C \).

OR

(B). Using Integration, find the area of the region bounded by the curve \( y^2=2x \) and the line \( x-2y=0 \). [5 Marks]

[Figure: A parabola \( y^2=2x \) opening rightwards and the line \( x=2y \), intersecting at (0,0) and (8,4); the shaded region lies between the parabola and the line.]

Answer:
1. \( y^2=2x \) is a parabola, and it meets the line \( x=2y \) at the points (0,0) and (8,4).
2. Area \( =\int_0^8\left(\sqrt{2x}-\frac{x}{2}\right)dx \).
3. \( =\left[\sqrt2\cdot\frac{2}{3}x^{3/2}-\frac{x^2}{4}\right]_0^8 \).
4. \( =\frac{64}{3}-16=\frac{16}{3} \) sq. units.

Teacher's Note:
a) Solve the parabola and line equations simultaneously to get the limits of integration.
b) Express \( y \) from both curves in terms of \( x \) before setting up the difference of areas.

For Visually Impaired Candidates (in lieu of Q.34(A))

Using Integration, find the area of the region enclosed within \( 25x^2+16y^2=400 \) [5 Marks]

Answer:
1. \( 25x^2+16y^2=400 \) is an ellipse with standard form \( \frac{x^2}{16}+\frac{y^2}{25}=1 \), giving \( y=\frac{5}{4}\sqrt{16-x^2} \) for the upper right quarter with \( a=4 \).
2. Area \( =4\int_0^4\frac{5}{4}\sqrt{16-x^2}\,dx=5\int_0^4\sqrt{16-x^2}\,dx \).
3. \( =5\left[\frac{x}{2}\sqrt{16-x^2}+8\sin^{-1}\left(\frac{x}{4}\right)\right]_0^4 \).
4. \( =5[4\pi]=20\pi \) sq. units.

Teacher's Note:
a) Convert the given equation into the standard ellipse form first to identify \( a \) and \( b \).
b) Use symmetry (4 times the first-quadrant area) to simplify the integration.

OR

Using Integration find the area of the region in the first quadrant bounded by the curve \( y^2=8x \), and the lines \( x=2, x=5 \) and x-axis. [5 Marks]

Answer:
1. \( y^2=8x \) is a parabola opening rightwards; the required area lies above the x-axis, below the curve, between \( x=2 \) and \( x=5 \).
2. Area \( =\int_2^5\sqrt{8x}\,dx \).
3. \( =\left[2\sqrt2\cdot\frac{2}{3}x^{3/2}\right]_2^5=\frac{4\sqrt2}{3}\left[5\sqrt5-2\sqrt2\right] \).
4. \( =\frac{4}{3}\left[5\sqrt{10}-4\right] \) sq. units.

Teacher's Note:
a) Express \( y \) as \( \sqrt{8x} \) since the region is in the first quadrant (positive y-values only).
b) Apply the limits directly since the boundaries \( x=2 \) and \( x=5 \) are already given.

Q34 (A). Using Integration, find the area of the region \( \{(x,y): 0 \leq y \leq \sqrt{4-x^2}, x \geq 2-y\} \). [5 Marks]

[Figure: A quarter/semicircle of radius 2 centred at origin and the line x + y = 2, with the shaded region between the curve and the line from (0,2) to (2,0)]

Answer:
1. \( y = \sqrt{4-x^2} \) is a semicircle of radius 2 centred at origin; \( x = 2-y \) i.e. \( x+y=2 \) is a straight line.
2. The curves intersect at \( (0,2) \) and \( (2,0) \).
3. Area \( = \int_0^2 \left(\sqrt{4-x^2} - (2-x)\right)dx \)
4. \( = \left[\dfrac{x}{2}\sqrt{4-x^2} + 2\sin^{-1}\left(\dfrac{x}{2}\right) - 2x + \dfrac{x^2}{2}\right]_0^2 \)
5. \( = \pi - 4 + 2 = (\pi - 2) \) sq. units

Teacher's Note:
a) Sketching both curves and finding intersection points is essential before integrating.
b) Use the standard formula for the area under a semicircle.
c) Substitute the limits carefully to avoid sign errors.

OR

Q34 (B). Using Integration, find the area of the region bounded by the curve \( y^2 = 2x \) and the line \( x - 2y = 0 \). [5 Marks]

[Figure: A rightward-opening parabola \( y^2=2x \) and the line \( x=2y \), intersecting at (0,0) and (8,4), with the shaded region between them]

Answer:
1. The parabola \( y^2=2x \) and the line \( x=2y \) intersect at \( (0,0) \) and \( (8,4) \).
2. Area \( = \int_0^8 \left(\sqrt{2x} - \dfrac{x}{2}\right)dx \)
3. \( = \left[\sqrt{2}\dfrac{2}{3}x^{3/2} - \dfrac{x^2}{4}\right]_0^8 \)
4. \( = \dfrac{64}{3} - 16 = \dfrac{16}{3} \) sq. units

Teacher's Note:
a) Find points of intersection by solving both equations simultaneously.
b) Take the curve above the line as the upper boundary while setting up the integral.
c) Double-check the power rule while integrating \( \sqrt{2x} \).

For Visually Impaired Candidates (in lieu of Q. 34)

Q34 (A). Using Integration, find the area of the region enclosed within \( 25x^2 + 16y^2 = 400 \) [5 Marks]

Answer:
1. The equation represents an ellipse: \( \dfrac{x^2}{16} + \dfrac{y^2}{25} = 1 \), so \( y = \dfrac{5}{4}\sqrt{16-x^2} \).
2. By symmetry, Area \( = 4\int_0^4 \dfrac{5}{4}\sqrt{16-x^2}\,dx = 5\int_0^4 \sqrt{16-x^2}\,dx \)
3. \( = 5\left[\dfrac{x}{2}\sqrt{16-x^2} + 8\sin^{-1}\left(\dfrac{x}{4}\right)\right]_0^4 \)
4. \( = 5[4\pi] = 20\pi \) sq. units

Teacher's Note:
a) Convert the given equation into the standard form of an ellipse first.
b) Use symmetry about both axes to simplify the integral to one quadrant multiplied by 4.

OR

Q34 (B). Using Integration find the area of the region in the first quadrant bounded by the curve \( y^2 = 8x \), and the lines \( x=2, x=5 \) and x-axis. [5 Marks]

Answer:
1. \( y^2=8x \) represents a parabola opening rightward.
2. Area \( = \int_2^5 \sqrt{8x}\,dx \)
3. \( = \left[2\sqrt{2}\dfrac{2}{3}x^{3/2}\right]_2^5 \)
4. \( = \sqrt{2}\left(\dfrac{4}{3}\right)[5\sqrt{5}-2\sqrt{2}] = \dfrac{4}{3}[5\sqrt{10}-4] \) sq. units

Teacher's Note:
a) Take y as a function of x, i.e., \( y=\sqrt{8x} \), for the area under the curve.
b) Substitute the limits 2 and 5 carefully and simplify surds neatly.

Q35. A fitness app syncs with four types of smartwatches: Alpha, Beta, Gamma and Delta. 45% of the users use Alpha, 30% use Beta, 15% use Gamma and 10% use Delta.
Due to device errors, the probabilities of getting a false heart-rate alert from the Alpha, Beta, Gamma and Delta smartwatches are respectively 0.01, 0.02, 0.05, and 0.01.
Based on the given information, answer the following questions:
(i) Find the probability that a randomly selected user gets a false alert.
(ii) If a false alert occurs, what is the probability that it came from a Beta smartwatch? [5 Marks]

Answer:
1. Let \( E_1, E_2, E_3, E_4 \) denote using Alpha, Beta, Gamma, Delta respectively, with \( P(E_1)=0.45, P(E_2)=0.3, P(E_3)=0.15, P(E_4)=0.1 \).
2. Given \( P(A|E_1)=0.01, P(A|E_2)=0.02, P(A|E_3)=0.05, P(A|E_4)=0.01 \), where A is the event of a false alert.
3. By the Theorem of Total Probability: \( P(A) = 0.45(0.01)+0.3(0.02)+0.15(0.05)+0.1(0.01) = 0.019 \)
4. By Bayes' Theorem: \( P(E_2|A) = \dfrac{P(E_2)P(A|E_2)}{P(A)} = \dfrac{0.3 \times 0.02}{0.019} = \dfrac{6}{19} \)
5. So the probability of a false alert is 0.019, and given a false alert, the probability it came from Beta is \( \dfrac{6}{19} \).

Teacher's Note:
a) Use the Theorem of Total Probability for part (i), summing over all four watch types.
b) Use Bayes' Theorem for part (ii), placing the required event's product term in the numerator.
c) Keep the value of \( P(A) \) exact (0.019) to avoid rounding errors in part (ii).

SECTION - E (3 x 4 = 12)

This section comprises 3 case-study/passage-based questions of 4 marks each with sub parts. The first two case study questions have three sub parts (i), (ii), (iii) of marks 1, 1, 2 respectively. The third case study question has two sub parts of 2 marks each.

Q36. Two security sensors in a multi-storey mall emit laser beams for movement detection. Three of these beams travel along the following lines:
\( l_1: \vec{r} = (\lambda+1)\hat{i} + (3\lambda+2)\hat{j} + (3-\lambda)\hat{k} \) and
\( l_2: \vec{r} = 2(\mu+1)\hat{i} + (3\mu-1)\hat{j} + (\mu+1)\hat{k} \)
\( l_3: \vec{r} = (\gamma+3)\hat{i} + (3\gamma-1)\hat{j} + (1-\gamma)\hat{k} \)
Based on the given information, answer the following questions:

[Figure: Photo of a multi-storey mall interior with red laser beams crossing between the floors and escalators, illustrating the paths of lines l1, l2 and l3]

(i) Show that the lines \( l_1 \) and \( l_2 \) are not parallel to each other. [1 Mark]

Answer: Writing \( l_1 = (\hat{i}+2\hat{j}+3\hat{k}) + \lambda(\hat{i}+3\hat{j}-\hat{k}) \) and \( l_2 = (2\hat{i}-\hat{j}+\hat{k}) + \mu(2\hat{i}+3\hat{j}+\hat{k}) \), the direction vectors are \( \vec{b}=(1,3,-1) \) and \( \vec{d}=(2,3,1) \). Since \( \vec{b} \neq \lambda\vec{d} \) for any scalar \( \lambda \), the lines are not parallel.

Teacher's Note:
a) Rewrite the given parametric equations in the standard form \( \vec{a}+\lambda\vec{b} \) first.
b) Two lines are parallel only if one direction vector is a scalar multiple of the other.

(ii) Show that the lines \( l_1 \) and \( l_3 \) are parallel to each other. [1 Mark]

Answer: Writing \( l_3 = (3\hat{i}-\hat{j}+\hat{k}) + \gamma(\hat{i}+3\hat{j}-\hat{k}) \), its direction vector is \( (1,3,-1) \), which is identical to the direction vector of \( l_1 \). Hence \( l_1 \) and \( l_3 \) are parallel.

Teacher's Note:
a) Compare the direction vectors directly; identical vectors mean the lines are parallel.
b) Note that parallel lines may still be distinct (non-intersecting) lines.

(iii) (a) Find the shortest distance between the lines \( l_1 \) and \( l_2 \). [2 Marks]

Answer:
1. \( \vec{c}-\vec{a} = \hat{i}-3\hat{j}-2\hat{k} \), and \( \vec{b}\times\vec{d} = 6\hat{i}-3\hat{j}-3\hat{k} = 3(2\hat{i}-\hat{j}-\hat{k}) \)
2. \( (\vec{c}-\vec{a})\cdot(\vec{b}\times\vec{d}) = 3(2+3+2) = 21 \neq 0 \), so the lines are skew.
3. \( |\vec{b}\times\vec{d}| = 3\sqrt{6} \), so shortest distance \( = \left|\dfrac{21}{3\sqrt{6}}\right| = \dfrac{7\sqrt{6}}{6} \) units

Teacher's Note:
a) Use the skew-lines formula \( D = \left|\dfrac{(\vec{c}-\vec{a})\cdot(\vec{b}\times\vec{d})}{|\vec{b}\times\vec{d}|}\right| \).
b) Always rationalise the final surd in the denominator.

OR

(iii) (b) Find the shortest distance between the lines \( l_1 \) and \( l_3 \). [2 Marks]

Answer:
1. Since \( l_1 \) and \( l_3 \) are parallel with the same direction vector \( \vec{b} \), use \( D = \dfrac{|\vec{b}\times(\vec{e}-\vec{a})|}{|\vec{b}|} \).
2. \( \vec{e}-\vec{a} = 2\hat{i}-3\hat{j}-2\hat{k} \), and \( \vec{b}\times(\vec{e}-\vec{a}) = -9\hat{i}-9\hat{k} \)
3. \( D = \dfrac{|-9\hat{i}-9\hat{k}|}{\sqrt{11}} = \dfrac{9\sqrt{22}}{11} \) units

Teacher's Note:
a) For parallel lines, use the formula based on the cross product of the direction vector and the vector joining the two points.
b) Compute the magnitude of the cross product carefully before dividing.

Q37. Water is essential for life, and rivers are vital water sources across the world. The Yamuna Action Plan (YAP), one of India's largest river restoration projects, was launched to clean the Yamuna, but its effectiveness must be measured through cost-benefit, economic and environmental analysis. Therefore, comprehensive pollutant treatment, prevention of further contamination, and community awareness are crucial for sustainable river management.
Implementing the same concept to the river Yamuna, if an initial base line WQI is 35, a quadratic expression is formed to measure the impact of investment of Rs. x crore on the WQI i.e., \( W(x) \) as
\( W(x) = 35 + 1.6x - 0.02x^2 \), where x is in Rupees in crores
Use the given \( W(x) \) to answer the following questions:

[Figure: Photo of a heavily polluted river with a bridge overhead and debris floating on the water, showing the state of the Yamuna]

(i) Find the value of \( x \) at the stationary point for the function \( W(x) \). [1 Mark]

Answer: \( W'(x) = 1.6 - 0.04x = 0 \Rightarrow x = 40 \). So the stationary point occurs at x = 40 crores.

Teacher's Note:
a) Set the first derivative equal to zero to find the stationary point.
b) The unit of x is crores of rupees, as stated in the question.

(ii) Find the maximum value of the WQI using the second derivative test? [1 Mark]

Answer: \( W''(x) = -0.04 \lt 0 \), so \( W(x) \) is maximum at x = 40. Maximum WQI \( = W(40) = 35+64-32 = 67 \).

Teacher's Note:
a) A negative second derivative at the stationary point confirms a maximum.
b) Substitute x = 40 back into W(x) to get the maximum value.

(iii) (a) If the investment is Rs. 35 crores, find by how much the WQI is short of its maximum achievable value. [2 Marks]

Answer:
1. \( W(35) = 35+1.6(35)-0.02(35)^2 = 91-24.5 = 66.5 \)
2. Difference from maximum: \( 67-66.5 = 0.5 \). So the WQI is short by 0.5 units of its maximum level.

Teacher's Note:
a) Substitute x = 35 directly into the given function to find W(35).
b) Subtract from the maximum value found in part (ii).

OR

(iii) (b) Evaluate the WQI when the investment is Rs. 60 crore. Determine whether this indicates an enhancement or a diminishing return in our objective when compared with the optimum WQI. [2 Marks]

Answer:
1. \( W(60) = 35+1.6(60)-0.02(60)^2 = 35+96-72 = 59 \)
2. Since 59 is less than the maximum WQI of 67 even after increasing the investment from 40 to 60 crores, this shows diminishing returns.

Teacher's Note:
a) Compare the value obtained with the maximum value found earlier.
b) A lower WQI despite higher investment beyond the optimum point indicates diminishing returns.

Q38. To celebrate their child's birthday, a family decided to distribute fruits among children in three orphanages. The fruit vendor packed three different types of baskets, according to the quantity required at each orphanage, details are given as follows:
Basket A contains 5 kg Apples, 10 kg Oranges and 5 kg Pomegranate
Basket B contains 10 kg Apples, 10 kg Oranges and 5kg Pomegranate
Basket C contains 8 kg Apples, 15 kg Oranges and 7kg Pomegranate
Cost of basket A, B and C are respectively Rs. 2750, Rs. 3500, and Rs. 4100.
Based on the given information, answer the following questions:

[Figure: Photo of a fruit vendor selling fruit baskets to a family with a child, at a market stall]

(i) Write the matrix form of the equations involved in this situation and find the adjoint of the coefficient matrix involved. [2 Marks]

Answer:
1. Taking price per kg of apples, oranges and pomegranate as x, y, z, the matrix form is \( AX = B \), where \( A = \begin{pmatrix} 5 & 10 & 5 \\ 10 & 10 & 5 \\ 8 & 15 & 7 \end{pmatrix} \), \( X = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \), \( B = \begin{pmatrix} 2750 \\ 3500 \\ 4100 \end{pmatrix} \)
2. \( |A| = 25 \neq 0 \), so a unique solution exists. \( adj(A) = \begin{pmatrix} -5 & 5 & 0 \\ -30 & -5 & 25 \\ 70 & 5 & -50 \end{pmatrix} \)

Teacher's Note:
a) Form the coefficient matrix carefully using the quantities of each fruit in every basket.
b) Check that the determinant is non-zero before proceeding to find the adjoint and inverse.

(ii) Using inverse of the matrix, calculate the price of each fruit per kg. [2 Marks]

Answer:
1. \( X = A^{-1}B = \dfrac{1}{25}(adj\,A)B \)
2. Solving gives \( x = 150, y = 100, z = 200 \). So Apples cost Rs. 150 per kg, Oranges cost Rs. 100 per kg, and Pomegranate costs Rs. 200 per kg.

Teacher's Note:
a) Use \( A^{-1} = \dfrac{1}{|A|}adj(A) \) and multiply by B to solve for X.
b) State the final prices with correct units (Rs. per kg) for full marks.