HSC Mathematics & Statistics (40) March 2026 Board Paper (J-165) Solution (Class 12 Maharashtra Board)

Prepare effectively for your upcoming board exams with our comprehensive, step-by-step HSC Mathematics & Statistics (40) 12th Standard Board Paper Solution - March 2026 (Paper Code: J-165). Designed specifically for Maharashtra State Board Class 12 Arts and Science students, this complete Board Question Paper Solution covers all sections from multiple-choice questions to 4-mark theorems and numericals, complete with marking-scheme steps, key formulas, and examiner tips to help you score 80/80 in Mathematics.

HSC Mathematics & Statistics (40) 12th Standard Board Paper Solution - March 2026 (J-165)

SECTION - A

Q. 1. Select and write the correct answer of the following multiple choice type of questions: [16]

(i) The converse of contrapositive of \(\sim p \rightarrow q\) is ____. (2)

• (a) \(q \rightarrow p\)
• (b) \(\sim q \rightarrow p\)
• (c) \(p \rightarrow \sim q\)
• (d) \(\sim q \rightarrow \sim p\)

Answer: (c) \(p \rightarrow \sim q\)

Solution:

• Given statement: \(\sim p \rightarrow q\)
• Contrapositive of \(\sim p \rightarrow q\) is \(\sim q \rightarrow \sim(\sim p)\), which simplifies to \(\sim q \rightarrow p\).
• Converse of \(\sim q \rightarrow p\) is \(p \rightarrow \sim q\).

Memory Trick: The converse of the contrapositive of any conditional statement is always logically equivalent to its inverse. For \(P \rightarrow Q\), the inverse is \(\sim P \rightarrow \sim Q\). Here, \(\sim(\sim p) \rightarrow \sim q\) directly gives \(p \rightarrow \sim q\).

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(ii) If \(A = \begin{bmatrix} 2 & -4 \\ 3 & 1 \end{bmatrix}\), then the adjoint of matrix \(A\) is ____. (2)

• (a) \(\begin{bmatrix} 1 & 3 \\ 4 & -2 \end{bmatrix}\)
• (b) \(\begin{bmatrix} -1 & 3 \\ -4 & 1 \end{bmatrix}\)
• (c) \(\begin{bmatrix} -1 & -3 \\ -4 & 2 \end{bmatrix}\)
• (d) \(\begin{bmatrix} 1 & 4 \\ -3 & 2 \end{bmatrix}\)

Answer: (d) \(\begin{bmatrix} 1 & 4 \\ -3 & 2 \end{bmatrix}\)

Solution:

• Formula: For a \(2 \times 2\) matrix \(A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\), the adjoint is obtained by interchanging the principal diagonal elements and changing the signs of the off-diagonal elements: \[\text{adj}(A) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}\]
• Calculation: For \(A = \begin{bmatrix} 2 & -4 \\ 3 & 1 \end{bmatrix}\), we get: \[\text{adj}(A) = \begin{bmatrix} 1 & 4 \\ -3 & 2 \end{bmatrix}\]

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(iii) If \(\tan^{-1}(2x) + \tan^{-1}(3x) = \frac{\pi}{4}\), then \(x =\) ____. (2)

• (a) \(-1\)
• (b) \(\frac{1}{6}\)
• (c) \(\frac{1}{3}\)
• (d) \(\frac{3}{2}\)

Answer: (b) \(\frac{1}{6}\)

Solution:

• Given: \(\tan^{-1}(2x) + \tan^{-1}(3x) = \frac{\pi}{4}\)
• Formula: \(\tan^{-1}A + \tan^{-1}B = \tan^{-1}\left(\frac{A + B}{1 - AB}\right)\) for \(AB < 1\)
• Calculation: \[\tan^{-1}\left(\frac{2x + 3x}{1 - (2x)(3x)}\right) = \frac{\pi}{4}\] \[\frac{5x}{1 - 6x^2} = \tan\left(\frac{\pi}{4}\right) = 1\] \[5x = 1 - 6x^2 \implies 6x^2 + 5x - 1 = 0\] \[(6x - 1)(x + 1) = 0 \implies x = \frac{1}{6} \text{ or } x = -1\]
• Since \(x = -1\) makes the LHS negative (which cannot equal \(\frac{\pi}{4} > 0\)), \(x = -1\) is rejected. Therefore, \(x = \frac{1}{6}\).

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(iv) The angle between the line \(\overline{r} = (\hat{i} + 2\hat{j} + \hat{k}) + \lambda(\hat{i} + \hat{j} + \hat{k})\) and the plane \(\overline{r} \cdot (2\hat{i} - \hat{j} + \hat{k}) = 8\) is ____. (2)

• (a) \(\sin^{-1}\left(\frac{\sqrt{2}}{3}\right)\)
• (b) \(\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)\)
• (c) \(\sin^{-1}\left(\frac{1}{2}\right)\)
• (d) \(\sin^{-1}\left(\frac{1}{\sqrt{2}}\right)\)

Answer: (a) \(\sin^{-1}\left(\frac{\sqrt{2}}{3}\right)\)

Solution:

• Given: Vector parallel to the line is \(\overline{b} = \hat{i} + \hat{j} + \hat{k}\) and normal vector to the plane is \(\overline{n} = 2\hat{i} - \hat{j} + \hat{k}\).
• Formula: The acute angle \(\theta\) between a line and a plane is given by: \[\sin\theta = \left|\frac{\overline{b} \cdot \overline{n}}{|\overline{b}| |\overline{n}|}\right|\]
• Calculation: \[\overline{b} \cdot \overline{n} = (1)(2) + (1)(-1) + (1)(1) = 2 - 1 + 1 = 2\] \[|\overline{b}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}\] \[|\overline{n}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}\] \[\sin\theta = \frac{2}{\sqrt{3}\sqrt{6}} = \frac{2}{\sqrt{18}} = \frac{2}{3\sqrt{2}} = \frac{\sqrt{2}}{3} \implies \theta = \sin^{-1}\left(\frac{\sqrt{2}}{3}\right)\]

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(v) If \(y = \sec(\tan^{-1}x)\), then \(\frac{dy}{dx}\) at \(x = 1\) is ____. (2)

• (a) \(\frac{1}{2}\)
• (b) \(1\)
• (c) \(\frac{1}{\sqrt{2}}\)\)
• (d) \(\sqrt{2}\)

Answer: (c) \(\frac{1}{\sqrt{2}}\)

Solution:

• Given: \(y = \sec(\tan^{-1}x)\)
• Simplification: Let \(\tan^{-1}x = \theta \implies \tan\theta = x\). Using the identity \(\sec\theta = \sqrt{1 + \tan^2\theta} = \sqrt{1 + x^2}\), we get: \[y = \sqrt{1 + x^2}\]
• Differentiation: \[\frac{dy}{dx} = \frac{1}{2\sqrt{1 + x^2}} \cdot (2x) = \frac{x}{\sqrt{1 + x^2}}\]
• At \(x = 1\): \[\left(\frac{dy}{dx}\right)_{x=1} = \frac{1}{\sqrt{1 + 1^2}} = \frac{1}{\sqrt{2}}\]

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(vi) The approximate value of the function \(f(x) = x^3 - 3x + 5\) at \(x = 1.99\) is ____. (2)

• (a) \(6.09\)
• (b) \(6.91\)
• (c) \(7.09\)
• (d) \(7.91\)

Answer: (b) \(6.91\)

Solution:

• Given: \(f(x) = x^3 - 3x + 5\), \(a = 2\), and \(h = -0.01\) such that \(a + h = 1.99\).
• Formula: \(f(a + h) \approx f(a) + h \cdot f'(a)\)
• Calculation:
  • \(f(a) = f(2) = (2)^3 - 3(2) + 5 = 8 - 6 + 5 = 7\)
  • \(f'(x) = 3x^2 - 3 \implies f'(2) = 3(2)^2 - 3 = 12 - 3 = 9\)
  • \(f(1.99) \approx 7 + (-0.01)(9) = 7 - 0.09 = 6.91\)

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(vii) \(\int_{1}^{2} \frac{1}{x^2} \cdot e^{\frac{1}{x}} dx =\) ____. (2)

• (a) \(\sqrt{e} + 1\)
• (b) \(\sqrt{e} - 1\)
• (c) \(\sqrt{e}(\sqrt{e} - 1)\)
• (d) \(\frac{\sqrt{e} - 1}{e}\)

Answer: (c) \(\sqrt{e}(\sqrt{e} - 1)\)

Solution:

• Substitution: Put \(\frac{1}{x} = t \implies -\frac{1}{x^2} dx = dt \implies \frac{1}{x^2} dx = -dt\).
• Limits: When \(x = 1\), \(t = 1\); when \(x = 2\), \(t = \frac{1}{2}\).
• Calculation: \[I = \int_{1}^{1/2} e^t (-dt) = \int_{1/2}^{1} e^t dt = \left[e^t\right]_{1/2}^{1} = e^1 - e^{1/2} = e - \sqrt{e} = \sqrt{e}(\sqrt{e} - 1)\]

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(viii) If the p.d.f. of a continuous r.v. \(X\) is \(f(x) = \frac{x + 2}{18}\) for \(-2 < x < 4\) and \(= 0\) otherwise, then \(P(|X| < 1) =\) ____. (2)

• (a) \(\frac{1}{9}\)
• (b) \(\frac{2}{9}\)
• (c) \(\frac{1}{27}\)
• (d) \(\frac{2}{27}\)

Answer: (b) \(\frac{2}{9}\)

Solution:

• Formula: \(P(|X| < 1) = P(-1 < X < 1) = \int_{-1}^{1} f(x) dx\)
• Calculation: \[P(|X| < 1) = \int_{-1}^{1} \frac{x + 2}{18} dx = \frac{1}{18} \left[\frac{x^2}{2} + 2x\right]_{-1}^{1}\] \[= \frac{1}{18} \left[\left(\frac{1}{2} + 2\right) - \left(\frac{1}{2} - 2\right)\right] = \frac{1}{18}(4) = \frac{4}{18} = \frac{2}{9}\]

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Q. 2. Answer the following questions: [4]

(i) Write the dual of \((p \vee q) \vee r \equiv p \vee (q \vee r)\). (1)

Answer:

To obtain the dual of a statement pattern, we replace \(\vee\) with \(\wedge\) and \(\wedge\) with \(\vee\):

Dual: \((p \wedge q) \wedge r \equiv p \wedge (q \wedge r)\)

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(ii) Evaluate: \(\cos^{-1}\left(\frac{1}{2}\right) + 2\sin^{-1}\left(\frac{1}{2}\right)\). (1)

Answer:

• Principal Values: \(\cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3}\) and \(\sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}\)
• Calculation: \[\cos^{-1}\left(\frac{1}{2}\right) + 2\sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} + 2\left(\frac{\pi}{6}\right) = \frac{\pi}{3} + \frac{\pi}{3} = \frac{2\pi}{3}\]
• Final Answer: \(\frac{2\pi}{3}\)

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(iii) Evaluate: \(\int \frac{2x}{1 + x^2} dx\). (1)

Answer:

• Formula: Using the standard corollary \(\int \frac{f'(x)}{f(x)} dx = \log|f(x)| + c\)
• Calculation: Since \(\frac{d}{dx}(1 + x^2) = 2x\), we get: \[\int \frac{2x}{1 + x^2} dx = \log(1 + x^2) + c\]

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(iv) Write the degree of the differential equation \((y''')^2 + 3(y'')^3 + 3xy' + 5y = 0\). (1)

Answer:

• The given differential equation is a polynomial equation in its derivatives \(y''', y''\), and \(y'\).
• The highest order derivative present in the equation is \(y'''\) (order \(= 3\)), and its exponent (power) is \(2\).
• Final Answer: Therefore, the degree of the differential equation is \(2\).

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SECTION - B

Attempt any EIGHT of the following questions: [16]

Q. 3. Construct the switching circuit of the statement pattern \((\sim p \wedge q) \vee (p \wedge \sim r)\). (2)

Answer:

Let the simple statements represent the switches as follows:

• \(p\): The switch \(S_1\) is closed.
• \(q\): The switch \(S_2\) is closed.
• \(r\): The switch \(S_3\) is closed.
• \(\sim p\): The switch \(S_1'\) is closed.
• \(\sim r\): The switch \(S_3'\) is closed.

Construction of Switching Circuit:

1. Upper Branch \((\sim p \wedge q)\): Switch \(S_1'\) and switch \(S_2\) are connected in series.
2. Lower Branch \((p \wedge \sim r)\): Switch \(S_1\) and switch \(S_3'\) are connected in series.
3. Complete Circuit: The upper branch and lower branch are connected in parallel with each other, and the combination is connected in series with a battery and a lamp \(L\).

Exam Tip: In the board exam, draw the two parallel branches neatly using a ruler, label the switches clearly as \(S_1', S_2\) in the top branch and \(S_1, S_3'\) in the bottom branch, and show the current flow arrows from the battery to the lamp \(L\).

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Q. 4. In \(\triangle ABC\), prove that \(a(b \cos C - c \cos B) = b^2 - c^2\). (2)

Answer:

Proof:

• Consider LHS: \[\text{LHS} = a(b \cos C - c \cos B) = ab \cos C - ac \cos B\]
• Formula (Cosine Rule): \[\cos C = \frac{a^2 + b^2 - c^2}{2ab} \quad \text{and} \quad \cos B = \frac{a^2 + c^2 - b^2}{2ac}\]
• Substitution & Calculation: \[\text{LHS} = ab\left(\frac{a^2 + b^2 - c^2}{2ab}\right) - ac\left(\frac{a^2 + c^2 - b^2}{2ac}\right)\] \[\text{LHS} = \frac{a^2 + b^2 - c^2}{2} - \frac{a^2 + c^2 - b^2}{2}\] \[\text{LHS} = \frac{a^2 + b^2 - c^2 - a^2 - c^2 + b^2}{2}\] \[\text{LHS} = \frac{2b^2 - 2c^2}{2} = \frac{2(b^2 - c^2)}{2} = b^2 - c^2 = \text{RHS}\]

Therefore, \(a(b \cos C - c \cos B) = b^2 - c^2\). Hence proved.

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Q. 5. Find the general solution of \(4\cos^2\theta = 3\). (2)

Answer:

• Given Equation: \[4\cos^2\theta = 3 \implies \cos^2\theta = \frac{3}{4}\]
• Formula: The general solution of \(\cos^2\theta = \cos^2\alpha\) is given by: \[\theta = n\pi \pm \alpha, \quad n \in \mathbb{Z}\]
• Calculation: \[\cos^2\theta = \left(\frac{\sqrt{3}}{2}\right)^2\] Since \(\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}\), we have: \[\cos^2\theta = \cos^2\left(\frac{\pi}{6}\right)\]
• Final Answer: The required general solution is \(\theta = n\pi \pm \frac{\pi}{6}\), where \(n \in \mathbb{Z}\).

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Q. 6. Find \(k\), if the sum of the slopes of the lines represented by \(x^2 + kxy - 3y^2 = 0\) is twice their product. (2)

Answer:

• Given: Homogeneous equation of lines is \(x^2 + kxy - 3y^2 = 0\). Comparing with \(ax^2 + 2hxy + by^2 = 0\), we get: \[a = 1, \quad 2h = k, \quad b = -3\]
• Formula: If \(m_1\) and \(m_2\) are the slopes of the lines, then:
  • Sum of slopes: \(m_1 + m_2 = -\frac{2h}{b} = -\frac{k}{-3} = \frac{k}{3}\)
  • Product of slopes: \(m_1 m_2 = \frac{a}{b} = \frac{1}{-3} = -\frac{1}{3}\)
• Substitution & Calculation: Given that \(m_1 + m_2 = 2(m_1 m_2)\): \[\frac{k}{3} = 2\left(-\frac{1}{3}\right) \implies \frac{k}{3} = -\frac{2}{3} \implies k = -2\]
• Final Answer: \(k = -2\).

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Q. 7. Find the value of \(p\), for which the vectors \(\overline{a} = 3\hat{i} + 2\hat{j} + 9\hat{k}\) and \(\overline{b} = \hat{i} + p\hat{j} + 3\hat{k}\) are perpendicular to each other. (2)

Answer:

• Given: \(\overline{a} = 3\hat{i} + 2\hat{j} + 9\hat{k}\) and \(\overline{b} = \hat{i} + p\hat{j} + 3\hat{k}\) are perpendicular (\(\overline{a} \perp \overline{b}\)).
• Formula: Two non-zero vectors \(\overline{a}\) and \(\overline{b}\) are perpendicular if and only if their scalar (dot) product is zero: \[\overline{a} \cdot \overline{b} = 0\]
• Substitution & Calculation: \[(3\hat{i} + 2\hat{j} + 9\hat{k}) \cdot (\hat{i} + p\hat{j} + 3\hat{k}) = 0\] \[(3)(1) + (2)(p) + (9)(3) = 0\] \[3 + 2p + 27 = 0 \implies 2p + 30 = 0 \implies p = -15\]
• Final Answer: \(p = -15\).

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Q. 8. Find the vector equation of the line passing through the points \(A(1, 2, 3)\) and \(B(2, 3, 4)\). (2)

Answer:

• Given: Position vectors of points \(A(1, 2, 3)\) and \(B(2, 3, 4)\) with respect to the origin are: \[\overline{a} = \hat{i} + 2\hat{j} + 3\hat{k} \quad \text{and} \quad \overline{b} = 2\hat{i} + 3\hat{j} + 4\hat{k}\]
• Formula: The vector equation of the line passing through \(A(\overline{a})\) and \(B(\overline{b})\) is: \[\overline{r} = \overline{a} + \lambda(\overline{b} - \overline{a})\]
• Substitution & Calculation: \[\overline{b} - \overline{a} = (2\hat{i} + 3\hat{j} + 4\hat{k}) - (\hat{i} + 2\hat{j} + 3\hat{k}) = \hat{i} + \hat{j} + \hat{k}\] \[\overline{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} + \hat{j} + \hat{k})\]
• Final Answer: The vector equation of the line is \(\overline{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} + \hat{j} + \hat{k})\).

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Q. 9. Find \(\frac{dy}{dx}\), if \(\sqrt{x} + \sqrt{y} = \sqrt{a}\). (2)

Answer:

• Given: \(\sqrt{x} + \sqrt{y} = \sqrt{a}\)
• Formula: \(\frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}}\) and \(\frac{d}{dx}(\text{constant}) = 0\)
• Differentiation: Differentiating both sides with respect to \(x\): \[\frac{d}{dx}(\sqrt{x}) + \frac{d}{dx}(\sqrt{y}) = \frac{d}{dx}(\sqrt{a})\] \[\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \cdot \frac{dy}{dx} = 0\] \[\frac{1}{2\sqrt{y}} \cdot \frac{dy}{dx} = -\frac{1}{2\sqrt{x}}\] \[\frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}} = -\sqrt{\frac{y}{x}}\]
• Final Answer: \(\frac{dy}{dx} = -\sqrt{\frac{y}{x}}\)

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Q. 10. Find \(\frac{d^2y}{dx^2}\), if \(y = x^3 + 7x^2 - 2x - 9\). (2)

Answer:

• Given: \(y = x^3 + 7x^2 - 2x - 9\)
• First Derivative: Differentiating with respect to \(x\): \[\frac{dy}{dx} = \frac{d}{dx}(x^3 + 7x^2 - 2x - 9) = 3x^2 + 14x - 2\]
• Second Derivative: Differentiating again with respect to \(x\): \[\frac{d^2y}{dx^2} = \frac{d}{dx}(3x^2 + 14x - 2) = 6x + 14\]
• Final Answer: \(\frac{d^2y}{dx^2} = 6x + 14\)

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Q. 11. Test whether the function \(f(x) = x^3 + 6x^2 + 12x - 7\) is increasing or decreasing for all \(x \in \mathbb{R}\). (2)

Answer:

• Given Function: \(f(x) = x^3 + 6x^2 + 12x - 7\)
• Condition for Increasing Function: A function \(f(x)\) is increasing on \(\mathbb{R}\) if \(f'(x) \ge 0\) for all \(x \in \mathbb{R}\).
• Calculation: Differentiating with respect to \(x\): \[f'(x) = 3x^2 + 12x + 12\] \[f'(x) = 3(x^2 + 4x + 4) = 3(x + 2)^2\]
• Since the square of any real number is always non-negative, \((x + 2)^2 \ge 0\) for all \(x \in \mathbb{R}\). Therefore: \[f'(x) = 3(x + 2)^2 \ge 0 \quad \forall x \in \mathbb{R}\]
• Final Answer: The function \(f(x) = x^3 + 6x^2 + 12x - 7\) is an increasing function for all \(x \in \mathbb{R}\).

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Q. 12. A stone is dropped into a quiet lake and waves in the form of circles are generated. Radius of the circular wave increases at the rate of \(3\text{ cm/sec}\). How fast the area enclosed is increasing when the radius is \(8\text{ cm}\)? (2)

Answer:

• Given: Let \(r\) be the radius and \(A\) be the area of the circular wave at time \(t\).
  • Rate of increase of radius: \(\frac{dr}{dt} = 3\text{ cm/sec}\)
  • Instantaneous radius: \(r = 8\text{ cm}\)
• Formula: Area of a circle is \(A = \pi r^2\). Differentiating with respect to \(t\): \[\frac{dA}{dt} = 2\pi r \frac{dr}{dt}\]
• Substitution & Calculation: \[\frac{dA}{dt} = 2\pi (8)(3) = 48\pi\text{ cm}^2\text{/sec}\]
• Final Answer: The area enclosed by the circular wave is increasing at the rate of \(48\pi\text{ cm}^2\text{/sec}\) (or \(48\pi\text{ sq.cm/sec}\)).

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Q. 13. Evaluate: \(\int \sqrt{1 + \sin 2x} \cdot dx\). (2)

Answer:

• Given: Let \(I = \int \sqrt{1 + \sin 2x} \, dx\)
• Formula: Using trigonometric identities \(\sin^2 x + \cos^2 x = 1\) and \(\sin 2x = 2\sin x \cos x\): \[1 + \sin 2x = \sin^2 x + \cos^2 x + 2\sin x \cos x = (\sin x + \cos x)^2\]
• Calculation: \[I = \int \sqrt{(\sin x + \cos x)^2} \, dx = \int (\sin x + \cos x) \, dx\] \[I = -\cos x + \sin x + c = \sin x - \cos x + c\]
• Final Answer: \(\int \sqrt{1 + \sin 2x} \cdot dx = \sin x - \cos x + c\)

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Q. 14. Given that, \(X \sim B(n, p)\), if \(n = 10\), \(E(X) = 8\) then find \(\text{Var}(X)\). (2)

Answer:

• Given: For binomial distribution \(X \sim B(n, p)\), \(n = 10\) and mean \(E(X) = 8\).
• Formulas:
  • Mean: \(E(X) = np\)
  • Variance: \(\text{Var}(X) = npq\), where \(q = 1 - p\)
• Substitution & Calculation: \[E(X) = np \implies 8 = 10p \implies p = \frac{8}{10} = 0.8\] \[q = 1 - p = 1 - 0.8 = 0.2\] \[\text{Var}(X) = npq = 10 \times 0.8 \times 0.2 = 1.6\]
• Final Answer: \(\text{Var}(X) = 1.6\) (or \(\frac{8}{5}\)).

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SECTION - C

Attempt any EIGHT of the following questions: [24]

Q. 15. Examine whether the statement pattern \((p \wedge q) \wedge (\sim p \vee \sim q)\) is a tautology or contradiction or contingency. (3)

Answer:

Truth Table for \((p \wedge q) \wedge (\sim p \vee \sim q)\):

\(p\)	\(q\)	\(\sim p\)	\(\sim q\)	\(p \wedge q\)	\(\sim p \vee \sim q\)	\((p \wedge q) \wedge (\sim p \vee \sim q)\)
T	T	F	F	T	F	F
T	F	F	T	F	T	F
F	T	T	F	F	T	F
F	F	T	T	F	T	F

Conclusion: Since all the truth values in the last column are F (False), the given statement pattern \((p \wedge q) \wedge (\sim p \vee \sim q)\) is a contradiction.

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Q. 16. In \(\triangle ABC\), if \(A = 45^{\circ}\), \(B = 60^{\circ}\) then find the ratio of its sides. (3)

Answer:

• Given: In \(\triangle ABC\), \(A = 45^{\circ}\) and \(B = 60^{\circ}\).
• Formula:
  • Sum of angles of a triangle: \(A + B + C = 180^{\circ}\)
  • Sine Rule: \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \implies a : b : c = \sin A : \sin B : \sin C\)
• Step 1: Find angle \(C\): \[45^{\circ} + 60^{\circ} + C = 180^{\circ} \implies C = 180^{\circ} - 105^{\circ} = 75^{\circ}\]
• Step 2: Find the values of \(\sin 45^{\circ}, \sin 60^{\circ}\), and \(\sin 75^{\circ}\):
  • \(\sin 45^{\circ} = \frac{1}{\sqrt{2}} = \frac{2}{2\sqrt{2}}\)
  • \(\sin 60^{\circ} = \frac{\sqrt{3}}{2} = \frac{\sqrt{6}}{2\sqrt{2}}\)
  • \(\sin 75^{\circ} = \sin(45^{\circ} + 30^{\circ}) = \sin 45^{\circ}\cos 30^{\circ} + \cos 45^{\circ}\sin 30^{\circ}\) \[\sin 75^{\circ} = \left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2}\right) = \frac{\sqrt{3} + 1}{2\sqrt{2}}\]
• Step 3: Ratio of sides: \[a : b : c = \sin 45^{\circ} : \sin 60^{\circ} : \sin 75^{\circ}\] \[a : b : c = \frac{1}{\sqrt{2}} : \frac{\sqrt{3}}{2} : \frac{\sqrt{3} + 1}{2\sqrt{2}}\] Multiplying each term of the ratio by \(2\sqrt{2}\): \[a : b : c = 2 : \sqrt{6} : (\sqrt{3} + 1)\]
• Final Answer: The ratio of the sides of \(\triangle ABC\) is \(2 : \sqrt{6} : (\sqrt{3} + 1)\).

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Q. 17. If two vertices of a triangle are \(A(3, 1, 4)\) and \(B(-4, 5, -3)\) and the centroid of the triangle is \(G(-1, 2, 1)\), then find the coordinates of the third vertex \(C\) of the triangle. (3)

Answer:

• Given: Let the vertices of \(\triangle ABC\) be \(A(x_1, y_1, z_1) \equiv (3, 1, 4)\), \(B(x_2, y_2, z_2) \equiv (-4, 5, -3)\), and \(C(x_3, y_3, z_3) \equiv (x, y, z)\). The centroid is \(G(-1, 2, 1)\).
• Formula (Centroid Formula): \[G(x, y, z) = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3}\right)\]
• Substitution & Calculation: \[(-1, 2, 1) = \left(\frac{3 - 4 + x}{3}, \frac{1 + 5 + y}{3}, \frac{4 - 3 + z}{3}\right)\] Equating corresponding coordinates:
  • \(\frac{-1 + x}{3} = -1 \implies -1 + x = -3 \implies x = -2\)
  • \(\frac{6 + y}{3} = 2 \implies 6 + y = 6 \implies y = 0\)
  • \(\frac{1 + z}{3} = 1 \implies 1 + z = 3 \implies z = 2\)
• Final Answer: The coordinates of the third vertex \(C\) are \((-2, 0, 2)\).

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Q. 18. If \(D, E, F\) are the mid-points of the sides \(BC, CA, AB\) respectively of \(\triangle ABC\), then prove that \(\overline{AD} + \overline{BE} + \overline{CF} = \overline{0}\). (3)

Answer:

Proof:

• Let \(\overline{a}, \overline{b}, \overline{c}\) be the position vectors of the vertices \(A, B, C\) of \(\triangle ABC\) and \(\overline{d}, \overline{e}, \overline{f}\) be the position vectors of the mid-points \(D, E, F\) of sides \(BC, CA, AB\) respectively.
• By Mid-point Formula: \[\overline{d} = \frac{\overline{b} + \overline{c}}{2}, \quad \overline{e} = \frac{\overline{c} + \overline{a}}{2}, \quad \overline{f} = \frac{\overline{a} + \overline{b}}{2}\]
• Consider LHS: \[\text{LHS} = \overline{AD} + \overline{BE} + \overline{CF}\] \[\text{LHS} = (\overline{d} - \overline{a}) + (\overline{e} - \overline{b}) + (\overline{f} - \overline{c})\] \[\text{LHS} = (\overline{d} + \overline{e} + \overline{f}) - (\overline{a} + \overline{b} + \overline{c})\] Substituting \(\overline{d}, \overline{e}, \overline{f}\): \[\text{LHS} = \left(\frac{\overline{b} + \overline{c}}{2} + \frac{\overline{c} + \overline{a}}{2} + \frac{\overline{a} + \overline{b}}{2}\right) - (\overline{a} + \overline{b} + \overline{c})\] \[\text{LHS} = \frac{2(\overline{a} + \overline{b} + \overline{c})}{2} - (\overline{a} + \overline{b} + \overline{c}) = (\overline{a} + \overline{b} + \overline{c}) - (\overline{a} + \overline{b} + \overline{c}) = \overline{0} = \text{RHS}\]

Therefore, \(\overline{AD} + \overline{BE} + \overline{CF} = \overline{0}\). Hence proved.

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Q. 19. Show that the lines \(\overline{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(2\hat{i} - 2\hat{j} + \hat{k})\) and \(\overline{r} = (4\hat{i} - 3\hat{j} + 2\hat{k}) + \mu(\hat{i} - 2\hat{j} + 2\hat{k})\) intersect each other. (3)

Answer:

• Given: Comparing the given lines with \(\overline{r} = \overline{a}_1 + \lambda\overline{b}_1\) and \(\overline{r} = \overline{a}_2 + \mu\overline{b}_2\):
  • \(\overline{a}_1 = \hat{i} + \hat{j} - \hat{k}, \quad \overline{b}_1 = 2\hat{i} - 2\hat{j} + \hat{k}\)
  • \(\overline{a}_2 = 4\hat{i} - 3\hat{j} + 2\hat{k}, \quad \overline{b}_2 = \hat{i} - 2\hat{j} + 2\hat{k}\)
• Condition for Intersection of Two Lines: Two non-parallel lines intersect each other if and only if: \[(\overline{a}_2 - \overline{a}_1) \cdot (\overline{b}_1 \times \overline{b}_2) = 0\]
• Calculation: \[\overline{a}_2 - \overline{a}_1 = (4 - 1)\hat{i} + (-3 - 1)\hat{j} + (2 - (-1))\hat{k} = 3\hat{i} - 4\hat{j} + 3\hat{k}\] \[(\overline{a}_2 - \overline{a}_1) \cdot (\overline{b}_1 \times \overline{b}_2) = \begin{vmatrix} 3 & -4 & 3 \\ 2 & -2 & 1 \\ 1 & -2 & 2 \end{vmatrix}\] \[= 3(-4 + 2) - (-4)(4 - 1) + 3(-4 + 2)\] \[= 3(-2) + 4(3) + 3(-2) = -6 + 12 - 6 = 0\]
• Conclusion: Since \((\overline{a}_2 - \overline{a}_1) \cdot (\overline{b}_1 \times \overline{b}_2) = 0\) and \(\overline{b}_1\) is not parallel to \(\overline{b}_2\), the given lines intersect each other.

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Q. 20. Find the cartesian equation of the plane \(\overline{r} = (\hat{i} - \hat{j}) + \lambda(\hat{i} + \hat{j} + \hat{k}) + \mu(\hat{i} - 2\hat{j} + 3\hat{k})\). (3)

Answer:

• Given: Comparing with the parametric vector equation \(\overline{r} = \overline{a} + \lambda\overline{b} + \mu\overline{c}\):
  • \(\overline{a} = \hat{i} - \hat{j} + 0\hat{k}\) (point \(A(1, -1, 0)\) on the plane)
  • \(\overline{b} = \hat{i} + \hat{j} + \hat{k}\) and \(\overline{c} = \hat{i} - 2\hat{j} + 3\hat{k}\) (vectors parallel to the plane)
• Formula: Normal to the plane is \(\overline{n} = \overline{b} \times \overline{c}\), and the vector equation of the plane is \(\overline{r} \cdot \overline{n} = \overline{a} \cdot \overline{n}\).
• Calculation: \[\overline{n} = \overline{b} \times \overline{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & -2 & 3 \end{vmatrix} = \hat{i}(3 + 2) - \hat{j}(3 - 1) + \hat{k}(-2 - 1) = 5\hat{i} - 2\hat{j} - 3\hat{k}\] \[\overline{a} \cdot \overline{n} = (\hat{i} - \hat{j} + 0\hat{k}) \cdot (5\hat{i} - 2\hat{j} - 3\hat{k}) = (1)(5) + (-1)(-2) + (0)(-3) = 5 + 2 + 0 = 7\] Put \(\overline{r} = x\hat{i} + y\hat{j} + z\hat{k}\) in \(\overline{r} \cdot \overline{n} = \overline{a} \cdot \overline{n}\): \[(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (5\hat{i} - 2\hat{j} - 3\hat{k}) = 7 \implies 5x - 2y - 3z = 7\]
• Final Answer: The Cartesian equation of the plane is \(5x - 2y - 3z = 7\) (or \(5x - 2y - 3z - 7 = 0\)).

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Q. 21. If \(y = f(u)\) is a differentiable function of \(u\) and \(u = g(x)\) is a differentiable function of \(x\) then prove that \(y\) is a differentiable function of \(x\) and \(\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}\). (3)

Answer:

Proof (Chain Rule):

• Let \(\delta x\) be a small increment in \(x\), and let \(\delta u\) and \(\delta y\) be the corresponding increments in \(u\) and \(y\) respectively, such that \(\delta x \neq 0\) and \(\delta u \neq 0\).
• Consider the incrementary ratio: \[\frac{\delta y}{\delta x} = \frac{\delta y}{\delta u} \times \frac{\delta u}{\delta x}\]
• Taking limits on both sides as \(\delta x \to 0\): \[\lim_{\delta x \to 0} \frac{\delta y}{\delta x} = \lim_{\delta x \to 0} \left(\frac{\delta y}{\delta u} \times \frac{\delta u}{\delta x}\right) = \left(\lim_{\delta x \to 0} \frac{\delta y}{\delta u}\right) \times \left(\lim_{\delta x \to 0} \frac{\delta u}{\delta x}\right)\]
• Since \(u = g(x)\) is differentiable, it is also continuous; hence as \(\delta x \to 0\), \(\delta u \to 0\): \[\lim_{\delta x \to 0} \frac{\delta y}{\delta x} = \left(\lim_{\delta u \to 0} \frac{\delta y}{\delta u}\right) \times \left(\lim_{\delta x \to 0} \frac{\delta u}{\delta x}\right) \quad \text{...(Equation 1)}\]
• Since \(y\) is a differentiable function of \(u\) and \(u\) is a differentiable function of \(x\), both limits on the RHS of Equation 1 exist and are finite: \[\lim_{\delta u \to 0} \frac{\delta y}{\delta u} = \frac{dy}{du} \quad \text{and} \quad \lim_{\delta x \to 0} \frac{\delta u}{\delta x} = \frac{du}{dx}\]
• Therefore, the limit on the LHS of Equation 1 also exists and is finite, meaning \(y\) is a differentiable function of \(x\) and \(\lim_{\delta x \to 0} \frac{\delta y}{\delta x} = \frac{dy}{dx}\).
• Substituting these into Equation 1, we get: \[\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}\]

Hence proved.

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Q. 22. Verify LMVT for the function \(f(x) = \log x\), on \([1, e]\). (3)

Answer:

• Step 1: Verify Conditions of LMVT:
  1. Since \(f(x) = \log x\) is defined and continuous for all \(x > 0\), \(f(x)\) is continuous on the closed interval \([1, e]\).
  2. \(f'(x) = \frac{1}{x}\) exists finitely for all \(x \in (1, e)\), so \(f(x)\) is differentiable on the open interval \((1, e)\).
  Thus, all conditions of Lagrange's Mean Value Theorem (LMVT) are satisfied.
• Step 2: Find \(c \in (1, e)\): By LMVT, there exists at least one \(c \in (1, e)\) such that: \[f'(c) = \frac{f(b) - f(a)}{b - a}\] Here \(a = 1, b = e\), \(f(1) = \log 1 = 0\), \(f(e) = \log e = 1\), and \(f'(c) = \frac{1}{c}\): \[\frac{1}{c} = \frac{1 - 0}{e - 1} \implies c = e - 1\]
• Step 3: Check Interval Membership: Since \(2 < e < 3 \implies 1 < e - 1 < 2 < e\), we have \(c = e - 1 \in (1, e)\).
• Final Answer: Lagrange's Mean Value Theorem (LMVT) is verified with \(c = e - 1\).

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Q. 23. Evaluate: \(\int \frac{\sin x}{\sin 3x} dx\). (3)

Answer:

• Given: Let \(I = \int \frac{\sin x}{\sin 3x} dx\)
• Formula: \(\sin 3x = 3\sin x - 4\sin^3 x\)
• Calculation: \[I = \int \frac{\sin x}{3\sin x - 4\sin^3 x} dx = \int \frac{1}{3 - 4\sin^2 x} dx\] Dividing numerator and denominator by \(\cos^2 x\): \[I = \int \frac{\sec^2 x}{3\sec^2 x - 4\tan^2 x} dx = \int \frac{\sec^2 x}{3(1 + \tan^2 x) - 4\tan^2 x} dx = \int \frac{\sec^2 x}{3 - \tan^2 x} dx\] Put \(\tan x = t \implies \sec^2 x \, dx = dt\): \[I = \int \frac{1}{(\sqrt{3})^2 - t^2} dt\] Using \(\int \frac{1}{a^2 - x^2} dx = \frac{1}{2a}\log\left|\frac{a + x}{a - x}\right| + c\): \[I = \frac{1}{2\sqrt{3}}\log\left|\frac{\sqrt{3} + t}{\sqrt{3} - t}\right| + c = \frac{1}{2\sqrt{3}}\log\left|\frac{\sqrt{3} + \tan x}{\sqrt{3} - \tan x}\right| + c\]
• Final Answer: \(\int \frac{\sin x}{\sin 3x} dx = \frac{1}{2\sqrt{3}}\log\left|\frac{\sqrt{3} + \tan x}{\sqrt{3} - \tan x}\right| + c\)

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Q. 24. Solve the D.E. \(3e^x \tan y \, dx + (1 + e^x)\sec^2 y \, dy = 0\). (3)

Answer:

• Given Differential Equation: \[3e^x \tan y \, dx + (1 + e^x)\sec^2 y \, dy = 0\]
• Separating Variables: Dividing throughout by \((1 + e^x)\tan y\): \[\frac{3e^x}{1 + e^x} dx + \frac{\sec^2 y}{\tan y} dy = 0\]
• Integration: Integrating both sides: \[3\int \frac{e^x}{1 + e^x} dx + \int \frac{\sec^2 y}{\tan y} dy = c_1\] Using \(\int \frac{f'(x)}{f(x)} dx = \log|f(x)|\): \[3\log(1 + e^x) + \log|\tan y| = \log c \quad (\text{where } c_1 = \log c)\] \[\log\left((1 + e^x)^3 \cdot \tan y\right) = \log c\] \[(1 + e^x)^3 \tan y = c\]
• Final Answer: The general solution is \((1 + e^x)^3 \tan y = c\).

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Q. 25. Find \(E(X)\) and \(V(X)\), where \(X\) is the number obtained on uppermost face, when a fair die is thrown. (3)

Answer:

• Given: Sample space \(S = \{1, 2, 3, 4, 5, 6\}\). Since the die is fair, \(P(X = x_i) = \frac{1}{6}\) for each \(x_i \in \{1, 2, 3, 4, 5, 6\}\).

Probability Distribution Table:

\(x_i\)	\(1\)	\(2\)	\(3\)	\(4\)	\(5\)	\(6\)	Total
\(p_i\)	\(\frac{1}{6}\)	\(\frac{1}{6}\)	\(\frac{1}{6}\)	\(\frac{1}{6}\)	\(\frac{1}{6}\)	\(\frac{1}{6}\)	\(1\)
\(x_i p_i\)	\(\frac{1}{6}\)	\(\frac{2}{6}\)	\(\frac{3}{6}\)	\(\frac{4}{6}\)	\(\frac{5}{6}\)	\(\frac{6}{6}\)	\(\frac{21}{6}\)
\(x_i^2 p_i\)	\(\frac{1}{6}\)	\(\frac{4}{6}\)	\(\frac{9}{6}\)	\(\frac{16}{6}\)	\(\frac{25}{6}\)	\(\frac{36}{6}\)	\(\frac{91}{6}\)

• 1. Expected Value \(E(X)\): \[E(X) = \sum_{i=1}^{6} x_i p_i = \frac{1 + 2 + 3 + 4 + 5 + 6}{6} = \frac{21}{6} = \frac{7}{2} = 3.5\]
• 2. Variance \(V(X)\): \[E(X^2) = \sum_{i=1}^{6} x_i^2 p_i = \frac{1 + 4 + 9 + 16 + 25 + 36}{6} = \frac{91}{6}\] \[V(X) = E(X^2) - [E(X)]^2 = \frac{91}{6} - \left(\frac{7}{2}\right)^2 = \frac{91}{6} - \frac{49}{4} = \frac{182 - 147}{12} = \frac{35}{12} \approx 2.92\]
• Final Answer: \(E(X) = \frac{7}{2}\) (or \(3.5\)) and \(V(X) = \frac{35}{12}\) (or \(2.92\)).

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Q. 26. A pair of dice is thrown 4 times. If getting a doublet is considered as success, find the probability of two successes. (3)

Answer:

• Given: Number of trials \(n = 4\). When a pair of dice is thrown, \(n(S) = 36\).
  • Doublets are \(\{(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)\} \implies 6\) outcomes.
  • Probability of getting a doublet in a single throw: \(p = \frac{6}{36} = \frac{1}{6}\)
  • Probability of failure: \(q = 1 - p = 1 - \frac{1}{6} = \frac{5}{6}\)
• Formula: Let \(X\) be the number of successes, so \(X \sim B\left(4, \frac{1}{6}\right)\). The p.m.f. is: \[P(X = x) = \binom{n}{x} p^x q^{n-x}\]
• Substitution & Calculation (for \(x = 2\)): \[P(X = 2) = \binom{4}{2} \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^{4-2} = 6 \times \frac{1}{36} \times \frac{25}{36} = \frac{25}{216}\]
• Final Answer: The probability of getting two successes is \(\frac{25}{216}\).

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SECTION - D

Attempt any FIVE of the following questions: [20]

Q. 27. If \(A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\), prove that \(A \cdot (\text{adj } A) = (\text{adj } A) \cdot A = |A| \cdot I\). (4)

Answer:

• Step 1: Find \(|A|\) and \(|A| \cdot I\): \[|A| = \begin{vmatrix} 1 & 2 \\ 3 & 4 \end{vmatrix} = (1)(4) - (2)(3) = 4 - 6 = -2\] \[|A| \cdot I = -2 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} -2 & 0 \\ 0 & -2 \end{bmatrix} \quad \text{...(Equation 1)}\]
• Step 2: Find \(\text{adj } A\): Cofactors of elements of matrix \(A\):
  • \(A_{11} = (-1)^{1+1}(4) = 4\)
  • \(A_{12} = (-1)^{1+2}(3) = -3\)
  • \(A_{21} = (-1)^{2+1}(2) = -2\)
  • \(A_{22} = (-1)^{2+2}(1) = 1\)
  \[\text{adj } A = \begin{bmatrix} A_{11} & A_{21} \\ A_{12} & A_{22} \end{bmatrix} = \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}\]
• Step 3: Compute \(A \cdot (\text{adj } A)\): \[A \cdot (\text{adj } A) = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix} = \begin{bmatrix} 4 - 6 & -2 + 2 \\ 12 - 12 & -6 + 4 \end{bmatrix} = \begin{bmatrix} -2 & 0 \\ 0 & -2 \end{bmatrix} \quad \text{...(Equation 2)}\]
• Step 4: Compute \((\text{adj } A) \cdot A\): \[(\text{adj } A) \cdot A = \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 4 - 6 & 8 - 8 \\ -3 + 3 & -6 + 4 \end{bmatrix} = \begin{bmatrix} -2 & 0 \\ 0 & -2 \end{bmatrix} \quad \text{...(Equation 3)}\]

Conclusion: From Equations 1, 2, and 3, \(A \cdot (\text{adj } A) = (\text{adj } A) \cdot A = |A| \cdot I\). Hence proved.

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Q. 28. Show that every homogeneous equation of degree two in \(x\) and \(y\), i.e., \(ax^2 + 2hxy + by^2 = 0\), represents a pair of lines passing through the origin, if \(h^2 - ab \ge 0\). (4)

Answer:

Proof:

Consider the homogeneous equation of degree two in \(x\) and \(y\):

\[ax^2 + 2hxy + by^2 = 0 \quad \text{...(Equation 1)}\]

where \(a, h, b\) are not all zero. We consider two cases:

• Case 1: When \(b = 0\) (so \(a\) and \(h\) are not both zero) Substituting \(b = 0\) in Equation 1: \[ax^2 + 2hxy = 0 \implies x(ax + 2hy) = 0\] \[\implies x = 0 \quad \text{and} \quad ax + 2hy = 0\] Both of these are linear equations in \(x\) and \(y\) without constant terms, so they represent two straight lines passing through the origin.
• Case 2: When \(b \neq 0\) Multiplying Equation 1 by \(b\): \[abx^2 + 2hbxy + b^2y^2 = 0 \implies b^2y^2 + 2hbxy = -abx^2\] Adding \(h^2x^2\) to both sides to complete the square: \[b^2y^2 + 2hbxy + h^2x^2 = h^2x^2 - abx^2\] \[(by + hx)^2 = (h^2 - ab)x^2\] Since \(h^2 - ab \ge 0\), taking the square root on both sides gives real linear factors: \[by + hx = \pm\sqrt{h^2 - ab}\,x\] \[\implies \left(h + \sqrt{h^2 - ab}\right)x + by = 0 \quad \text{and} \quad \left(h - \sqrt{h^2 - ab}\right)x + by = 0\] Both equations are of the first degree in \(x\) and \(y\) with no constant term, so they represent two straight lines passing through the origin.

Conclusion: In both cases, \(ax^2 + 2hxy + by^2 = 0\) represents a pair of lines passing through the origin if \(h^2 - ab \ge 0\). Hence proved.

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Q. 29. If \(A(\overline{a})\) and \(B(\overline{b})\) are any two points in space and \(R(\overline{r})\) be a point on the line segment \(AB\) dividing internally in the ratio \(m : n\) then prove that \(\overline{r} = \frac{m\overline{b} + n\overline{a}}{m + n}\). (4)

Answer:

Proof (Section Formula for Internal Division):

• Let \(O\) be the origin, and let \(\overline{OA} = \overline{a}\), \(\overline{OB} = \overline{b}\), and \(\overline{OR} = \overline{r}\) be the position vectors of points \(A, B\), and \(R\) respectively.
• Since point \(R\) divides segment \(AB\) internally in the ratio \(m : n\), we have: \[\frac{l(AR)}{l(RB)} = \frac{m}{n} \implies n \cdot l(AR) = m \cdot l(RB)\]
• Since \(R\) lies between \(A\) and \(B\), the directed vectors \(\overline{AR}\) and \(\overline{RB}\) have the same direction and sense: \[n\overline{AR} = m\overline{RB}\]
• Expressing \(\overline{AR}\) and \(\overline{RB}\) in terms of position vectors \(\overline{a}, \overline{b}, \overline{r}\): \[n(\overline{r} - \overline{a}) = m(\overline{b} - \overline{r})\] \[n\overline{r} - n\overline{a} = m\overline{b} - m\overline{r}\] \[m\overline{r} + n\overline{r} = m\overline{b} + n\overline{a}\] \[(m + n)\overline{r} = m\overline{b} + n\overline{a}\] \[\overline{r} = \frac{m\overline{b} + n\overline{a}}{m + n}\]

Hence proved.

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Q. 30. Solve the L.P.P. graphically: Minimize: \(z = 5x + 2y\), Subject to, \(5x + y \ge 10\), \(x + y \ge 6\), \(x \ge 0\), \(y \ge 0\). (4)

Answer:

Step 1: Table of Boundary Lines and Half-Planes:

Inequality	Boundary Equation	Points on Coordinate Axes	Origin Test \((0, 0)\)	Feasible Half-Plane
\(5x + y \ge 10\)	\(5x + y = 10\)	\(A(0, 10)\), \(C(2, 0)\)	\(0 \ge 10\) (False)	Non-origin side
\(x + y \ge 6\)	\(x + y = 6\)	\(D(0, 6)\), \(B(6, 0)\)	\(0 \ge 6\) (False)	Non-origin side

Since \(x \ge 0\) and \(y \ge 0\), the feasible region lies in the first quadrant.

Step 2: Point of Intersection \(P\) of \(5x + y = 10\) and \(x + y = 6\):

• Subtracting \(x + y = 6\) from \(5x + y = 10\): \[4x = 4 \implies x = 1\]
• Substituting \(x = 1\) in \(x + y = 6\): \[1 + y = 6 \implies y = 5\]
• Thus, the point of intersection is \(P(1, 5)\).

Step 3: Corner Points and Evaluation of Objective Function \(z = 5x + 2y\):

The unbounded feasible region is shaded above the segments \(AP\) and \(PB\) in the first quadrant, with corner points \(A(0, 10)\), \(P(1, 5)\), and \(B(6, 0)\):

Corner Point	Value of Objective Function \(z = 5x + 2y\)
\(A(0, 10)\)	\(z(A) = 5(0) + 2(10) = 20\)
\(P(1, 5)\)	\(z(P) = 5(1) + 2(5) = 5 + 10 = \mathbf{15}\) (Minimum)
\(B(6, 0)\)	\(z(B) = 5(6) + 2(0) = 30\)

Final Answer: The minimum value of \(z\) is \(15\), which occurs at \(x = 1, y = 5\).

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Q. 31. Evaluate: \(\int \frac{3x^2 + 4x - 5}{(x^2 - 1)(x + 2)} \cdot dx\). (4)

Answer:

• Given: Let \(I = \int \frac{3x^2 + 4x - 5}{(x^2 - 1)(x + 2)} dx = \int \frac{3x^2 + 4x - 5}{(x - 1)(x + 1)(x + 2)} dx\)
• Step 1: Partial Fraction Decomposition: Let: \[\frac{3x^2 + 4x - 5}{(x - 1)(x + 1)(x + 2)} = \frac{A}{x - 1} + \frac{B}{x + 1} + \frac{C}{x + 2}\] Multiplying both sides by \((x - 1)(x + 1)(x + 2)\): \[3x^2 + 4x - 5 = A(x + 1)(x + 2) + B(x - 1)(x + 2) + C(x - 1)(x + 1)\]
  • Put \(x = 1\): \[3(1)^2 + 4(1) - 5 = A(1 + 1)(1 + 2) \implies 2 = 6A \implies A = \frac{1}{3}\]
  • Put \(x = -1\): \[3(-1)^2 + 4(-1) - 5 = B(-1 - 1)(-1 + 2) \implies -6 = -2B \implies B = 3\]
  • Put \(x = -2\): \[3(-2)^2 + 4(-2) - 5 = C(-2 - 1)(-2 + 1) \implies 12 - 8 - 5 = 3C \implies -1 = 3C \implies C = -\frac{1}{3}\]
• Step 2: Integration: Substituting \(A = \frac{1}{3}, B = 3\), and \(C = -\frac{1}{3}\): \[I = \int \left(\frac{1/3}{x - 1} + \frac{3}{x + 1} - \frac{1/3}{x + 2}\right) dx\] \[I = \frac{1}{3}\int \frac{1}{x - 1} dx + 3\int \frac{1}{x + 1} dx - \frac{1}{3}\int \frac{1}{x + 2} dx\] \[I = \frac{1}{3}\log|x - 1| + 3\log|x + 1| - \frac{1}{3}\log|x + 2| + c\]
• Final Answer: \(\int \frac{3x^2 + 4x - 5}{(x^2 - 1)(x + 2)} dx = \frac{1}{3}\log|x - 1| + 3\log|x + 1| - \frac{1}{3}\log|x + 2| + c\)

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Q. 32. Prove that: \(\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a + b - x) \cdot dx\). Hence, find \(\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sin^2 x \cdot dx\). (4)

Answer:

Part 1: Proof of Property:

• Consider \(\text{RHS} = \int_{a}^{b} f(a + b - x) dx\).
• Put \(a + b - x = t \implies -dx = dt \implies dx = -dt\).
• Change of Limits:
  • When \(x = a\), \(t = a + b - a = b\)
  • When \(x = b\), \(t = a + b - b = a\)
• Substituting in RHS: \[\text{RHS} = \int_{b}^{a} f(t)(-dt) = -\int_{b}^{a} f(t) dt\] Using the property \(\int_{a}^{b} f(t) dt = -\int_{b}^{a} f(t) dt\) and \(\int_{a}^{b} f(t) dt = \int_{a}^{b} f(x) dx\): \[\text{RHS} = \int_{a}^{b} f(t) dt = \int_{a}^{b} f(x) dx = \text{LHS}\] Hence proved.

Part 2: Evaluation of \(\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sin^2 x \cdot dx\):

• Let: \[I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sin^2 x \, dx \quad \text{...(Equation 1)}\]
• Using \(\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a + b - x) dx\), with \(a + b = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}\): \[I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sin^2\left(\frac{\pi}{2} - x\right) dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \cos^2 x \, dx \quad \text{...(Equation 2)}\]
• Adding Equation 1 and Equation 2: \[2I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} (\sin^2 x + \cos^2 x) dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1 \, dx = [x]_{\frac{\pi}{6}}^{\frac{\pi}{3}} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}\] \[I = \frac{\pi}{12}\]
• Final Answer: \(\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sin^2 x \cdot dx = \frac{\pi}{12}\)

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Q. 33. Find the area enclosed between the circle \(x^2 + y^2 = 1\) and the line \(x + y = 1\) lying in the first quadrant. (4)

Answer:

• Given Curves:
  • Circle: \(x^2 + y^2 = 1 \implies y_1 = \sqrt{1 - x^2}\) (in the first quadrant)
  • Line: \(x + y = 1 \implies y_2 = 1 - x\)
• Points of Intersection: The circle \(x^2 + y^2 = 1\) and the line \(x + y = 1\) intersect at \(A(1, 0)\) on the X-axis and \(B(0, 1)\) on the Y-axis.
• Formula: Required area \(A\) enclosed between the circle and the line from \(x = 0\) to \(x = 1\) is: \[A = \int_{0}^{1} (y_1 - y_2) dx = \int_{0}^{1} \sqrt{1 - x^2} \, dx - \int_{0}^{1} (1 - x) \, dx\]
• Substitution & Calculation: Using \(\int \sqrt{a^2 - x^2} dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)\): \[A = \left[\frac{x}{2}\sqrt{1 - x^2} + \frac{1}{2}\sin^{-1}(x)\right]_{0}^{1} - \left[x - \frac{x^2}{2}\right]_{0}^{1}\] \[A = \left[\left(0 + \frac{1}{2}\sin^{-1}(1)\right) - (0 + 0)\right] - \left[\left(1 - \frac{1}{2}\right) - 0\right]\] \[A = \left(\frac{1}{2} \cdot \frac{\pi}{2}\right) - \frac{1}{2} = \frac{\pi}{4} - \frac{1}{2} = \left(\frac{\pi - 2}{4}\right)\text{ sq. units}\]
• Final Answer: The required enclosed area is \(\left(\frac{\pi}{4} - \frac{1}{2}\right)\text{ sq. units}\) (or \(\frac{\pi - 2}{4}\text{ sq. units}\)).

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Q. 34. Solve the differential equation \(x^2 \cdot \frac{dy}{dx} = x^2 + xy + y^2\). (4)

Answer:

• Given Differential Equation: \[x^2 \frac{dy}{dx} = x^2 + xy + y^2\] Dividing both sides by \(x^2\): \[\frac{dy}{dx} = 1 + \frac{y}{x} + \left(\frac{y}{x}\right)^2 \quad \text{...(Equation 1)}\] This is a homogeneous differential equation.
• Substitution: Put \(y = vx \implies v = \frac{y}{x}\). Differentiating with respect to \(x\): \[\frac{dy}{dx} = v + x \frac{dv}{dx} \quad \text{...(Equation 2)}\]
• Separating Variables & Integrating: Substituting Equation 2 and \(\frac{y}{x} = v\) into Equation 1: \[v + x \frac{dv}{dx} = 1 + v + v^2 \implies x \frac{dv}{dx} = 1 + v^2\] \[\frac{1}{1 + v^2} dv = \frac{1}{x} dx\] Integrating both sides: \[\int \frac{1}{1 + v^2} dv = \int \frac{1}{x} dx\] \[\tan^{-1}(v) = \log|x| + c\] Resubstituting \(v = \frac{y}{x}\): \[\tan^{-1}\left(\frac{y}{x}\right) = \log|x| + c\]
• Final Answer: The general solution of the differential equation is \(\tan^{-1}\left(\frac{y}{x}\right) = \log|x| + c\).

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HSC Mathematics & Statistics (40) 12th Standard March 2026 Board Question Paper