Roll No.
Q.P. Code 65/1/1
Mathematics
CBSE Class XII Board Examination 2026 — with answers
Time allowed: 3 hoursMaximum Marks: 80
General Instructions:
- The paper has 38 questions in five sections (A to E); all of them must be answered.
- Section A: Q1–18 are multiple choice and Q19–20 are Assertion-Reason questions, 1 mark each.
- Section B: Q21–25, very short answer (VSA), 2 marks each.
- Section C: Q26–31, short answer (SA), 3 marks each.
- Section D: Q32–35, long answer (LA), 5 marks each.
- Section E: Q36–38, case-study based, 4 marks each.
- There is no overall choice, but some questions in Sections B to E offer an internal choice (OR). Calculators are not allowed.
SECTION A
Q1–20 carry 1 mark each (MCQ; Q19–20 are Assertion-Reason).
1.
Given \(2\cos^{-1}x = y\), the values that y can take lie in
- (A) \(0 \le y \le \pi\)
- (B) \(-\pi \le y \le \pi\)
- (C) \(0 \le y \le 2\pi\)
- (D) \(-\pi \le y \le 0\)
Answer
(C) 0 ≤ y ≤ 2π
Principal value branch of cos⁻¹x is [0, π], so 0 ≤ cos⁻¹x ≤ π.
Multiply by 2: 0 ≤ 2cos⁻¹x ≤ 2π, i.e. 0 ≤ y ≤ 2π.
Common mistake: answering [0, π] – that is the range of cos⁻¹x itself, not of 2cos⁻¹x.
Multiply by 2: 0 ≤ 2cos⁻¹x ≤ 2π, i.e. 0 ≤ y ≤ 2π.
Common mistake: answering [0, π] – that is the range of cos⁻¹x itself, not of 2cos⁻¹x.
1
2.
Which one of these can NOT be the order of a row matrix?
- (A) 2 × 1
- (B) 1 × 2
- (C) 1 × 1
- (D) 1 × n
Answer
(A) 2 × 1
A row matrix has exactly one row, so its order is 1 × n.
1 × 2, 1 × 1 and 1 × n all have one row; 2 × 1 has two rows (it is a column matrix).
1 × 2, 1 × 1 and 1 × n all have one row; 2 × 1 has two rows (it is a column matrix).
1
3.
For matrices A and B of suitable orders, which of these properties hold?
(i) (A + B)′ = A′ + B′
(ii) (A − B)′ = B′ − A′
(iii) (AB)′ = A′B′
(iv) (kAB)′ = kB′A′, k a scalar
(i) (A + B)′ = A′ + B′
(ii) (A − B)′ = B′ − A′
(iii) (AB)′ = A′B′
(iv) (kAB)′ = kB′A′, k a scalar
- (A) (i) only
- (B) (i), (ii) and (iii)
- (C) (i) and (ii)
- (D) (i) and (iv)
Answer
(D) (i) and (iv)
(i) Transpose of a sum is the sum of transposes – true.
(ii) (A − B)′ = A′ − B′, not B′ − A′ – false.
(iii) Reversal law: (AB)′ = B′A′, not A′B′ – false.
(iv) (kAB)′ = k(AB)′ = kB′A′ – true.
So only (i) and (iv) are correct.
(ii) (A − B)′ = A′ − B′, not B′ − A′ – false.
(iii) Reversal law: (AB)′ = B′A′, not A′B′ – false.
(iv) (kAB)′ = k(AB)′ = kB′A′ – true.
So only (i) and (iv) are correct.
1
4.
Let \(\Delta_1 = \begin{vmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{vmatrix}\) and \(\Delta_2 = \begin{vmatrix} 0 & 2 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 6 \end{vmatrix}\). Then
- (A) \(\Delta_1 = 2\Delta_2\)
- (B) \(\Delta_2 = -2\Delta_1\)
- (C) \(\Delta_1 = \Delta_2\)
- (D) \(\Delta_2 = -\Delta_1\)
Answer
(B) Δ₂ = −2Δ₁
Δ₁ is a diagonal determinant: Δ₁ = 1 × 2 × 3 = 6.
Δ₂: expand along R₃ (only the 6 is non-zero): Δ₂ = 6 × (0·0 − 2·1) = 6 × (−2) = −12.
So Δ₂ = −12 = −2 × 6 = −2Δ₁.
Δ₂: expand along R₃ (only the 6 is non-zero): Δ₂ = 6 × (0·0 − 2·1) = 6 × (−2) = −12.
So Δ₂ = −12 = −2 × 6 = −2Δ₁.
1
5.
One value of x satisfying \(\begin{vmatrix} \cos x & \sin x \\ -\cos x & \sin x \end{vmatrix} = 1\) is
- (A) 0
- (B) \(\frac{\pi}{4}\)
- (C) \(\frac{\pi}{3}\)
- (D) \(\frac{\pi}{2}\)
Answer
(B) x = π/4
Determinant = cos x · sin x − (sin x)(−cos x) = 2 sin x cos x = sin 2x.
sin 2x = 1 ⇒ 2x = π/2 ⇒ x = π/4.
Check others: x = 0 gives 0, x = π/3 gives √3/2, x = π/2 gives 0.
sin 2x = 1 ⇒ 2x = π/2 ⇒ x = π/4.
Check others: x = 0 gives 0, x = π/3 gives √3/2, x = π/2 gives 0.
1
6.
A and B are skew-symmetric matrices of the same order. Which of the following is also skew-symmetric?
- (A) AB
- (B) AB + BA
- (C) \((A + B)^2\)
- (D) A − B
Answer
(D) A − B
Given A′ = −A, B′ = −B.
(A − B)′ = A′ − B′ = −A + B = −(A − B) ⇒ A − B is skew-symmetric.
For the others: (AB)′ = B′A′ = BA (not −AB in general); (AB + BA)′ = BA + AB, symmetric; ((A+B)²)′ = (A+B)², symmetric.
(A − B)′ = A′ − B′ = −A + B = −(A − B) ⇒ A − B is skew-symmetric.
For the others: (AB)′ = B′A′ = BA (not −AB in general); (AB + BA)′ = BA + AB, symmetric; ((A+B)²)′ = (A+B)², symmetric.
1
7.
On the interval [0, 3], the smallest value of \(f(x) = x^3 - 12x\) is
- (A) −16
- (B) −9
- (C) 0
- (D) 16
Answer
(A) −16
f′(x) = 3x² − 12 = 0 ⇒ x = ±2; only x = 2 lies in [0, 3].
Compare values at the critical point and the end points:
f(0) = 0, f(2) = 8 − 24 = −16, f(3) = 27 − 36 = −9.
Least value = −16 (at x = 2).
Compare values at the critical point and the end points:
f(0) = 0, f(2) = 8 − 24 = −16, f(3) = 27 − 36 = −9.
Least value = −16 (at x = 2).
1
8.
If \(\int \frac{3ax}{b^2 + c^2x^2}\,dx = A\log|b^2 + c^2x^2| + K\), then A equals
- (A) 3a
- (B) \(\frac{3a}{2b^2}\)
- (C) \(\frac{3a}{b^2c^2}\)
- (D) \(\frac{3a}{2c^2}\)
Answer
(D) A = 3a/(2c²)
Put t = b² + c²x² ⇒ dt = 2c²x dx ⇒ x dx = dt/(2c²).
∫ 3a x dx/(b² + c²x²) = (3a/(2c²)) ∫ dt/t = (3a/(2c²)) log|b² + c²x²| + K.
So A = 3a/(2c²).
∫ 3a x dx/(b² + c²x²) = (3a/(2c²)) ∫ dt/t = (3a/(2c²)) log|b² + c²x²| + K.
So A = 3a/(2c²).
1
9.
Evaluate \(\displaystyle\int_{-1}^{1} \frac{x^3}{x^2 + 2|x| + 1}\,dx\).
- (A) 0
- (B) log 2
- (C) 2 log 2
- (D) \(\frac{1}{2}\log 2\)
Answer
(A) 0
Let f(x) = x³/(x² + 2|x| + 1). Then f(−x) = −x³/(x² + 2|x| + 1) = −f(x), so f is odd.
For an odd function, ∫ from −a to a of f(x) dx = 0.
Tip: always test for odd/even before integrating over a symmetric interval.
For an odd function, ∫ from −a to a of f(x) dx = 0.
Tip: always test for odd/even before integrating over a symmetric interval.
1
10.
Find the area enclosed by the curve y = x|x|, the x-axis and the lines x = −1 and x = 1.
- (A) 0
- (B) \(\frac{1}{3}\)
- (C) \(\frac{2}{3}\)
- (D) \(\frac{4}{3}\)
Answer
(C) 2/3 square units
y = x|x| equals x² for x ≥ 0 and −x² for x < 0; the curve is symmetric about the origin.
Area = ∫₋₁⁰ |−x²| dx + ∫₀¹ x² dx = 1/3 + 1/3 = 2/3.
Common mistake: ∫₋₁¹ x|x| dx = 0 is the signed integral, not the area.
Area = ∫₋₁⁰ |−x²| dx + ∫₀¹ x² dx = 1/3 + 1/3 = 2/3.
Common mistake: ∫₋₁¹ x|x| dx = 0 is the signed integral, not the area.
1
11.
For the differential equation \(R\frac{dx}{dy} + Px = Q\), where P, Q, R are functions of y only, the integrating factor is
- (A) \(e^{\int \frac{P}{Q}dy}\)
- (B) \(e^{\int P\,dy}\)
- (C) \(e^{\int \frac{P}{R}dy}\)
- (D) \(e^{\int \frac{P}{R}dx}\)
Answer
(C) e^(∫(P/R) dy)
Divide by R to get standard form: dx/dy + (P/R)x = Q/R.
This is linear in x with coefficient P/R (a function of y), so IF = e^(∫(P/R) dy).
Mistake to avoid: forgetting to make the coefficient of dx/dy equal to 1.
This is linear in x with coefficient P/R (a function of y), so IF = e^(∫(P/R) dy).
Mistake to avoid: forgetting to make the coefficient of dx/dy equal to 1.
1
12.
Give the order and the degree (in that order) of the differential equation \(\frac{d}{dx}(e^{y}) = 0\).
- (A) 0, 1
- (B) 1, 1
- (C) 2, 1
- (D) 1, not defined
Answer
(B) 1, 1
Carry out the differentiation: e^y · dy/dx = 0.
Highest derivative is dy/dx ⇒ order 1; it appears to power 1 and the equation is a polynomial in dy/dx ⇒ degree 1.
Highest derivative is dy/dx ⇒ order 1; it appears to power 1 and the equation is a polynomial in dy/dx ⇒ degree 1.
1
13.
For what value of p are the vectors \(\hat i + 2\hat j + 3\hat k\) and \(2\hat i - p\hat j + \hat k\) perpendicular?
- (A) 0
- (B) 1
- (C) \(\frac{5}{2}\)
- (D) \(-\frac{5}{2}\)
Answer
(C) p = 5/2
Perpendicular ⇒ dot product = 0.
(1)(2) + (2)(−p) + (3)(1) = 0 ⇒ 5 − 2p = 0 ⇒ p = 5/2.
(1)(2) + (2)(−p) + (3)(1) = 0 ⇒ 5 − 2p = 0 ⇒ p = 5/2.
1
14.
Points with position vectors \(-\hat i - \hat j + 2\hat k\), \(2\hat i + m\hat j + 5\hat k\) and \(3\hat i + 11\hat j + 6\hat k\) lie on one line. Find m.
- (A) 8
- (B) −8
- (C) 2
- (D) \(\frac{5}{2}\)
Answer
(A) m = 8
Let A(−1, −1, 2), B(2, m, 5), C(3, 11, 6).
AB = (3, m + 1, 3), AC = (4, 12, 4).
Collinear ⇒ AB ∥ AC ⇒ 3/4 = (m + 1)/12 = 3/4 ⇒ m + 1 = 9 ⇒ m = 8.
AB = (3, m + 1, 3), AC = (4, 12, 4).
Collinear ⇒ AB ∥ AC ⇒ 3/4 = (m + 1)/12 = 3/4 ⇒ m + 1 = 9 ⇒ m = 8.
1
15.
Given \(|\vec a| = 8\), \(|\vec b| = 3\) and \(|\vec a \times \vec b| = 12\), find \(|\vec a \cdot \vec b|\).
- (A) \(6\sqrt3\)
- (B) \(8\sqrt3\)
- (C) \(12\sqrt3\)
- (D) \(3\sqrt{12}\)
Answer
(C) 12√3
|a × b| = |a||b| sin θ ⇒ 12 = 24 sin θ ⇒ sin θ = 1/2 ⇒ |cos θ| = √3/2.
|a · b| = |a||b||cos θ| = 24 × √3/2 = 12√3.
Shortcut: |a·b|² + |a×b|² = |a|²|b|² ⇒ |a·b|² = 576 − 144 = 432 ⇒ 12√3.
|a · b| = |a||b||cos θ| = 24 × √3/2 = 12√3.
Shortcut: |a·b|² + |a×b|² = |a|²|b|² ⇒ |a·b|² = 576 − 144 = 432 ⇒ 12√3.
1
16.
Find the length of the perpendicular from the point (2, 5, 7) to the line \(\frac{x}{1} = \frac{y}{0} = \frac{z}{0}\).
- (A) 2
- (B) 5
- (C) \(\sqrt{74}\)
- (D) \(\sqrt{78}\)
Answer
(C) √74
Direction ratios 1, 0, 0 through the origin ⇒ the line is the x-axis.
Foot of perpendicular from (2, 5, 7) is (2, 0, 0).
Distance = √(5² + 7²) = √(25 + 49) = √74.
Foot of perpendicular from (2, 5, 7) is (2, 0, 0).
Distance = √(5² + 7²) = √(25 + 49) = √74.
1
17.
An LPP has objective function Z = 5x + 7y. Its feasible region (shown in a figure) is the quadrilateral with corners O(0, 0), (0, 2), (3, 4) and (7, 0). Find (maximum value of Z) − (minimum value of Z).
- (A) 8
- (B) 29
- (C) 35
- (D) 43
Answer
(D) 43
Evaluate Z at each corner:
(0, 0): 0; (0, 2): 14; (3, 4): 15 + 28 = 43; (7, 0): 35.
Max = 43, min = 0 ⇒ difference = 43.
(0, 0): 0; (0, 2): 14; (3, 4): 15 + 28 = 43; (7, 0): 35.
Max = 43, min = 0 ⇒ difference = 43.
1
18.
What is the degree of the objective function of a linear programming problem?
- (A) 0
- (B) 1
- (C) 2
- (D) Any natural number
Answer
(B) 1
In an LPP the objective function Z = ax + by is linear, i.e. a polynomial of degree 1 in the decision variables.
1
19.
Assertion (A): When a fair die is thrown, the probability of getting a prime number, given that the number shown is odd, is \(\frac{2}{3}\).
Reason (R): For any two events A and B, \(P(A|B) = \frac{P(A \cup B)}{P(B)}\).
Reason (R): For any two events A and B, \(P(A|B) = \frac{P(A \cup B)}{P(B)}\).
- (A) Both A and R are true and R is the correct explanation of A
- (B) Both A and R are true but R is not the correct explanation of A
- (C) A is true but R is false
- (D) A is false but R is true
Answer
(C) A true, R false
Odd outcomes B = {1, 3, 5}; primes among them = {3, 5}.
P(prime | odd) = 2/3 ⇒ A is true.
The correct formula is P(A|B) = P(A ∩ B)/P(B), not P(A ∪ B)/P(B) ⇒ R is false.
P(prime | odd) = 2/3 ⇒ A is true.
The correct formula is P(A|B) = P(A ∩ B)/P(B), not P(A ∪ B)/P(B) ⇒ R is false.
1
20.
Assertion (A): The lines x = py + q, z = ry + s and x = p′y + q′, z = r′y + s′ are perpendicular when pp′ + rr′ = 1.
Reason (R): The lines \(\vec r = \vec a_1 + \lambda \vec b_1\) and \(\vec r = \vec a_2 + \mu \vec b_2\) are perpendicular if \(\vec b_1 \cdot \vec b_2 = 0\).
Reason (R): The lines \(\vec r = \vec a_1 + \lambda \vec b_1\) and \(\vec r = \vec a_2 + \mu \vec b_2\) are perpendicular if \(\vec b_1 \cdot \vec b_2 = 0\).
- (A) Both A and R are true and R is the correct explanation of A
- (B) Both A and R are true but R is not the correct explanation of A
- (C) A is true but R is false
- (D) A is false but R is true
Answer
(D) A false, R true
First line: (x − q)/p = y/1 = (z − s)/r ⇒ direction ratios p, 1, r. Second line: p′, 1, r′.
Perpendicular ⇔ pp′ + 1·1 + rr′ = 0 ⇔ pp′ + rr′ = −1, not 1 ⇒ A is false.
R is the standard condition for perpendicular lines ⇒ true.
Perpendicular ⇔ pp′ + 1·1 + rr′ = 0 ⇔ pp′ + rr′ = −1, not 1 ⇒ A is false.
R is the standard condition for perpendicular lines ⇒ true.
1
SECTION B
Q21–25 are very short answer (VSA) questions of 2 marks each.
21.
A function is defined by \(f(x) = \frac{|x-3|}{2(x-3)}\) for x < 3 and \(f(x) = \frac{x-6}{6}\) for x ≥ 3. Is f continuous at x = 3?
Answer
Yes – LHL = RHL = f(3) = −1/2, so f is continuous at x = 3
For x < 3, |x − 3| = −(x − 3), so f(x) = −(x − 3)/(2(x − 3)) = −1/2.
LHL = lim(x→3⁻) (−1/2) = −1/2.
RHL = lim(x→3⁺) (x − 6)/6 = −3/6 = −1/2.
f(3) = (3 − 6)/6 = −1/2.
LHL = RHL = f(3) ⇒ f is continuous at x = 3.
LHL = lim(x→3⁻) (−1/2) = −1/2.
RHL = lim(x→3⁺) (x − 6)/6 = −3/6 = −1/2.
f(3) = (3 − 6)/6 = −1/2.
LHL = RHL = f(3) ⇒ f is continuous at x = 3.
OR
Given \(\sqrt3\,(x^2 + y^2) = 4xy\), find \(\frac{dy}{dx}\) at the point \(\left(\frac12, \frac{\sqrt3}{2}\right)\).
Answer
dy/dx = √3
Differentiate both sides w.r.t. x: √3(2x + 2y y′) = 4(y + x y′).
Collect y′: y′(2√3 y − 4x) = 4y − 2√3 x ⇒ y′ = (2y − √3x)/(√3y − 2x).
At (1/2, √3/2): numerator = √3 − √3/2 = √3/2; denominator = 3/2 − 1 = 1/2.
y′ = (√3/2)/(1/2) = √3.
Collect y′: y′(2√3 y − 4x) = 4y − 2√3 x ⇒ y′ = (2y − √3x)/(√3y − 2x).
At (1/2, √3/2): numerator = √3 − √3/2 = √3/2; denominator = 3/2 − 1 = 1/2.
y′ = (√3/2)/(1/2) = √3.
2
22.
A room-freshener bottle is an inverted cone with semi-vertical angle \(\frac{\pi}{6}\). The volume of perfume in it falls steadily at 1 mm³/min. How fast is the perfume level falling when the level (height of liquid) is 10 mm?
Answer
Level falls at 3/(100π) mm/min
Let h = height and r = radius of the liquid surface. tan(π/6) = r/h ⇒ r = h/√3.
V = (1/3)πr²h = (1/3)π(h²/3)h = πh³/9.
dV/dt = (π/3)h² dh/dt.
Given dV/dt = −1: dh/dt = −3/(πh²).
At h = 10: dh/dt = −3/(100π) mm/min.
So the level drops at 3/(100π) mm/min (the minus sign shows a decrease).
V = (1/3)πr²h = (1/3)π(h²/3)h = πh³/9.
dV/dt = (π/3)h² dh/dt.
Given dV/dt = −1: dh/dt = −3/(πh²).
At h = 10: dh/dt = −3/(100π) mm/min.
So the level drops at 3/(100π) mm/min (the minus sign shows a decrease).
2
23.
P and Q are the points (1, 3, 2) and (−1, 0, 8). Find the vector of magnitude 14 along \(\overrightarrow{QP}\).
Answer
4î + 6ĵ − 12k̂
QP = P − Q = (1 + 1)î + (3 − 0)ĵ + (2 − 8)k̂ = 2î + 3ĵ − 6k̂.
|QP| = √(4 + 9 + 36) = 7.
Unit vector = (2î + 3ĵ − 6k̂)/7.
Required vector = 14 × unit vector = 4î + 6ĵ − 12k̂.
Careful: QP means from Q to P, i.e. P − Q.
|QP| = √(4 + 9 + 36) = 7.
Unit vector = (2î + 3ĵ − 6k̂)/7.
Required vector = 14 × unit vector = 4î + 6ĵ − 12k̂.
Careful: QP means from Q to P, i.e. P − Q.
2
24.
The vectors \(\vec a = 3\hat i - 2\hat j + 2\hat k\) and \(\vec b = \hat i + 2\hat k\) are adjacent sides of a parallelogram. Find vectors along its diagonals and their lengths.
Answer
Diagonals 4î − 2ĵ + 4k̂ (length 6) and 2î − 2ĵ (length 2√2)
Diagonals of a parallelogram with sides a, b are a + b and a − b.
a + b = 4î − 2ĵ + 4k̂, length √(16 + 4 + 16) = 6.
a − b = 2î − 2ĵ + 0k̂, length √(4 + 4) = 2√2.
a + b = 4î − 2ĵ + 4k̂, length √(16 + 4 + 16) = 6.
a − b = 2î − 2ĵ + 0k̂, length √(4 + 4) = 2√2.
2
25.
Write in simplest form: \(\tan^{-1}\left(\frac{\cos 2x - \sin 2x}{\cos 2x + \sin 2x}\right)\), where \(0 < x < \frac{\pi}{4}\).
Answer
π/4 − 2x
Divide numerator and denominator by cos 2x: (1 − tan 2x)/(1 + tan 2x) = tan(π/4 − 2x).
For 0 < x < π/4, π/4 − 2x lies in (−π/4, π/4), inside the principal range (−π/2, π/2).
So tan⁻¹[tan(π/4 − 2x)] = π/4 − 2x.
For 0 < x < π/4, π/4 − 2x lies in (−π/4, π/4), inside the principal range (−π/2, π/2).
So tan⁻¹[tan(π/4 − 2x)] = π/4 − 2x.
OR
Find the value of \(\tan\left(\sin^{-1}1 - \cos^{-1}\left(-\frac12\right)\right)\).
Answer
−1/√3
sin⁻¹1 = π/2.
cos⁻¹(−1/2) = π − π/3 = 2π/3 (principal range [0, π]).
tan(π/2 − 2π/3) = tan(−π/6) = −1/√3.
Common mistake: taking cos⁻¹(−1/2) = −π/3, which is outside [0, π].
cos⁻¹(−1/2) = π − π/3 = 2π/3 (principal range [0, π]).
tan(π/2 − 2π/3) = tan(−π/6) = −1/√3.
Common mistake: taking cos⁻¹(−1/2) = −π/3, which is outside [0, π].
2
SECTION C
Q26–31 are short answer (SA) questions of 3 marks each.
26.
Evaluate \(\displaystyle\int_0^1 x\tan^{-1}x\,dx\).
Answer
π/4 − 1/2
Integrate by parts with u = tan⁻¹x, dv = x dx:
∫₀¹ x tan⁻¹x dx = [(x²/2) tan⁻¹x]₀¹ − ∫₀¹ (x²/2)·1/(1 + x²) dx
= π/8 − (1/2)∫₀¹ [1 − 1/(1 + x²)] dx
= π/8 − (1/2)[x − tan⁻¹x]₀¹
= π/8 − (1/2)(1 − π/4) = π/8 − 1/2 + π/8 = π/4 − 1/2.
∫₀¹ x tan⁻¹x dx = [(x²/2) tan⁻¹x]₀¹ − ∫₀¹ (x²/2)·1/(1 + x²) dx
= π/8 − (1/2)∫₀¹ [1 − 1/(1 + x²)] dx
= π/8 − (1/2)[x − tan⁻¹x]₀¹
= π/8 − (1/2)(1 − π/4) = π/8 − 1/2 + π/8 = π/4 − 1/2.
3
27.
Find \(\displaystyle\int \sqrt{\frac{x+2}{x-2}}\,dx\).
Answer
√(x² − 4) + 2 log|x + √(x² − 4)| + C
Multiply inside the root by (x + 2)/(x + 2): integrand = (x + 2)/√(x² − 4).
Split: ∫ x/√(x² − 4) dx + 2∫ dx/√(x² − 4).
First: put t = x² − 4, dt = 2x dx ⇒ (1/2)∫ t^(−1/2) dt = √t = √(x² − 4).
Second: standard form ∫ dx/√(x² − a²) = log|x + √(x² − a²)| ⇒ 2 log|x + √(x² − 4)|.
Answer: √(x² − 4) + 2 log|x + √(x² − 4)| + C.
Split: ∫ x/√(x² − 4) dx + 2∫ dx/√(x² − 4).
First: put t = x² − 4, dt = 2x dx ⇒ (1/2)∫ t^(−1/2) dt = √t = √(x² − 4).
Second: standard form ∫ dx/√(x² − a²) = log|x + √(x² − a²)| ⇒ 2 log|x + √(x² − 4)|.
Answer: √(x² − 4) + 2 log|x + √(x² − 4)| + C.
OR
Find \(\displaystyle\int \frac{x^2}{(x^2+9)(x^2+16)}\,dx\).
Answer
−(3/7) tan⁻¹(x/3) + (4/7) tan⁻¹(x/4) + C
Put x² = t only for the partial fractions: t/((t + 9)(t + 16)) = A/(t + 9) + B/(t + 16).
t = A(t + 16) + B(t + 9). t = −9 ⇒ A = −9/7; t = −16 ⇒ B = 16/7.
Integral = −(9/7)∫ dx/(x² + 9) + (16/7)∫ dx/(x² + 16)
= −(9/7)(1/3)tan⁻¹(x/3) + (16/7)(1/4)tan⁻¹(x/4) + C
= −(3/7)tan⁻¹(x/3) + (4/7)tan⁻¹(x/4) + C.
t = A(t + 16) + B(t + 9). t = −9 ⇒ A = −9/7; t = −16 ⇒ B = 16/7.
Integral = −(9/7)∫ dx/(x² + 9) + (16/7)∫ dx/(x² + 16)
= −(9/7)(1/3)tan⁻¹(x/3) + (16/7)(1/4)tan⁻¹(x/4) + C
= −(3/7)tan⁻¹(x/3) + (4/7)tan⁻¹(x/4) + C.
3
28.
Let \(I_1 = \displaystyle\int_{-\pi/4}^{\pi/4} \frac{dx}{1 + \cos 2x}\) and \(I_2 = \displaystyle\int_{-1/2}^{1/2} |x|\,dx\). Show that \(I_1 - 4I_2 = 0\).
Answer
I₁ = 1, I₂ = 1/4, so I₁ − 4I₂ = 0
I₁: 1/(1 + cos 2x) is even, so I₁ = 2∫₀^(π/4) dx/(2cos²x) = ∫₀^(π/4) sec²x dx = [tan x]₀^(π/4) = 1.
I₂: |x| is even, so I₂ = 2∫₀^(1/2) x dx = [x²]₀^(1/2) = 1/4.
I₁ − 4I₂ = 1 − 4(1/4) = 0. Hence proved.
I₂: |x| is even, so I₂ = 2∫₀^(1/2) x dx = [x²]₀^(1/2) = 1/4.
I₁ − 4I₂ = 1 − 4(1/4) = 0. Hence proved.
3
29.
Solve (general solution): \(x^2\frac{dy}{dx} = x^2 + xy + y^2\).
Answer
tan⁻¹(y/x) = log|x| + C
Divide by x²: dy/dx = 1 + y/x + (y/x)² – homogeneous.
Put y = vx ⇒ dy/dx = v + x dv/dx.
v + x dv/dx = 1 + v + v² ⇒ x dv/dx = 1 + v².
Separate: dv/(1 + v²) = dx/x ⇒ tan⁻¹v = log|x| + C.
Replace v = y/x: tan⁻¹(y/x) = log|x| + C.
Put y = vx ⇒ dy/dx = v + x dv/dx.
v + x dv/dx = 1 + v + v² ⇒ x dv/dx = 1 + v².
Separate: dv/(1 + v²) = dx/x ⇒ tan⁻¹v = log|x| + C.
Replace v = y/x: tan⁻¹(y/x) = log|x| + C.
OR
Find the particular solution of \(xy\frac{dy}{dx} = (x+2)(y+2)\) given that y(1) = −1.
Answer
y − 2 log|y + 2| = x + 2 log|x| − 2
Separate variables: y/(y + 2) dy = (x + 2)/x dx.
Write y/(y + 2) = 1 − 2/(y + 2) and (x + 2)/x = 1 + 2/x.
Integrate: y − 2 log|y + 2| = x + 2 log|x| + C.
Use x = 1, y = −1: −1 − 2 log 1 = 1 + 2 log 1 + C ⇒ C = −2.
Particular solution: y − 2 log|y + 2| = x + 2 log|x| − 2.
Write y/(y + 2) = 1 − 2/(y + 2) and (x + 2)/x = 1 + 2/x.
Integrate: y − 2 log|y + 2| = x + 2 log|x| + C.
Use x = 1, y = −1: −1 − 2 log 1 = 1 + 2 log 1 + C ⇒ C = −2.
Particular solution: y − 2 log|y + 2| = x + 2 log|x| − 2.
3
30.
Solve graphically: minimise Z = 13x − 15y subject to x + y ≤ 7, 2x − 3y + 6 ≥ 0, x ≥ 0, y ≥ 0.
Answer
Minimum Z = −30 at (0, 2)
Draw x + y = 7 (through (7, 0), (0, 7)) and 2x − 3y + 6 = 0 (through (0, 2), (−3, 0)).
Feasible region: below x + y = 7, on the origin side of 2x − 3y + 6 = 0, in the first quadrant – bounded.
Intersection of the two lines: y = 7 − x ⇒ 2x − 21 + 3x + 6 = 0 ⇒ x = 3, y = 4.
Corners and Z: (0, 0) → 0; (7, 0) → 91; (3, 4) → 39 − 60 = −21; (0, 2) → −30.
Minimum Z = −30 at x = 0, y = 2.
Feasible region: below x + y = 7, on the origin side of 2x − 3y + 6 = 0, in the first quadrant – bounded.
Intersection of the two lines: y = 7 − x ⇒ 2x − 21 + 3x + 6 = 0 ⇒ x = 3, y = 4.
Corners and Z: (0, 0) → 0; (7, 0) → 91; (3, 4) → 39 − 60 = −21; (0, 2) → −30.
Minimum Z = −30 at x = 0, y = 2.
3
31.
Bag I has 3 red and 4 white balls; bag II has 8 red and 6 white balls. A die is rolled: if it shows a number less than 3, a ball is drawn from bag I, otherwise from bag II. Find the probability that the ball drawn is red.
Answer
P(red) = 11/21
E₁: die shows 1 or 2 ⇒ P(E₁) = 2/6 = 1/3; E₂: die shows 3–6 ⇒ P(E₂) = 4/6 = 2/3.
A: red ball. P(A|E₁) = 3/7, P(A|E₂) = 8/14 = 4/7.
Total probability: P(A) = (1/3)(3/7) + (2/3)(4/7) = 3/21 + 8/21 = 11/21.
A: red ball. P(A|E₁) = 3/7, P(A|E₂) = 8/14 = 4/7.
Total probability: P(A) = (1/3)(3/7) + (2/3)(4/7) = 3/21 + 8/21 = 11/21.
OR
For two events X and Y, the probability that at least one of them happens is a, and the probability that exactly one of them happens is b. Prove that P(X′) + P(Y′) = 2 − 2a + b.
Answer
P(X′) + P(Y′) = 2 − 2a + b (proved)
At least one: P(X ∪ Y) = a.
Exactly one: P(X ∪ Y) − P(X ∩ Y) = b ⇒ P(X ∩ Y) = a − b.
P(X′) + P(Y′) = 2 − [P(X) + P(Y)] = 2 − [P(X ∪ Y) + P(X ∩ Y)]
= 2 − [a + (a − b)] = 2 − 2a + b. Hence proved.
Exactly one: P(X ∪ Y) − P(X ∩ Y) = b ⇒ P(X ∩ Y) = a − b.
P(X′) + P(Y′) = 2 − [P(X) + P(Y)] = 2 − [P(X ∪ Y) + P(X ∩ Y)]
= 2 − [a + (a − b)] = 2 − 2a + b. Hence proved.
3
SECTION D
Q32–35 are long answer (LA) questions of 5 marks each.
32.
On the set Z of integers, a relation is defined by R = {(x, y) : |x − y| is divisible by a fixed prime p}. Determine whether R is an equivalence relation.
Answer
Yes, R is reflexive, symmetric and transitive – an equivalence relation
Reflexive: for any x ∈ Z, |x − x| = 0 and 0 is divisible by p ⇒ (x, x) ∈ R.
Symmetric: if (x, y) ∈ R, p divides |x − y| = |y − x| ⇒ (y, x) ∈ R.
Transitive: if (x, y), (y, z) ∈ R then p | (x − y) and p | (y − z) ⇒ p | (x − y) + (y − z) = x − z ⇒ p divides |x − z| ⇒ (x, z) ∈ R.
All three hold, so R is an equivalence relation.
Symmetric: if (x, y) ∈ R, p divides |x − y| = |y − x| ⇒ (y, x) ∈ R.
Transitive: if (x, y), (y, z) ∈ R then p | (x − y) and p | (y − z) ⇒ p | (x − y) + (y − z) = x − z ⇒ p divides |x − z| ⇒ (x, z) ∈ R.
All three hold, so R is an equivalence relation.
OR
A function \(f: \mathbb{R} - \{\frac35\} \to \mathbb{R} - \{\frac35\}\) is given by \(f(x) = \frac{3x+2}{5x-3}\). Prove that f is both one-one and onto.
Answer
f is one-one and onto (it is its own inverse: x = (3y + 2)/(5y − 3))
One-one: let f(x₁) = f(x₂): (3x₁ + 2)(5x₂ − 3) = (3x₂ + 2)(5x₁ − 3).
Expand: 15x₁x₂ − 9x₁ + 10x₂ − 6 = 15x₁x₂ − 9x₂ + 10x₁ − 6 ⇒ 19x₂ = 19x₁ ⇒ x₁ = x₂.
Onto: take any y ≠ 3/5. Solve y = (3x + 2)/(5x − 3): 5xy − 3y = 3x + 2 ⇒ x(5y − 3) = 3y + 2 ⇒ x = (3y + 2)/(5y − 3).
This x is real, and x ≠ 3/5 (x = 3/5 would give 15y + 10 = 15y − 9, i.e. 10 = −9, impossible), and f(x) = y.
So every element of the codomain has a pre-image ⇒ f is onto.
Expand: 15x₁x₂ − 9x₁ + 10x₂ − 6 = 15x₁x₂ − 9x₂ + 10x₁ − 6 ⇒ 19x₂ = 19x₁ ⇒ x₁ = x₂.
Onto: take any y ≠ 3/5. Solve y = (3x + 2)/(5x − 3): 5xy − 3y = 3x + 2 ⇒ x(5y − 3) = 3y + 2 ⇒ x = (3y + 2)/(5y − 3).
This x is real, and x ≠ 3/5 (x = 3/5 would give 15y + 10 = 15y − 9, i.e. 10 = −9, impossible), and f(x) = y.
So every element of the codomain has a pre-image ⇒ f is onto.
5
33.
Let \(A = \begin{bmatrix} 0 & 2 & 1 \\ -2 & -1 & -2 \\ 1 & -1 & 0 \end{bmatrix}\). Find \(A^{-1}\) and use it to solve: −2y + z = 7, 2x − y − z = 8, x − 2y = 10.
Answer
A⁻¹ = [[2, 1, 3], [2, 1, 2], [−3, −2, −4]]; x = 0, y = −5, z = −3
|A| = 0(0 − 2) − 2(0 + 2) + 1(2 + 1) = −4 + 3 = −1 ≠ 0.
Cofactors: C₁₁ = −2, C₁₂ = −2, C₁₃ = 3; C₂₁ = −1, C₂₂ = −1, C₂₃ = 2; C₃₁ = −3, C₃₂ = −2, C₃₃ = 4.
adj A = [[−2, −1, −3], [−2, −1, −2], [3, 2, 4]] ⇒ A⁻¹ = adj A/|A| = [[2, 1, 3], [2, 1, 2], [−3, −2, −4]].
The coefficient matrix of the system has rows (0, −2, 1), (2, −1, −1), (1, −2, 0) – this is A′. So A′X = B with B = (7, 8, 10).
X = (A′)⁻¹B = (A⁻¹)′B = [[2, 2, −3], [1, 1, −2], [3, 2, −4]](7, 8, 10).
x = 14 + 16 − 30 = 0; y = 7 + 8 − 20 = −5; z = 21 + 16 − 40 = −3.
Check: −2(−5) − 3 = 7 ✓, 0 + 5 + 3 = 8 ✓, 0 + 10 = 10 ✓.
Key point: notice the system uses the transpose of A.
Cofactors: C₁₁ = −2, C₁₂ = −2, C₁₃ = 3; C₂₁ = −1, C₂₂ = −1, C₂₃ = 2; C₃₁ = −3, C₃₂ = −2, C₃₃ = 4.
adj A = [[−2, −1, −3], [−2, −1, −2], [3, 2, 4]] ⇒ A⁻¹ = adj A/|A| = [[2, 1, 3], [2, 1, 2], [−3, −2, −4]].
The coefficient matrix of the system has rows (0, −2, 1), (2, −1, −1), (1, −2, 0) – this is A′. So A′X = B with B = (7, 8, 10).
X = (A′)⁻¹B = (A⁻¹)′B = [[2, 2, −3], [1, 1, −2], [3, 2, −4]](7, 8, 10).
x = 14 + 16 − 30 = 0; y = 7 + 8 − 20 = −5; z = 21 + 16 − 40 = −3.
Check: −2(−5) − 3 = 7 ✓, 0 + 5 + 3 = 8 ✓, 0 + 10 = 10 ✓.
Key point: notice the system uses the transpose of A.
OR
The matrix \(\begin{bmatrix} 3 & -1 & \sin 3x \\ -7 & 4 & \cos 2x \\ -11 & 7 & 2 \end{bmatrix}\) is singular. Find all x in \(\left[0, \frac{\pi}{2}\right]\).
Answer
x = 0 or x = π/6
Singular ⇒ determinant = 0. Expand along R₁:
3(8 − 7cos 2x) + 1(−14 + 11cos 2x) + sin 3x(−49 + 44) = 0
⇒ 24 − 21cos 2x − 14 + 11cos 2x − 5 sin 3x = 0 ⇒ 10 − 10cos 2x − 5 sin 3x = 0 ⇒ 2cos 2x + sin 3x = 2.
Use cos 2x = 1 − 2sin²x and sin 3x = 3sin x − 4sin³x:
2 − 4sin²x + 3sin x − 4sin³x = 2 ⇒ sin x(4sin²x + 4sin x − 3) = 0 ⇒ sin x(2sin x − 1)(2sin x + 3) = 0.
sin x = 0 ⇒ x = 0; sin x = 1/2 ⇒ x = π/6; sin x = −3/2 impossible.
In [0, π/2]: x = 0, π/6.
3(8 − 7cos 2x) + 1(−14 + 11cos 2x) + sin 3x(−49 + 44) = 0
⇒ 24 − 21cos 2x − 14 + 11cos 2x − 5 sin 3x = 0 ⇒ 10 − 10cos 2x − 5 sin 3x = 0 ⇒ 2cos 2x + sin 3x = 2.
Use cos 2x = 1 − 2sin²x and sin 3x = 3sin x − 4sin³x:
2 − 4sin²x + 3sin x − 4sin³x = 2 ⇒ sin x(4sin²x + 4sin x − 3) = 0 ⇒ sin x(2sin x − 1)(2sin x + 3) = 0.
sin x = 0 ⇒ x = 0; sin x = 1/2 ⇒ x = π/6; sin x = −3/2 impossible.
In [0, π/2]: x = 0, π/6.
5
34.
Given x = cos t and y = cos mt, prove that \((1 - x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} + m^2y = 0\).
Answer
(1 − x²)y″ − xy′ + m²y = 0 (proved)
dx/dt = −sin t, dy/dt = −m sin mt ⇒ dy/dx = m sin mt/sin t, i.e. sin t · dy/dx = m sin mt.
Differentiate w.r.t. x (using dt/dx = −1/sin t):
sin t · y″ + cos t · (dt/dx) · y′ = m² cos mt · (dt/dx).
Multiply by sin t: sin²t · y″ − cos t · y′ = −m² cos mt.
Substitute sin²t = 1 − x², cos t = x, cos mt = y:
(1 − x²)y″ − x y′ = −m²y ⇒ (1 − x²)y″ − x y′ + m²y = 0.
Differentiate w.r.t. x (using dt/dx = −1/sin t):
sin t · y″ + cos t · (dt/dx) · y′ = m² cos mt · (dt/dx).
Multiply by sin t: sin²t · y″ − cos t · y′ = −m² cos mt.
Substitute sin²t = 1 − x², cos t = x, cos mt = y:
(1 − x²)y″ − x y′ = −m²y ⇒ (1 − x²)y″ − x y′ + m²y = 0.
5
35.
Are the lines \(\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}\) and \(\frac{x-4}{5} = \frac{y-1}{2} = z\) parallel? If they are, find the distance between them; if not, find where they meet (if they intersect).
Answer
Not parallel; they intersect at (−1, −1, −1)
Direction ratios: 2, 3, 4 and 5, 2, 1 – not proportional ⇒ not parallel.
General points: (2λ + 1, 3λ + 2, 4λ + 3) and (5μ + 4, 2μ + 1, μ).
Equate: 2λ + 1 = 5μ + 4 and 3λ + 2 = 2μ + 1 ⇒ 2λ − 5μ = 3, 3λ − 2μ = −1.
Solving: λ = −1, μ = −1.
Third coordinate: 4(−1) + 3 = −1 and μ = −1 ✓ – consistent, so the lines intersect.
Point: (2(−1) + 1, 3(−1) + 2, 4(−1) + 3) = (−1, −1, −1).
General points: (2λ + 1, 3λ + 2, 4λ + 3) and (5μ + 4, 2μ + 1, μ).
Equate: 2λ + 1 = 5μ + 4 and 3λ + 2 = 2μ + 1 ⇒ 2λ − 5μ = 3, 3λ − 2μ = −1.
Solving: λ = −1, μ = −1.
Third coordinate: 4(−1) + 3 = −1 and μ = −1 ✓ – consistent, so the lines intersect.
Point: (2(−1) + 1, 3(−1) + 2, 4(−1) + 3) = (−1, −1, −1).
5
SECTION E
Q36–38 are case-study based questions of 4 marks each.
36.
An online delivery company has 5000 subscribers, each paying ₹300 a year for unlimited free deliveries. It plans to raise the annual fee; for every ₹1 increase, 10 subscribers are expected to leave. Suppose the fee is raised by ₹x.
(i) How many subscribers leave after a rise of ₹x? (1 mark)
(ii) Write the total revenue R(x) after the rise as a function of x. (1 mark)
(iii)(a) Find the value of x for which R(x) is maximum. (2 marks)
(i) How many subscribers leave after a rise of ₹x? (1 mark)
(ii) Write the total revenue R(x) after the rise as a function of x. (1 mark)
(iii)(a) Find the value of x for which R(x) is maximum. (2 marks)
Answer
(i) 10x; (ii) R(x) = (5000 − 10x)(300 + x); (iii) x = 100
(i) 10 subscribers per ₹1 ⇒ 10x subscribers leave.
(ii) Remaining subscribers 5000 − 10x, new fee 300 + x ⇒ R(x) = (5000 − 10x)(300 + x) = 1500000 + 2000x − 10x².
(iii) R′(x) = 2000 − 20x = 0 ⇒ x = 100. R″(x) = −20 < 0 ⇒ maximum at x = 100.
(Maximum revenue = 4900 × 400 = ₹19,60,000.)
(ii) Remaining subscribers 5000 − 10x, new fee 300 + x ⇒ R(x) = (5000 − 10x)(300 + x) = 1500000 + 2000x − 10x².
(iii) R′(x) = 2000 − 20x = 0 ⇒ x = 100. R″(x) = −20 < 0 ⇒ maximum at x = 100.
(Maximum revenue = 4900 × 400 = ₹19,60,000.)
OR
(iii)(b)
Find the sub-intervals of (0, 5000) on which R(x) is increasing and on which it is decreasing.
Answer
Increasing on (0, 100), decreasing on (100, 5000)
R(x) = (5000 − 10x)(300 + x) = 1500000 + 2000x − 10x².
R′(x) = 2000 − 20x = 20(100 − x); R′(x) = 0 at x = 100.
For 0 < x < 100, R′(x) > 0 ⇒ R increasing on (0, 100).
For 100 < x < 5000, R′(x) < 0 ⇒ R decreasing on (100, 5000).
R′(x) = 2000 − 20x = 20(100 − x); R′(x) = 0 at x = 100.
For 0 < x < 100, R′(x) > 0 ⇒ R increasing on (0, 100).
For 100 < x < 5000, R′(x) < 0 ⇒ R decreasing on (100, 5000).
4
37.
An online jackpot has one first prize of ₹3,00,000, two second prizes of ₹2,00,000 each and three third prizes of ₹50,000 each. In all 1,00,000 tickets were sold at ₹100 each. Rohan buys one ticket.
(i) What amounts can he possibly win? (1 mark)
(ii)(a) Find the probability that he wins at least ₹2,00,000. (2 marks)
(iii) Rohan also buys a ticket in another jackpot with a single prize of ₹5,00,000, where the chance of winning is 1 in 1,00,000. Find the probability that he wins on exactly one of his two tickets. (1 mark)
(i) What amounts can he possibly win? (1 mark)
(ii)(a) Find the probability that he wins at least ₹2,00,000. (2 marks)
(iii) Rohan also buys a ticket in another jackpot with a single prize of ₹5,00,000, where the chance of winning is 1 in 1,00,000. Find the probability that he wins on exactly one of his two tickets. (1 mark)
Answer
(i) ₹3,00,000, ₹2,00,000, ₹50,000 (or nothing); (ii) 3/100000; (iii) 699988/10¹⁰ = 174997/2500000000
(i) Possible winnings: ₹3,00,000, ₹2,00,000 or ₹50,000 (otherwise nothing).
(ii) Tickets worth at least ₹2,00,000 = 1 first + 2 second = 3 ⇒ P = 3/100000.
(iii) First jackpot: P(win) = 6/100000, P(lose) = 99994/100000. Second: P(win) = 1/100000, P(lose) = 99999/100000. The two are independent.
P(exactly one) = (6/100000)(99999/100000) + (99994/100000)(1/100000)
= (599994 + 99994)/10¹⁰ = 699988/10¹⁰ = 174997/2500000000.
(ii) Tickets worth at least ₹2,00,000 = 1 first + 2 second = 3 ⇒ P = 3/100000.
(iii) First jackpot: P(win) = 6/100000, P(lose) = 99994/100000. Second: P(win) = 1/100000, P(lose) = 99999/100000. The two are independent.
P(exactly one) = (6/100000)(99999/100000) + (99994/100000)(1/100000)
= (599994 + 99994)/10¹⁰ = 699988/10¹⁰ = 174997/2500000000.
OR
(ii)(b)
Find the probability that he does not win any amount.
Answer
99994/100000 = 49997/50000
Winning tickets = 1 + 2 + 3 = 6 out of 1,00,000.
P(no prize) = 1 − 6/100000 = 99994/100000 = 49997/50000.
P(no prize) = 1 − 6/100000 = 99994/100000 = 49997/50000.
4
38.
A roundabout on a busy road has its outer boundary along the circle C₁: x² + y² = 64; a circular pond with a fountain at its centre is C₂: x² + y² = 4.
(i) Draw C₁ and C₂ in one diagram. (1 mark)
(ii) Write y as a function of x for C₁ and for C₂. (1 mark)
(iii)(a) Using integration, find the area covered by the roundabout (region inside C₁). (2 marks)
(i) Draw C₁ and C₂ in one diagram. (1 mark)
(ii) Write y as a function of x for C₁ and for C₂. (1 mark)
(iii)(a) Using integration, find the area covered by the roundabout (region inside C₁). (2 marks)
Answer
(i) concentric circles at O, radii 8 and 2; (ii) y = ±√(64 − x²), y = ±√(4 − x²); (iii) 64π sq units
(i) Both circles have centre (0, 0); C₁ has radius 8 (cuts axes at ±8), C₂ radius 2 (cuts axes at ±2).
(ii) C₁: y = √(64 − x²) (upper half) or y = −√(64 − x²) (lower half); C₂: y = ±√(4 − x²).
(iii) By symmetry, area = 4∫₀⁸ √(64 − x²) dx = 4[(x/2)√(64 − x²) + (64/2) sin⁻¹(x/8)]₀⁸ = 4[0 + 32 · π/2] = 64π sq units.
(ii) C₁: y = √(64 − x²) (upper half) or y = −√(64 − x²) (lower half); C₂: y = ±√(4 − x²).
(iii) By symmetry, area = 4∫₀⁸ √(64 − x²) dx = 4[(x/2)√(64 − x²) + (64/2) sin⁻¹(x/8)]₀⁸ = 4[0 + 32 · π/2] = 64π sq units.
OR
(iii)(b)
Using integration, find the area of the circular pond.
Answer
4π sq units
Area of pond = 4∫₀² √(4 − x²) dx = 4[(x/2)√(4 − x²) + 2 sin⁻¹(x/2)]₀² = 4[2 · π/2] = 4π sq units.
4
— END —
Questions restated from the CBSE Class XII Board Examination 2026 (set 65/1/1) for practice. Reading passages and map items are described, not reproduced.