Plus Two Mathematics Answer Key
December 2025 Question Paper
Section 1: Answer any 6 questions from 1 to 8 (6 × 3 = 18 marks)
Question 1
(i) Let R be a relation defined on A = {1, 2, 3} by R = {(1,1), (2,2), (3,3), (1,3)}. Then R is:
Answer: (d) reflexive and transitive
Explanation:
- Reflexive: Yes, because (1,1), (2,2), (3,3) are all present.
- Symmetric: No, because (1,3) is present but (3,1) is not.
- Transitive: Yes, because the only nontrivial case is (1,3) and (3,3) → (1,3) is already present.
(ii) Make the relation R equivalence by adding minimum number of ordered pairs.
Answer: Add (3,1) to make it symmetric.
New equivalence relation: \( R_{eq} = \{(1,1), (2,2), (3,3), (1,3), (3,1)\} \)
(iii) Write the equivalence class [1].
Answer: \( [1] = \{1, 3\} \)
Question 2
(i) \(\sin^{-1}x : [-1, 1] \rightarrow A\)
Write an example of A other than principal value branch.
Answer: \( A = \left[\frac{\pi}{2}, \frac{3\pi}{2}\right] \)
(Any interval of length π where sin is bijective)
(ii) Simplify: \(\tan^{-1} \left( \frac{\cos x}{1 - \sin x} \right), \quad -\frac{\pi}{2} < x < \frac{3\pi}{2}\)
Answer:
For \(-\pi/2 < x < \pi/2\): \( \frac{x}{2} + \frac{\pi}{4} \)
For \(\pi/2 < x < 3\pi/2\): \( \frac{x}{2} + \frac{\pi}{4} - \pi \)
Explanation: \(\frac{\cos x}{1 - \sin x} = \tan\left(\frac{\pi}{4} + \frac{x}{2}\right)\) after simplification.
Question 3
(i) \(A = \begin{bmatrix} a & c & 0 \\ b & d & 0 \\ 0 & 0 & b \end{bmatrix}\) is a scalar matrix. Find \(a + 2b + 3c + d\).
Answer: 4
Explanation: For a scalar matrix, all diagonal elements are equal and off-diagonal elements are 0. So \(a = d = b\) and \(c = 0\). Let \(k = a = b = d\), then \(a + 2b + 3c + d = k + 2k + 0 + k = 4k\). If we assume the scalar is 1, then answer is 4.
(ii) \(A = \begin{bmatrix} -1 \\ 2 \\ 3 \end{bmatrix}, B = [-2 \quad -1 \quad -4]\)
Find \(B^T A^T\).
Answer: \(\begin{bmatrix} 2 & -4 & -6 \\ 1 & -2 & -3 \\ 4 & -8 & -12 \end{bmatrix}\)
Calculation:
\(B^T = \begin{bmatrix} -2 \\ -1 \\ -4 \end{bmatrix}\), \(A^T = [-1, 2, 3]\)
\(B^T A^T = \begin{bmatrix} (-2)(-1) & (-2)(2) & (-2)(3) \\ (-1)(-1) & (-1)(2) & (-1)(3) \\ (-4)(-1) & (-4)(2) & (-4)(3) \end{bmatrix}\)
Question 4
Solve the system of equations using matrix method: \(2x + 3y = 4; \quad 4x + 5y = 6\)
Answer: \(x = -1, y = 2\)
Solution:
Matrix form: \(\begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 4 \\ 6 \end{bmatrix}\)
Determinant = \(10 - 12 = -2\)
\(x = \frac{\begin{vmatrix} 4 & 3 \\ 6 & 5 \end{vmatrix}}{-2} = \frac{20 - 18}{-2} = \frac{2}{-2} = -1\)
\(y = \frac{\begin{vmatrix} 2 & 4 \\ 4 & 6 \end{vmatrix}}{-2} = \frac{12 - 16}{-2} = \frac{-4}{-2} = 2\)
Question 6
The surface area of a cube increases at the rate of 72 cm²/sec. Find the rate of change of its volume when the edge is 3 cm.
Answer: 54 cm³/sec
Solution:
Surface area \(S = 6s^2\) ⇒ \(\frac{dS}{dt} = 12s \frac{ds}{dt}\)
Given \(\frac{dS}{dt} = 72\), \(s = 3\):
\(72 = 12(3)\frac{ds}{dt}\) ⇒ \(\frac{ds}{dt} = 2\) cm/s
Volume \(V = s^3\) ⇒ \(\frac{dV}{dt} = 3s^2 \frac{ds}{dt} = 3(9)(2) = 54\) cm³/s
Question 7
(i) Evaluate \(\int \frac{x^2 \tan^{-1}(x^3)}{1 + x^6} \, dx\)
Answer: \(\frac{(\tan^{-1}(x^3))^2}{6} + C\)
Solution: Let \(t = x^3\), \(dt = 3x^2 dx\)
Integral becomes \(\frac{1}{3} \int \frac{\tan^{-1} t}{1 + t^2} dt\)
Let \(u = \tan^{-1} t\), \(du = \frac{dt}{1 + t^2}\)
⇒ \(\frac{1}{3} \int u du = \frac{u^2}{6} + C\)
(ii) If \(\int \frac{1}{4 + x^2} \, dx = \frac{\pi}{6}\), then find the value of \(a\).
Answer: \(a = 2\sqrt{3}\)
Solution: Assuming limits are from 0 to \(a\):
\(\int_0^a \frac{1}{4+x^2} dx = \frac{1}{2} \tan^{-1}\frac{x}{2} \Big|_0^a = \frac{1}{2} \tan^{-1}\frac{a}{2} = \frac{\pi}{6}\)
⇒ \(\tan^{-1}\frac{a}{2} = \frac{\pi}{3}\) ⇒ \(\frac{a}{2} = \sqrt{3}\) ⇒ \(a = 2\sqrt{3}\)
Question 8
If the vector \(8\hat{i} + a\hat{j}\) is of magnitude 10 in the direction of \(4\hat{i} - 3\hat{j}\), find \(a\).
Answer: \(a = -6\)
Solution: Direction of \(4\hat{i} - 3\hat{j}\): unit vector = \(\frac{4}{5}\hat{i} - \frac{3}{5}\hat{j}\)
So \(8\hat{i} + a\hat{j} = k\left(\frac{4}{5}\hat{i} - \frac{3}{5}\hat{j}\right)\) for some \(k > 0\)
Comparing: \(8 = \frac{4k}{5}\) ⇒ \(k = 10\)
\(a = -\frac{3k}{5} = -\frac{30}{5} = -6\)
Check: \(|8\hat{i} - 6\hat{j}| = \sqrt{64 + 36} = 10\) ✓
Section 2: Answer any 6 questions from 9 to 16 (6 × 4 = 24 marks)
Question 9
(i) The function \(f(x) = |x| + |x + 2|\) is:
Answer: (c) continuous, but not differentiable at \(x = 0\) and \(x = -2\)
Explanation: The function is continuous everywhere (sum of continuous functions). It has corners at \(x = 0\) and \(x = -2\) where the derivative doesn't exist.
(ii) If \(y^x = xy\), find \(\frac{dy}{dx}\).
Answer: \(\frac{dy}{dx} = y \cdot \frac{x-1 - x\ln x}{x(x-1)^2}\)
Solution: Take ln: \(x \ln y = \ln x + \ln y\)
Rearrange: \((x-1)\ln y = \ln x\) ⇒ \(\ln y = \frac{\ln x}{x-1}\)
Differentiate: \(\frac{y'}{y} = \frac{\frac{1}{x}(x-1) - \ln x}{(x-1)^2}\)
⇒ \(y' = y \cdot \frac{x-1 - x\ln x}{x(x-1)^2}\)
Question 10
(i) \(f(x) = (x-1)e^x + 1\) is increasing for:
Answer: (b) \(x \geq 0\)
Solution: \(f'(x) = e^x + (x-1)e^x = xe^x\)
\(f'(x) \geq 0\) when \(x \geq 0\) (since \(e^x > 0\))
(ii) Find intervals where \(f(x) = \sin x + \cos x\), \(0 \leq x \leq 2\pi\) is increasing/decreasing.
Answer:
- Increasing: \((0, \pi/4) \cup (5\pi/4, 2\pi)\)
- Decreasing: \((\pi/4, 5\pi/4)\)
Solution: \(f'(x) = \cos x - \sin x\)
\(f'(x) = 0\) ⇒ \(\tan x = 1\) ⇒ \(x = \pi/4, 5\pi/4\) in \([0, 2\pi]\)
Test intervals to determine sign of \(f'(x)\).
Question 11
Show that among all rectangles inscribed in a given circle, the square has maximum area.
Proof: Let circle radius = \(R\), rectangle sides = \(2a, 2b\)
Constraint: \(a^2 + b^2 = R^2\)
Area \(A = 4ab = 4a\sqrt{R^2 - a^2}\)
Maximize: \(\frac{dA}{da} = 4\sqrt{R^2-a^2} - \frac{4a^2}{\sqrt{R^2-a^2}} = 0\)
⇒ \(R^2 - a^2 = a^2\) ⇒ \(a^2 = R^2/2\) ⇒ \(a = R/\sqrt{2}\)
Then \(b = \sqrt{R^2 - R^2/2} = R/\sqrt{2}\) ⇒ \(a = b\) ⇒ square.
Question 12
(i) Draw rough sketch of \(y^2 = 16 - x^2\).
Answer: Circle with center at origin, radius 4.
(ii) Find area bounded by \(y^2 = 16 - x^2\) in III quadrant using integration.
Answer: \(4\pi\) square units
Solution: In III quadrant: \(x < 0, y < 0\)
Area = \(\int_{-4}^0 \sqrt{16-x^2} dx\) (since \(y = -\sqrt{16-x^2}\), we take positive for area)
This is quarter circle area = \(\frac{1}{4} \pi (4^2) = 4\pi\)
Question 13
Evaluate \(\int_0^a \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}} \, dx\)
Answer: \(\frac{a}{2}\)
Proof: Let \(I = \int_0^a \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a-x}} dx\)
Also \(I = \int_0^a \frac{\sqrt{a-x}}{\sqrt{a-x} + \sqrt{x}} dx\) (substitute \(x \rightarrow a-x\))
Add: \(2I = \int_0^a 1 dx = a\) ⇒ \(I = a/2\)
Question 14
(i) Find order and degree: \(\frac{d^3y}{dx^3} + x \left( \frac{dy}{dx} \right)^5 = 4 \log \left( \frac{d^4y}{dx^4} \right)\)
Answer: Order = 4, Degree not defined
Explanation: Highest derivative is \(d^4y/dx^4\) (order 4). Due to log term, the equation is not a polynomial in derivatives, so degree is not defined.
(ii) Principal increases at 5% per year. In how many years will ₹1000 double itself?
Answer: Approximately 13.86 years
Solution: \(\frac{dP}{dt} = 0.05P\) ⇒ \(P = P_0 e^{0.05t}\)
\(2P_0 = P_0 e^{0.05t}\) ⇒ \(e^{0.05t} = 2\) ⇒ \(t = \frac{\ln 2}{0.05} \approx 13.86\) years
Section 3: Answer any 3 questions from 17 to 20 (3 × 6 = 18 marks)
Question 17
(i) Integrate: \(\frac{5x + 3}{\sqrt{x^2 + 4x + 10}}\)
Answer: \(5\sqrt{(x+2)^2+6} - 7 \ln|(x+2) + \sqrt{(x+2)^2+6}| + C\)
Solution: Complete square: \(x^2+4x+10 = (x+2)^2+6\)
Let \(u = x+2\), then \(5x+3 = 5u-7\)
Integral = \(\int \frac{5u-7}{\sqrt{u^2+6}} du\)
= \(5\sqrt{u^2+6} - 7 \ln|u+\sqrt{u^2+6}| + C\)
(ii) Integrate: \(\sin^{-1}x\)
Answer: \(x\sin^{-1}x + \sqrt{1-x^2} + C\)
Solution: Integration by parts: \(u = \sin^{-1}x, dv = dx\)
⇒ \(x\sin^{-1}x - \int \frac{x}{\sqrt{1-x^2}} dx\)
= \(x\sin^{-1}x + \sqrt{1-x^2} + C\)
Question 19
(i) Angle between \(\vec{a} \times \vec{b}\) and \(\vec{b} \times \vec{a}\) is:
Answer: \(\pi\) (180°)
Explanation: \(\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a})\), so they are opposite vectors.
(ii) Find vector \(\vec{d}\) perpendicular to both \(\vec{a} = \hat{i} - \hat{j}\) and \(\vec{b} = 3\hat{j} - \hat{k}\), with \(\vec{c} \cdot \vec{d} = 1\) where \(\vec{c} = 7\hat{i} - \hat{k}\).
Answer: \(\vec{d} = \frac{1}{4}\hat{i} + \frac{1}{4}\hat{j} + \frac{3}{4}\hat{k}\)
Solution: \(\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 0 \\ 0 & 3 & -1 \end{vmatrix} = \hat{i} + \hat{j} + 3\hat{k}\)
So \(\vec{d} = k(\hat{i}+\hat{j}+3\hat{k})\)
\(\vec{c} \cdot \vec{d} = (7\hat{i} - \hat{k}) \cdot (k\hat{i}+k\hat{j}+3k\hat{k}) = 7k - 3k = 4k = 1\) ⇒ \(k = 1/4\)
(iii) Area of triangle with vertices A(1,1,2), B(2,3,5), C(1,5,5).
Answer: \(\frac{\sqrt{61}}{2}\) square units
Solution: \(\vec{AB} = \hat{i}+2\hat{j}+3\hat{k}\), \(\vec{AC} = 0\hat{i}+4\hat{j}+3\hat{k}\)
\(\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 0 & 4 & 3 \end{vmatrix} = -6\hat{i} - 3\hat{j} + 4\hat{k}\)
Magnitude = \(\sqrt{36+9+16} = \sqrt{61}\)
Area = \(\frac{1}{2} \sqrt{61}\)
Question 20
(ii) Find shortest distance between lines:
\(\vec{r} = \hat{i}+2\hat{j}+\hat{k} + \lambda(\hat{i}-\hat{j}+\hat{k})\) and
\(\vec{r} = 2\hat{i}-\hat{j}-\hat{k} + \mu(2\hat{i}+\hat{j}+2\hat{k})\)
Answer: \(\frac{3\sqrt{2}}{2}\) units
Solution:
Line 1: Point A(1,2,1), direction \(\vec{d_1} = (1,-1,1)\)
Line 2: Point B(2,-1,-1), direction \(\vec{d_2} = (2,1,2)\)
\(\vec{AB} = (1,-3,-2)\)
\(\vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 2 & 1 & 2 \end{vmatrix} = -3\hat{i} + 0\hat{j} + 3\hat{k}\)
Magnitude = \(\sqrt{9+9} = 3\sqrt{2}\)
\(\vec{AB} \cdot (\vec{d_1} \times \vec{d_2}) = (1)(-3)+(-3)(0)+(-2)(3) = -9\)
Shortest distance = \(\frac{|\vec{AB} \cdot (\vec{d_1} \times \vec{d_2})|}{|\vec{d_1} \times \vec{d_2}|} = \frac{9}{3\sqrt{2}} = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2}\)