Roll No.
Q.P. Code 65/1/1
Mathematics
CBSE Class XII Board Examination 2024 — 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, 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 internal choices (OR) are given in 2 questions of Section B, 3 of Section C, 2 of Section D and 2 of Section E. Calculators are not allowed.
SECTION A
Q1–20 carry 1 mark each (Q1–18 MCQ; Q19–20 Assertion-Reason).
1.
The function \( f : \mathbb{R}_+ \to \mathbb{R} \) (where \( \mathbb{R}_+ \) is the set of all non-negative real numbers) is given by \( f(x) = 4x + 3 \). Then f is:
- (A) one-one but not onto
- (B) onto but not one-one
- (C) both one-one and onto
- (D) neither one-one nor onto
Answer
(A) one-one but not onto
One-one: \( 4x_1 + 3 = 4x_2 + 3 \Rightarrow x_1 = x_2 \), so f is one-one.
Onto: for x ≥ 0, f(x) ≥ 3. So values like 0 or −5 in the codomain ℝ are never reached. Range = [3, ∞) ≠ ℝ, so f is not onto.
Common mistake: forgetting that the domain is only x ≥ 0 and the codomain is all of ℝ.
Onto: for x ≥ 0, f(x) ≥ 3. So values like 0 or −5 in the codomain ℝ are never reached. Range = [3, ∞) ≠ ℝ, so f is not onto.
Common mistake: forgetting that the domain is only x ≥ 0 and the codomain is all of ℝ.
1
2.
A matrix has 36 elements. How many different orders can it have?
- (A) 13
- (B) 3
- (C) 5
- (D) 9
Answer
(D) 9
An order m × n needs mn = 36, so each order corresponds to a divisor m of 36.
Divisors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36 → 9 divisors.
Orders: 1×36, 2×18, 3×12, 4×9, 6×6, 9×4, 12×3, 18×2, 36×1 = 9 orders.
Divisors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36 → 9 divisors.
Orders: 1×36, 2×18, 3×12, 4×9, 6×6, 9×4, 12×3, 18×2, 36×1 = 9 orders.
1
3.
Consider \( f(x) = \begin{cases} x^2 + 3, & x \ne 0 \\ 1, & x = 0 \end{cases} \). Which statement is true?
- (A) f(x) is continuous and differentiable for all x ∈ ℝ
- (B) f(x) is continuous for all x ∈ ℝ
- (C) f(x) is continuous and differentiable for all x ∈ ℝ − {0}
- (D) f(x) is discontinuous at infinitely many points
Answer
(C) continuous and differentiable on ℝ − {0}
At x = 0: \( \lim_{x\to 0} (x^2 + 3) = 3 \) but f(0) = 1, so f is discontinuous (hence not differentiable) at 0.
For x ≠ 0, f(x) = x² + 3 is a polynomial, so it is continuous and differentiable there.
So f is continuous and differentiable on ℝ − {0}, and has only one point of discontinuity.
For x ≠ 0, f(x) = x² + 3 is a polynomial, so it is continuous and differentiable there.
So f is continuous and differentiable on ℝ − {0}, and has only one point of discontinuity.
1
4.
f(x) is continuous on [a, b] and differentiable on (a, b). f(x) is strictly increasing on (a, b) if:
- (A) f′(x) < 0 for all x ∈ (a, b)
- (B) f′(x) > 0 for all x ∈ (a, b)
- (C) f′(x) = 0 for all x ∈ (a, b)
- (D) f(x) > 0 for all x ∈ (a, b)
Answer
(B) f′(x) > 0 on (a, b)
Standard result: if f′(x) > 0 at every point of (a, b), f is strictly increasing on (a, b).
f′ < 0 gives strictly decreasing, f′ = 0 gives a constant function, and f(x) > 0 says nothing about increasing.
f′ < 0 gives strictly decreasing, f′ = 0 gives a constant function, and f(x) > 0 says nothing about increasing.
1
5.
Given \( \begin{bmatrix} x+y & 2 \\ 5 & xy \end{bmatrix} = \begin{bmatrix} 6 & 2 \\ 5 & 8 \end{bmatrix} \), find \( \dfrac{24}{x} + \dfrac{24}{y} \).
- (A) 7
- (B) 6
- (C) 8
- (D) 18
Answer
(D) 18
Equal matrices → equal entries: x + y = 6 and xy = 8.
\( \dfrac{24}{x} + \dfrac{24}{y} = \dfrac{24(x+y)}{xy} = \dfrac{24 \times 6}{8} = 18 \).
No need to find x and y separately.
\( \dfrac{24}{x} + \dfrac{24}{y} = \dfrac{24(x+y)}{xy} = \dfrac{24 \times 6}{8} = 18 \).
No need to find x and y separately.
1
6.
\( \int_a^b f(x)\,dx \) equals:
- (A) \( \int_a^b f(a - x)\,dx \)
- (B) \( \int_a^b f(a + b - x)\,dx \)
- (C) \( \int_a^b f(x - (a + b))\,dx \)
- (D) \( \int_a^b f((a - x) + (b - x))\,dx \)
Answer
(B) \( \int_a^b f(a+b-x)\,dx \)
Property of definite integrals: put t = a + b − x, dt = −dx; x = a → t = b, x = b → t = a.
\( \int_a^b f(a+b-x)\,dx = \int_b^a f(t)(-dt) = \int_a^b f(t)\,dt \).
\( \int_a^b f(a+b-x)\,dx = \int_b^a f(t)(-dt) = \int_a^b f(t)\,dt \).
1
7.
θ is the angle between two unit vectors \( \hat{a} \) and \( \hat{b} \), and sin θ = 3/5. Then \( \hat{a} \cdot \hat{b} \) equals:
- (A) ± 3/5
- (B) ± 3/4
- (C) ± 4/5
- (D) ± 4/3
Answer
(C) ± 4/5
\( \hat{a} \cdot \hat{b} = |\hat a||\hat b|\cos\theta = \cos\theta \).
cos θ = ±√(1 − 9/25) = ±4/5 (θ may be acute or obtuse).
cos θ = ±√(1 − 9/25) = ±4/5 (θ may be acute or obtuse).
1
8.
Find the integrating factor of \( (1 - x^2)\dfrac{dy}{dx} + xy = ax \), where −1 < x < 1.
- (A) \( \dfrac{1}{x^2 - 1} \)
- (B) \( \dfrac{1}{\sqrt{x^2 - 1}} \)
- (C) \( \dfrac{1}{1 - x^2} \)
- (D) \( \dfrac{1}{\sqrt{1 - x^2}} \)
Answer
(D) \( 1/\sqrt{1-x^2} \)
Standard form: \( \dfrac{dy}{dx} + \dfrac{x}{1-x^2}\,y = \dfrac{ax}{1-x^2} \), so P = x/(1 − x²).
\( \int P\,dx = -\tfrac12 \log(1 - x^2) \).
IF = \( e^{-\frac12\log(1-x^2)} = \dfrac{1}{\sqrt{1-x^2}} \).
\( \int P\,dx = -\tfrac12 \log(1 - x^2) \).
IF = \( e^{-\frac12\log(1-x^2)} = \dfrac{1}{\sqrt{1-x^2}} \).
1
9.
A line has direction cosines √3 k, √3 k, √3 k. Find k.
- (A) ± 1
- (B) ± √3
- (C) ± 3
- (D) ± 1/3
Answer
(D) k = ± 1/3
Sum of squares of direction cosines = 1: 3k² + 3k² + 3k² = 1 ⇒ 9k² = 1 ⇒ k = ± 1/3.
1
10.
A linear programming problem is about optimising (maximising or minimising) a:
- (A) logarithmic function
- (B) linear function
- (C) quadratic function
- (D) exponential function
Answer
(B) linear function
In an LPP the objective function Z = ax + by is linear, and the constraints are linear inequalities.
1
11.
If P(A | B) = P(A′ | B), which statement is true?
- (A) P(A) = P(A′)
- (B) P(A) = 2 P(B)
- (C) P(A ∩ B) = ½ P(B)
- (D) P(A ∩ B) = 2 P(B)
Answer
(C) P(A ∩ B) = ½ P(B)
P(A | B) + P(A′ | B) = 1. If they are equal, each is ½.
So P(A ∩ B)/P(B) = ½ ⇒ P(A ∩ B) = ½ P(B).
So P(A ∩ B)/P(B) = ½ ⇒ P(A ∩ B) = ½ P(B).
1
12.
Evaluate \( \begin{vmatrix} x+1 & x-1 \\ x^2+x+1 & x^2-x+1 \end{vmatrix} \).
- (A) 2x³
- (B) 2
- (C) 0
- (D) 2x³ − 2
Answer
(B) 2
Value = (x + 1)(x² − x + 1) − (x − 1)(x² + x + 1).
Using a³ + b³ and a³ − b³: = (x³ + 1) − (x³ − 1) = 2.
Using a³ + b³ and a³ − b³: = (x³ + 1) − (x³ − 1) = 2.
1
13.
Find the derivative of sin(x²) with respect to x at \( x = \sqrt{\pi} \).
- (A) 1
- (B) −1
- (C) \( -2\sqrt{\pi} \)
- (D) \( 2\sqrt{\pi} \)
Answer
(C) \( -2\sqrt{\pi} \)
Chain rule: d/dx sin(x²) = 2x cos(x²).
At x = √π: 2√π · cos π = 2√π · (−1) = −2√π.
At x = √π: 2√π · cos π = 2√π · (−1) = −2√π.
1
14.
Find the order and degree of \( \left[1 + \left(\dfrac{dy}{dx}\right)^2\right]^3 = \dfrac{d^2y}{dx^2} \).
- (A) 1, 2
- (B) 2, 3
- (C) 2, 1
- (D) 2, 6
Answer
(C) order 2, degree 1
Highest derivative present is d²y/dx² → order 2.
The equation is already a polynomial in derivatives and d²y/dx² appears with power 1 → degree 1.
Common mistake: taking the power 3 (or 6) of dy/dx as the degree; degree is the power of the HIGHEST order derivative.
The equation is already a polynomial in derivatives and d²y/dx² appears with power 1 → degree 1.
Common mistake: taking the power 3 (or 6) of dy/dx as the degree; degree is the power of the HIGHEST order derivative.
1
15.
Find the vector whose initial point is B(3, −4, 7) and terminal point is A(2, −3, 5).
- (A) \( \hat i - \hat j + 2\hat k \)
- (B) \( \hat i + \hat j + 2\hat k \)
- (C) \( -\hat i - \hat j - 2\hat k \)
- (D) \( -\hat i + \hat j - 2\hat k \)
Answer
(D) \( -\hat i + \hat j - 2\hat k \)
Vector from B to A = (position of A) − (position of B) = (2 − 3)î + (−3 + 4)ĵ + (5 − 7)k̂ = −î + ĵ − 2k̂.
1
16.
The distance of the point P(a, b, c) from the y-axis is:
- (A) b
- (B) b²
- (C) \( \sqrt{a^2 + c^2} \)
- (D) a² + c²
Answer
(C) \( \sqrt{a^2+c^2} \)
Foot of perpendicular from P on the y-axis is (0, b, 0).
Distance = √((a − 0)² + (b − b)² + (c − 0)²) = √(a² + c²).
Distance = √((a − 0)² + (b − b)² + (c − 0)²) = √(a² + c²).
1
17.
How many corner points does the feasible region of x ≥ 0, y ≥ 0, x + y ≥ 4 have?
- (A) 0
- (B) 1
- (C) 2
- (D) 3
Answer
(C) 2
The region lies above the line x + y = 4 in the first quadrant and is unbounded.
Its corner points are where x + y = 4 meets the axes: (4, 0) and (0, 4). The origin is not in the region. So 2 corner points.
Its corner points are where x + y = 4 meets the axes: (4, 0) and (0, 4). The origin is not in the region. So 2 corner points.
1
18.
A and B are non-zero square matrices of the same order with (A + B)² = A² + B². Then:
- (A) AB = O
- (B) AB = −BA
- (C) BA = O
- (D) AB = BA
Answer
(B) AB = −BA
(A + B)² = A² + AB + BA + B² (matrix multiplication is not commutative, so AB + BA, not 2AB).
Given it equals A² + B² ⇒ AB + BA = O ⇒ AB = −BA.
Given it equals A² + B² ⇒ AB + BA = O ⇒ AB = −BA.
1
19.
Assertion (A): For \( A = \begin{bmatrix} 1 & \cos\theta & 1 \\ -\cos\theta & 1 & \cos\theta \\ -1 & -\cos\theta & 1 \end{bmatrix} \), θ ∈ [0, 2π], we have |A| ∈ [2, 4].
Reason (R): cos θ ∈ [−1, 1] for all θ ∈ [0, 2π].
Reason (R): cos θ ∈ [−1, 1] for all θ ∈ [0, 2π].
- (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
(A) Both true, R explains A
Let c = cos θ. Expanding along R1: |A| = 1(1 + c²) − c(−c + c) + 1(c² + 1) = 2 + 2c².
Since c ∈ [−1, 1], c² ∈ [0, 1], so |A| ∈ [2, 4]. A is true.
R is true and it is exactly the fact used to get the range of |A|, so R explains A.
Since c ∈ [−1, 1], c² ∈ [0, 1], so |A| ∈ [2, 4]. A is true.
R is true and it is exactly the fact used to get the range of |A|, so R explains A.
1
20.
Assertion (A): No line in space can be perpendicular to all three of the x, y and z axes at the same time.
Reason (R): If a line makes angles α, β, γ with the positive x, y, z axes, then cos²α + cos²β + cos²γ = 1.
Reason (R): If a line makes angles α, β, γ with the positive x, y, z axes, then cos²α + cos²β + cos²γ = 1.
- (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
(A) Both true, R explains A
If a line were perpendicular to all three axes, α = β = γ = 90°, so cos²α + cos²β + cos²γ = 0, contradicting R (sum must be 1).
So A is true, R is a standard true identity, and R is the reason A holds.
So A is true, R is a standard true identity, and R is the reason A holds.
1
SECTION B
Q21–25 are very short answer (VSA) questions, 2 marks each.
21.
Is the function f(x) = x²|x| differentiable at x = 0? Justify.
Answer
Yes, f is differentiable at x = 0 and f′(0) = 0.
f(x) = x³ for x ≥ 0 and f(x) = −x³ for x < 0; f(0) = 0.
RHD = \( \lim_{h\to0^+} \dfrac{h^3 - 0}{h} = \lim h^2 = 0 \).
LHD = \( \lim_{h\to0^+} \dfrac{f(-h) - f(0)}{-h} = \lim_{h\to0^+} \dfrac{h^3}{-h} = \lim(-h^2) = 0 \).
RHD = LHD = 0, so f is differentiable at x = 0.
RHD = \( \lim_{h\to0^+} \dfrac{h^3 - 0}{h} = \lim h^2 = 0 \).
LHD = \( \lim_{h\to0^+} \dfrac{f(-h) - f(0)}{-h} = \lim_{h\to0^+} \dfrac{h^3}{-h} = \lim(-h^2) = 0 \).
RHD = LHD = 0, so f is differentiable at x = 0.
OR
Given \( y = \sqrt{\tan\sqrt{x}} \), prove that \( \sqrt{x}\,\dfrac{dy}{dx} = \dfrac{1 + y^4}{4y} \).
Answer
Proved: \( \sqrt x\,y' = (1+y^4)/(4y) \)
Chain rule: \( \dfrac{dy}{dx} = \dfrac{1}{2\sqrt{\tan\sqrt x}} \cdot \sec^2\sqrt x \cdot \dfrac{1}{2\sqrt x} \).
So \( \sqrt x\,\dfrac{dy}{dx} = \dfrac{\sec^2\sqrt x}{4\sqrt{\tan\sqrt x}} = \dfrac{1 + \tan^2\sqrt x}{4\sqrt{\tan\sqrt x}} \).
Since y² = tan√x, tan²√x = y⁴ and √(tan√x) = y, giving \( \dfrac{1 + y^4}{4y} \).
So \( \sqrt x\,\dfrac{dy}{dx} = \dfrac{\sec^2\sqrt x}{4\sqrt{\tan\sqrt x}} = \dfrac{1 + \tan^2\sqrt x}{4\sqrt{\tan\sqrt x}} \).
Since y² = tan√x, tan²√x = y⁴ and √(tan√x) = y, giving \( \dfrac{1 + y^4}{4y} \).
2
22.
Show that f(x) = 4x³ − 18x² + 27x − 7 has no maximum or minimum value.
Answer
f′(x) = 3(2x − 3)² ≥ 0, so f is increasing on ℝ: no maxima/minima.
f′(x) = 12x² − 36x + 27 = 3(4x² − 12x + 9) = 3(2x − 3)².
f′(x) ≥ 0 for all x, and f′ = 0 only at x = 3/2, where f′ does not change sign.
So f is increasing on ℝ and has no local maximum or minimum.
Common mistake: calling x = 3/2 an extreme point just because f′(3/2) = 0; check the sign change.
f′(x) ≥ 0 for all x, and f′ = 0 only at x = 3/2, where f′ does not change sign.
So f is increasing on ℝ and has no local maximum or minimum.
Common mistake: calling x = 3/2 an extreme point just because f′(3/2) = 0; check the sign change.
2
23.
Find \( \int x\sqrt{1 + 2x}\,dx \).
Answer
\( \dfrac{(1+2x)^{5/2}}{10} - \dfrac{(1+2x)^{3/2}}{6} + C \)
Put 1 + 2x = t², so x = (t² − 1)/2 and dx = t dt.
Integral = \( \int \dfrac{t^2 - 1}{2}\cdot t \cdot t\,dt = \dfrac12\int (t^4 - t^2)\,dt = \dfrac12\left(\dfrac{t^5}{5} - \dfrac{t^3}{3}\right) + C \).
Back-substitute t = √(1 + 2x): \( \dfrac{(1+2x)^{5/2}}{10} - \dfrac{(1+2x)^{3/2}}{6} + C \).
Integral = \( \int \dfrac{t^2 - 1}{2}\cdot t \cdot t\,dt = \dfrac12\int (t^4 - t^2)\,dt = \dfrac12\left(\dfrac{t^5}{5} - \dfrac{t^3}{3}\right) + C \).
Back-substitute t = √(1 + 2x): \( \dfrac{(1+2x)^{5/2}}{10} - \dfrac{(1+2x)^{3/2}}{6} + C \).
OR
Evaluate \( \displaystyle\int_0^{\pi^2/4} \dfrac{\sin\sqrt x}{\sqrt x}\,dx \).
Answer
2
Put √x = t ⇒ dx = 2t dt; limits x = 0 → t = 0, x = π²/4 → t = π/2.
Integral = \( \int_0^{\pi/2} \dfrac{\sin t}{t}\cdot 2t\,dt = 2\int_0^{\pi/2}\sin t\,dt = 2[-\cos t]_0^{\pi/2} = 2(0 + 1) = 2 \).
Integral = \( \int_0^{\pi/2} \dfrac{\sin t}{t}\cdot 2t\,dt = 2\int_0^{\pi/2}\sin t\,dt = 2[-\cos t]_0^{\pi/2} = 2(0 + 1) = 2 \).
2
24.
\( \vec a \) and \( \vec b \) are non-zero vectors with \( (\vec a + \vec b) \perp \vec a \) and \( (2\vec a + \vec b) \perp \vec b \). Prove that \( |\vec b| = \sqrt2\,|\vec a| \).
Answer
Proved: |b|² = 2|a|² ⇒ |b| = √2 |a|
Perpendicular ⇒ dot product zero.
(1) \( (\vec a + \vec b)\cdot\vec a = 0 \Rightarrow |\vec a|^2 + \vec a\cdot\vec b = 0 \Rightarrow \vec a\cdot\vec b = -|\vec a|^2 \).
(2) \( (2\vec a + \vec b)\cdot\vec b = 0 \Rightarrow 2\,\vec a\cdot\vec b + |\vec b|^2 = 0 \).
Substitute (1) in (2): \( -2|\vec a|^2 + |\vec b|^2 = 0 \Rightarrow |\vec b|^2 = 2|\vec a|^2 \Rightarrow |\vec b| = \sqrt2|\vec a| \).
(1) \( (\vec a + \vec b)\cdot\vec a = 0 \Rightarrow |\vec a|^2 + \vec a\cdot\vec b = 0 \Rightarrow \vec a\cdot\vec b = -|\vec a|^2 \).
(2) \( (2\vec a + \vec b)\cdot\vec b = 0 \Rightarrow 2\,\vec a\cdot\vec b + |\vec b|^2 = 0 \).
Substitute (1) in (2): \( -2|\vec a|^2 + |\vec b|^2 = 0 \Rightarrow |\vec b|^2 = 2|\vec a|^2 \Rightarrow |\vec b| = \sqrt2|\vec a| \).
2
25.
ABCD is a parallelogram (figure: AB is the top side, DC the bottom side, and the diagonal DB is drawn from D to B). Given \( \vec{AB} = 2\hat i - 4\hat j + 5\hat k \) and \( \vec{DB} = 3\hat i - 6\hat j + 2\hat k \), find \( \vec{AD} \) and then the area of the parallelogram.
Answer
\( \vec{AD} = -\hat i + 2\hat j + 3\hat k \); area = 11√5 sq units
Triangle law: \( \vec{AD} + \vec{DB} = \vec{AB} \Rightarrow \vec{AD} = (2-3)\hat i + (-4+6)\hat j + (5-2)\hat k = -\hat i + 2\hat j + 3\hat k \).
Area = \( |\vec{AD}\times\vec{AB}| \).
\( \vec{AD}\times\vec{AB} = \hat i(2\cdot5 - 3\cdot(-4)) - \hat j((-1)\cdot5 - 3\cdot2) + \hat k((-1)(-4) - 2\cdot2) = 22\hat i + 11\hat j + 0\hat k \).
Area = √(484 + 121) = √605 = 11√5 square units.
Area = \( |\vec{AD}\times\vec{AB}| \).
\( \vec{AD}\times\vec{AB} = \hat i(2\cdot5 - 3\cdot(-4)) - \hat j((-1)\cdot5 - 3\cdot2) + \hat k((-1)(-4) - 2\cdot2) = 22\hat i + 11\hat j + 0\hat k \).
Area = √(484 + 121) = √605 = 11√5 square units.
2
SECTION C
Q26–31 are short answer (SA) questions, 3 marks each.
26.
On A = {1, 2, 3, 4, 5}, the relation R = {(x, y) : |x² − y²| < 8}. Check whether R is reflexive, symmetric and transitive.
Answer
R is reflexive and symmetric but not transitive.
Reflexive: |x² − x²| = 0 < 8 for every x ∈ A, so (x, x) ∈ R. ✔
Symmetric: if |x² − y²| < 8 then |y² − x²| = |x² − y²| < 8, so (y, x) ∈ R. ✔
Transitive: (1, 2) ∈ R since |1 − 4| = 3 < 8; (2, 3) ∈ R since |4 − 9| = 5 < 8; but |1 − 9| = 8 is not < 8, so (1, 3) ∉ R. ✘
Hence R is reflexive, symmetric, not transitive.
Symmetric: if |x² − y²| < 8 then |y² − x²| = |x² − y²| < 8, so (y, x) ∈ R. ✔
Transitive: (1, 2) ∈ R since |1 − 4| = 3 < 8; (2, 3) ∈ R since |4 − 9| = 5 < 8; but |1 − 9| = 8 is not < 8, so (1, 3) ∉ R. ✘
Hence R is reflexive, symmetric, not transitive.
OR
f : ℝ → ℝ is given by f(x) = ax + b with f(1) = 1 and f(2) = 3. Find f(x) and check whether f is one-one and onto.
Answer
f(x) = 2x − 1; f is one-one and onto (bijective).
a + b = 1 and 2a + b = 3 ⇒ a = 2, b = −1, so f(x) = 2x − 1.
One-one: 2x₁ − 1 = 2x₂ − 1 ⇒ x₁ = x₂. ✔
Onto: for any y ∈ ℝ take x = (y + 1)/2 ∈ ℝ; then f(x) = 2·(y + 1)/2 − 1 = y. ✔
So f is both one-one and onto.
One-one: 2x₁ − 1 = 2x₂ − 1 ⇒ x₁ = x₂. ✔
Onto: for any y ∈ ℝ take x = (y + 1)/2 ∈ ℝ; then f(x) = 2·(y + 1)/2 − 1 = y. ✔
So f is both one-one and onto.
3
27.
Given \( \sqrt{1 - x^2} + \sqrt{1 - y^2} = a(x - y) \), prove that \( \dfrac{dy}{dx} = \sqrt{\dfrac{1 - y^2}{1 - x^2}} \).
Answer
Proved: \( \sin^{-1}x - \sin^{-1}y = 2\cot^{-1}a \) ⇒ \( y' = \sqrt{(1-y^2)/(1-x^2)} \)
Put x = sin θ, y = sin φ. Then cos θ + cos φ = a(sin θ − sin φ).
Using sum-to-product: \( 2\cos\frac{\theta+\phi}{2}\cos\frac{\theta-\phi}{2} = 2a\cos\frac{\theta+\phi}{2}\sin\frac{\theta-\phi}{2} \).
Cancel ⇒ \( \cot\frac{\theta-\phi}{2} = a \Rightarrow \theta - \phi = 2\cot^{-1}a \) (a constant).
So \( \sin^{-1}x - \sin^{-1}y = 2\cot^{-1}a \). Differentiate: \( \dfrac{1}{\sqrt{1-x^2}} - \dfrac{1}{\sqrt{1-y^2}}\dfrac{dy}{dx} = 0 \).
Hence \( \dfrac{dy}{dx} = \dfrac{\sqrt{1-y^2}}{\sqrt{1-x^2}} \).
Using sum-to-product: \( 2\cos\frac{\theta+\phi}{2}\cos\frac{\theta-\phi}{2} = 2a\cos\frac{\theta+\phi}{2}\sin\frac{\theta-\phi}{2} \).
Cancel ⇒ \( \cot\frac{\theta-\phi}{2} = a \Rightarrow \theta - \phi = 2\cot^{-1}a \) (a constant).
So \( \sin^{-1}x - \sin^{-1}y = 2\cot^{-1}a \). Differentiate: \( \dfrac{1}{\sqrt{1-x^2}} - \dfrac{1}{\sqrt{1-y^2}}\dfrac{dy}{dx} = 0 \).
Hence \( \dfrac{dy}{dx} = \dfrac{\sqrt{1-y^2}}{\sqrt{1-x^2}} \).
OR
Find dy/dx if y = (tan x)^x.
Answer
\( \dfrac{dy}{dx} = (\tan x)^x\left[\log\tan x + \dfrac{x\sec^2x}{\tan x}\right] \)
Take logs: log y = x log(tan x).
Differentiate: \( \dfrac1y\dfrac{dy}{dx} = \log\tan x + x\cdot\dfrac{\sec^2 x}{\tan x} \).
So \( \dfrac{dy}{dx} = (\tan x)^x\left[\log\tan x + \dfrac{x\sec^2x}{\tan x}\right] \) (the bracket can also be written log tan x + 2x cosec 2x).
Differentiate: \( \dfrac1y\dfrac{dy}{dx} = \log\tan x + x\cdot\dfrac{\sec^2 x}{\tan x} \).
So \( \dfrac{dy}{dx} = (\tan x)^x\left[\log\tan x + \dfrac{x\sec^2x}{\tan x}\right] \) (the bracket can also be written log tan x + 2x cosec 2x).
3
28.
Find \( \displaystyle\int \dfrac{x^2}{(x^2 + 4)(x^2 + 9)}\,dx \).
Answer
\( -\dfrac25\tan^{-1}\dfrac x2 + \dfrac35\tan^{-1}\dfrac x3 + C \)
Put x² = t only for partial fractions: \( \dfrac{t}{(t+4)(t+9)} = \dfrac{A}{t+4} + \dfrac{B}{t+9} \).
t = −4: A = −4/5; t = −9: B = 9/5.
So integrand = \( -\dfrac45\cdot\dfrac{1}{x^2+2^2} + \dfrac95\cdot\dfrac{1}{x^2+3^2} \).
Integrate: \( -\dfrac45\cdot\dfrac12\tan^{-1}\dfrac x2 + \dfrac95\cdot\dfrac13\tan^{-1}\dfrac x3 + C = -\dfrac25\tan^{-1}\dfrac x2 + \dfrac35\tan^{-1}\dfrac x3 + C \).
t = −4: A = −4/5; t = −9: B = 9/5.
So integrand = \( -\dfrac45\cdot\dfrac{1}{x^2+2^2} + \dfrac95\cdot\dfrac{1}{x^2+3^2} \).
Integrate: \( -\dfrac45\cdot\dfrac12\tan^{-1}\dfrac x2 + \dfrac95\cdot\dfrac13\tan^{-1}\dfrac x3 + C = -\dfrac25\tan^{-1}\dfrac x2 + \dfrac35\tan^{-1}\dfrac x3 + C \).
OR
Evaluate \( \displaystyle\int_1^3 \big(|x - 1| + |x - 2| + |x - 3|\big)\,dx \).
Answer
5
On [1, 3]: |x − 1| = x − 1 and |x − 3| = 3 − x, so their sum is 2.
|x − 2| = 2 − x on [1, 2] and x − 2 on [2, 3].
Integral = \( \int_1^3 2\,dx + \int_1^2 (2-x)\,dx + \int_2^3 (x-2)\,dx = 4 + \tfrac12 + \tfrac12 = 5 \).
|x − 2| = 2 − x on [1, 2] and x − 2 on [2, 3].
Integral = \( \int_1^3 2\,dx + \int_1^2 (2-x)\,dx + \int_2^3 (x-2)\,dx = 4 + \tfrac12 + \tfrac12 = 5 \).
3
29.
Find the particular solution of \( x^2\dfrac{dy}{dx} - xy = x^2\cos^2\left(\dfrac{y}{2x}\right) \), given y = π/2 when x = 1.
Answer
\( 2\tan\left(\dfrac{y}{2x}\right) = \log|x| + 2 \)
Divide by x²: \( \dfrac{dy}{dx} = \dfrac yx + \cos^2\dfrac{y}{2x} \) (homogeneous).
Put y = vx: \( v + x\dfrac{dv}{dx} = v + \cos^2\dfrac v2 \Rightarrow \sec^2\dfrac v2\,dv = \dfrac{dx}{x} \).
Integrate: \( 2\tan\dfrac v2 = \log|x| + C \Rightarrow 2\tan\dfrac{y}{2x} = \log|x| + C \).
x = 1, y = π/2: 2 tan(π/4) = 0 + C ⇒ C = 2.
Particular solution: \( 2\tan\dfrac{y}{2x} = \log|x| + 2 \).
Put y = vx: \( v + x\dfrac{dv}{dx} = v + \cos^2\dfrac v2 \Rightarrow \sec^2\dfrac v2\,dv = \dfrac{dx}{x} \).
Integrate: \( 2\tan\dfrac v2 = \log|x| + C \Rightarrow 2\tan\dfrac{y}{2x} = \log|x| + C \).
x = 1, y = π/2: 2 tan(π/4) = 0 + C ⇒ C = 2.
Particular solution: \( 2\tan\dfrac{y}{2x} = \log|x| + 2 \).
3
30.
Solve graphically: maximise Z = 500x + 300y subject to x + 2y ≤ 12, 2x + y ≤ 12, 4x + 5y ≥ 20, x ≥ 0, y ≥ 0.
Answer
Max Z = 3200 at x = 4, y = 4
Draw the lines x + 2y = 12, 2x + y = 12 and 4x + 5y = 20. The feasible region is bounded, lying below the first two lines and above 4x + 5y = 20 in the first quadrant.
Corner points: (5, 0), (6, 0), (4, 4) [intersection of x + 2y = 12 and 2x + y = 12], (0, 6), (0, 4).
Z values: (5,0) → 2500; (6,0) → 3000; (4,4) → 3200; (0,6) → 1800; (0,4) → 1200.
Maximum Z = 3200 at (4, 4).
Corner points: (5, 0), (6, 0), (4, 4) [intersection of x + 2y = 12 and 2x + y = 12], (0, 6), (0, 4).
Z values: (5,0) → 2500; (6,0) → 3000; (4,4) → 3200; (0,6) → 1800; (0,4) → 1200.
Maximum Z = 3200 at (4, 4).
3
31.
E and F are independent events with P(Ē) = 0.6 and P(E ∪ F) = 0.6. Find P(F) and P(Ē ∪ F̄).
Answer
P(F) = 1/3; P(Ē ∪ F̄) = 13/15
P(E) = 1 − 0.6 = 0.4.
Independent ⇒ P(E ∩ F) = 0.4 P(F).
P(E ∪ F) = P(E) + P(F) − P(E ∩ F): 0.6 = 0.4 + P(F) − 0.4P(F) ⇒ 0.6 P(F) = 0.2 ⇒ P(F) = 1/3.
P(Ē ∪ F̄) = 1 − P(E ∩ F) (De Morgan) = 1 − 0.4 × 1/3 = 1 − 2/15 = 13/15.
Independent ⇒ P(E ∩ F) = 0.4 P(F).
P(E ∪ F) = P(E) + P(F) − P(E ∩ F): 0.6 = 0.4 + P(F) − 0.4P(F) ⇒ 0.6 P(F) = 0.2 ⇒ P(F) = 1/3.
P(Ē ∪ F̄) = 1 − P(E ∩ F) (De Morgan) = 1 − 0.4 × 1/3 = 1 − 2/15 = 13/15.
3
SECTION D
Q32–35 are long answer (LA) questions, 5 marks each.
32.
For \( A = \begin{bmatrix} 1 & -2 & 0 \\ 2 & -1 & -1 \\ 0 & -2 & 1 \end{bmatrix} \), find A⁻¹ and use it to solve x − 2y = 10, 2x − y − z = 8, −2y + z = 7.
Answer
\( A^{-1} = \begin{bmatrix} -3 & 2 & 2 \\ -2 & 1 & 1 \\ -4 & 2 & 3 \end{bmatrix} \); x = 0, y = −5, z = −3
|A| = 1(−1 − 2) + 2(2 − 0) + 0 = −3 + 4 = 1 ≠ 0, so A⁻¹ exists.
Cofactors: A₁₁ = −3, A₁₂ = −2, A₁₃ = −4; A₂₁ = 2, A₂₂ = 1, A₂₃ = 2; A₃₁ = 2, A₃₂ = 1, A₃₃ = 3.
adj A (transpose of cofactor matrix) = \( \begin{bmatrix} -3 & 2 & 2 \\ -2 & 1 & 1 \\ -4 & 2 & 3 \end{bmatrix} \), and A⁻¹ = adj A / 1 = the same matrix.
The system is AX = B with B = [10, 8, 7]ᵀ (its coefficient matrix is exactly A).
X = A⁻¹B: x = −30 + 16 + 14 = 0; y = −20 + 8 + 7 = −5; z = −40 + 16 + 21 = −3.
Check: 0 + 10 = 10 ✔, 0 + 5 + 3 = 8 ✔, 10 − 3 = 7 ✔.
Cofactors: A₁₁ = −3, A₁₂ = −2, A₁₃ = −4; A₂₁ = 2, A₂₂ = 1, A₂₃ = 2; A₃₁ = 2, A₃₂ = 1, A₃₃ = 3.
adj A (transpose of cofactor matrix) = \( \begin{bmatrix} -3 & 2 & 2 \\ -2 & 1 & 1 \\ -4 & 2 & 3 \end{bmatrix} \), and A⁻¹ = adj A / 1 = the same matrix.
The system is AX = B with B = [10, 8, 7]ᵀ (its coefficient matrix is exactly A).
X = A⁻¹B: x = −30 + 16 + 14 = 0; y = −20 + 8 + 7 = −5; z = −40 + 16 + 21 = −3.
Check: 0 + 10 = 10 ✔, 0 + 5 + 3 = 8 ✔, 10 − 3 = 7 ✔.
OR
Given \( A = \begin{bmatrix} -1 & a & 2 \\ 1 & 2 & x \\ 3 & 1 & 1 \end{bmatrix} \) and \( A^{-1} = \begin{bmatrix} 1 & -1 & 1 \\ -8 & 7 & -5 \\ b & y & 3 \end{bmatrix} \), find (a + x) − (b + y).
Answer
a = 1, x = 3, b = 5, y = −4; (a + x) − (b + y) = 3
Use AA⁻¹ = I and compare entries.
Row 3 of A × column 1 of A⁻¹: 3(1) + 1(−8) + 1(b) = 0 ⇒ b = 5.
Row 3 × column 2: 3(−1) + 7 + y = 0 ⇒ y = −4.
Row 1 × column 3: −1 − 5a + 6 = 0 ⇒ a = 1.
Row 2 × column 3: 1 − 10 + 3x = 0 ⇒ x = 3.
(Check row 1 × column 1: −1 − 8a + 2b = −1 − 8 + 10 = 1 ✔.)
(a + x) − (b + y) = (1 + 3) − (5 − 4) = 3.
Row 3 of A × column 1 of A⁻¹: 3(1) + 1(−8) + 1(b) = 0 ⇒ b = 5.
Row 3 × column 2: 3(−1) + 7 + y = 0 ⇒ y = −4.
Row 1 × column 3: −1 − 5a + 6 = 0 ⇒ a = 1.
Row 2 × column 3: 1 − 10 + 3x = 0 ⇒ x = 3.
(Check row 1 × column 1: −1 − 8a + 2b = −1 − 8 + 10 = 1 ✔.)
(a + x) − (b + y) = (1 + 3) − (5 − 4) = 3.
5
33.
Evaluate \( \displaystyle\int_0^{\pi/4} \dfrac{\sin x + \cos x}{9 + 16\sin 2x}\,dx \).
Answer
\( \dfrac{1}{20}\log 3 \)
Put t = sin x − cos x ⇒ dt = (cos x + sin x)dx. Then t² = 1 − sin 2x ⇒ sin 2x = 1 − t².
Limits: x = 0 → t = −1; x = π/4 → t = 0.
Denominator: 9 + 16(1 − t²) = 25 − 16t².
I = \( \int_{-1}^{0} \dfrac{dt}{25 - 16t^2} = \dfrac{1}{16}\int_{-1}^0 \dfrac{dt}{(5/4)^2 - t^2} = \dfrac{1}{16}\cdot\dfrac{1}{2\cdot\frac54}\left[\log\left|\dfrac{\frac54 + t}{\frac54 - t}\right|\right]_{-1}^0 \).
= \( \dfrac{1}{40}\left[\log 1 - \log\dfrac{1}{9}\right] = \dfrac{1}{40}\log 9 = \dfrac{1}{20}\log 3 \).
Limits: x = 0 → t = −1; x = π/4 → t = 0.
Denominator: 9 + 16(1 − t²) = 25 − 16t².
I = \( \int_{-1}^{0} \dfrac{dt}{25 - 16t^2} = \dfrac{1}{16}\int_{-1}^0 \dfrac{dt}{(5/4)^2 - t^2} = \dfrac{1}{16}\cdot\dfrac{1}{2\cdot\frac54}\left[\log\left|\dfrac{\frac54 + t}{\frac54 - t}\right|\right]_{-1}^0 \).
= \( \dfrac{1}{40}\left[\log 1 - \log\dfrac{1}{9}\right] = \dfrac{1}{40}\log 9 = \dfrac{1}{20}\log 3 \).
OR
Evaluate \( \displaystyle\int_0^{\pi/2} \sin 2x\,\tan^{-1}(\sin x)\,dx \).
Answer
\( \dfrac{\pi}{2} - 1 \)
sin 2x = 2 sin x cos x. Put sin x = t, cos x dx = dt; limits 0 → 0, π/2 → 1.
I = \( 2\int_0^1 t\tan^{-1}t\,dt \).
By parts (u = tan⁻¹t, dv = t dt): \( \int t\tan^{-1}t\,dt = \dfrac{t^2}{2}\tan^{-1}t - \dfrac12\int\dfrac{t^2}{1+t^2}dt = \dfrac{t^2}{2}\tan^{-1}t - \dfrac12(t - \tan^{-1}t) \).
At 1: \( \dfrac{\pi}{8} - \dfrac12 + \dfrac{\pi}{8} = \dfrac{\pi}{4} - \dfrac12 \); at 0: 0.
I = 2(π/4 − 1/2) = π/2 − 1.
I = \( 2\int_0^1 t\tan^{-1}t\,dt \).
By parts (u = tan⁻¹t, dv = t dt): \( \int t\tan^{-1}t\,dt = \dfrac{t^2}{2}\tan^{-1}t - \dfrac12\int\dfrac{t^2}{1+t^2}dt = \dfrac{t^2}{2}\tan^{-1}t - \dfrac12(t - \tan^{-1}t) \).
At 1: \( \dfrac{\pi}{8} - \dfrac12 + \dfrac{\pi}{8} = \dfrac{\pi}{4} - \dfrac12 \); at 0: 0.
I = 2(π/4 − 1/2) = π/2 − 1.
5
34.
Using integration, find the area of the part of the ellipse \( \dfrac{x^2}{16} + \dfrac{y^2}{4} = 1 \) that lies between the lines x = −2 and x = 2.
Answer
\( 4\sqrt3 + \dfrac{8\pi}{3} \) sq units
For the ellipse, y = ½√(16 − x²). The region is symmetric about both axes.
Area = \( 4\int_0^2 \dfrac12\sqrt{16 - x^2}\,dx = 2\left[\dfrac x2\sqrt{16-x^2} + \dfrac{16}{2}\sin^{-1}\dfrac x4\right]_0^2 \).
At x = 2: \( \dfrac22\sqrt{12} + 8\cdot\dfrac{\pi}{6} = 2\sqrt3 + \dfrac{4\pi}{3} \).
Area = 2(2√3 + 4π/3) = 4√3 + 8π/3 square units.
(Sketch: ellipse with vertices (±4, 0), (0, ±2), and the strip between x = −2 and x = 2 shaded.)
Area = \( 4\int_0^2 \dfrac12\sqrt{16 - x^2}\,dx = 2\left[\dfrac x2\sqrt{16-x^2} + \dfrac{16}{2}\sin^{-1}\dfrac x4\right]_0^2 \).
At x = 2: \( \dfrac22\sqrt{12} + 8\cdot\dfrac{\pi}{6} = 2\sqrt3 + \dfrac{4\pi}{3} \).
Area = 2(2√3 + 4π/3) = 4√3 + 8π/3 square units.
(Sketch: ellipse with vertices (±4, 0), (0, ±2), and the strip between x = −2 and x = 2 shaded.)
5
35.
The image of a point P(x, y, z) in the line \( \dfrac x1 = \dfrac{y-1}{2} = \dfrac{z-2}{3} \) is P′(1, 0, 7). Find P.
Answer
P(1, 6, 3)
The foot of the perpendicular M from P′ (same as from P) is the mid-point of PP′ and lies on the line: M = (λ, 2λ + 1, 3λ + 2).
Direction of P′M: (λ − 1, 2λ + 1, 3λ − 5). It is perpendicular to (1, 2, 3):
(λ − 1) + 2(2λ + 1) + 3(3λ − 5) = 0 ⇒ 14λ − 14 = 0 ⇒ λ = 1, so M = (1, 3, 5).
M is the mid-point of P and P′: (x + 1)/2 = 1, (y + 0)/2 = 3, (z + 7)/2 = 5 ⇒ x = 1, y = 6, z = 3.
P = (1, 6, 3).
Direction of P′M: (λ − 1, 2λ + 1, 3λ − 5). It is perpendicular to (1, 2, 3):
(λ − 1) + 2(2λ + 1) + 3(3λ − 5) = 0 ⇒ 14λ − 14 = 0 ⇒ λ = 1, so M = (1, 3, 5).
M is the mid-point of P and P′: (x + 1)/2 = 1, (y + 0)/2 = 3, (z + 7)/2 = 5 ⇒ x = 1, y = 6, z = 3.
P = (1, 6, 3).
5
SECTION E
Q36–38 are case-study based questions, 4 marks each.
36.
Case Study 1: Traffic police use speed cameras that can catch a speeding vehicle from up to 300 m away. One camera is fixed on a pole 5 m high, and a car C moves away from the pole at 20 m/s. When the car is x m from the foot of the pole, the angle of elevation of the camera from the car is θ.
(i) Express θ in terms of the height of the camera and x. (1)
(ii) Find dθ/dx. (1)
(iii)(a) Find the rate of change of the angle of elevation with respect to time when the car is 50 m from the pole. (2)
(ii) Find dθ/dx. (1)
(iii)(a) Find the rate of change of the angle of elevation with respect to time when the car is 50 m from the pole. (2)
Answer
(i) θ = tan⁻¹(5/x) (ii) dθ/dx = −5/(25 + x²) (iii) dθ/dt = −4/101 rad/s
(i) tan θ = height/distance = 5/x ⇒ θ = tan⁻¹(5/x).
(ii) \( \dfrac{d\theta}{dx} = \dfrac{1}{1 + 25/x^2}\cdot\left(-\dfrac{5}{x^2}\right) = -\dfrac{5}{x^2 + 25} \).
(iii) \( \dfrac{d\theta}{dt} = \dfrac{d\theta}{dx}\cdot\dfrac{dx}{dt} = -\dfrac{5}{25 + x^2}\times 20 \). At x = 50: −100/2525 = −4/101 rad/s.
The negative sign means the angle decreases as the car moves away.
(ii) \( \dfrac{d\theta}{dx} = \dfrac{1}{1 + 25/x^2}\cdot\left(-\dfrac{5}{x^2}\right) = -\dfrac{5}{x^2 + 25} \).
(iii) \( \dfrac{d\theta}{dt} = \dfrac{d\theta}{dx}\cdot\dfrac{dx}{dt} = -\dfrac{5}{25 + x^2}\times 20 \). At x = 50: −100/2525 = −4/101 rad/s.
The negative sign means the angle decreases as the car moves away.
OR
(iii)(b)
(iii)(b) For another car 50 m from the foot of the pole, the angle of elevation is changing at the rate of 3/101 rad/s (in magnitude). Find the speed of this car. (2)
Answer
Speed = 15 m/s
\( \dfrac{d\theta}{dt} = -\dfrac{5}{25 + x^2}\cdot\dfrac{dx}{dt} \).
At x = 50: 25 + 2500 = 2525, so magnitude of dθ/dt = \( \dfrac{5}{2525}\left|\dfrac{dx}{dt}\right| = \dfrac{1}{505}\left|\dfrac{dx}{dt}\right| \).
Set equal to 3/101: |dx/dt| = 3 × 505/101 = 15.
Speed of the car = 15 m/s.
At x = 50: 25 + 2500 = 2525, so magnitude of dθ/dt = \( \dfrac{5}{2525}\left|\dfrac{dx}{dt}\right| = \dfrac{1}{505}\left|\dfrac{dx}{dt}\right| \).
Set equal to 3/101: |dx/dt| = 3 × 505/101 = 15.
Speed of the car = 15 m/s.
4
37.
Case Study 2: Climate change has made air turbulence more common. An airplane meets severe, moderate or light turbulence with equal probability, and the chance of arriving late is 55%, 37% and 17% respectively for these three.
(i) Find the probability that an airplane reaches its destination late. (2)
(ii) Given that the airplane reached late, find the probability that this was due to moderate turbulence. (2)
(ii) Given that the airplane reached late, find the probability that this was due to moderate turbulence. (2)
Answer
(i) 109/300 (ii) 37/109
Let E₁, E₂, E₃ = severe, moderate, light turbulence; P(Eᵢ) = 1/3 each. A = reaching late.
P(A|E₁) = 0.55, P(A|E₂) = 0.37, P(A|E₃) = 0.17.
(i) Total probability: P(A) = (1/3)(0.55 + 0.37 + 0.17) = (1/3)(1.09) = 109/300.
(ii) Bayes' theorem: P(E₂|A) = \( \dfrac{\frac13 \times 0.37}{109/300} = \dfrac{37/300}{109/300} = \dfrac{37}{109} \).
P(A|E₁) = 0.55, P(A|E₂) = 0.37, P(A|E₃) = 0.17.
(i) Total probability: P(A) = (1/3)(0.55 + 0.37 + 0.17) = (1/3)(1.09) = 109/300.
(ii) Bayes' theorem: P(E₂|A) = \( \dfrac{\frac13 \times 0.37}{109/300} = \dfrac{37/300}{109/300} = \dfrac{37}{109} \).
4
38.
Case Study 3: A function has an inverse only when it is one-one and onto. Sine on ℝ is neither (its graph is the familiar wave between −1 and 1), so sine is restricted to an interval A, mapped onto [−1, 1], so that sin⁻¹ : [−1, 1] → A exists.
(i) Give an example of such an interval A other than the principal value branch. (1)
(ii) Using the principal value branch, find \( \sin^{-1}\left(-\tfrac12\right) - \sin^{-1}(1) \). (1)
(iii)(a) Draw the graph of sin⁻¹x on [−1, 1] for its principal value branch. (2)
(ii) Using the principal value branch, find \( \sin^{-1}\left(-\tfrac12\right) - \sin^{-1}(1) \). (1)
(iii)(a) Draw the graph of sin⁻¹x on [−1, 1] for its principal value branch. (2)
Answer
(i) e.g. [π/2, 3π/2] (ii) −2π/3 (iii) increasing curve from (−1, −π/2) through (0, 0) to (1, π/2)
(i) Any interval of length π on which sine is one-one and takes all values in [−1, 1], e.g. [π/2, 3π/2] or [−3π/2, −π/2].
(ii) sin⁻¹(−1/2) = −π/6 and sin⁻¹(1) = π/2, so the value = −π/6 − π/2 = −2π/3.
(iii) Domain [−1, 1], range [−π/2, π/2]. Plot points (−1, −π/2), (−1/2, −π/6), (0, 0), (1/2, π/6), (1, π/2) and join them with a smooth increasing curve (the mirror image of y = sin x on [−π/2, π/2] in the line y = x).
(ii) sin⁻¹(−1/2) = −π/6 and sin⁻¹(1) = π/2, so the value = −π/6 − π/2 = −2π/3.
(iii) Domain [−1, 1], range [−π/2, π/2]. Plot points (−1, −π/2), (−1/2, −π/6), (0, 0), (1/2, π/6), (1, π/2) and join them with a smooth increasing curve (the mirror image of y = sin x on [−π/2, π/2] in the line y = x).
OR
(iii)(b)
(iii)(b) Find the domain and range of f(x) = 2 sin⁻¹(1 − x). (2)
Answer
Domain [0, 2]; range [−π, π]
sin⁻¹ needs −1 ≤ 1 − x ≤ 1 ⇒ −2 ≤ −x ≤ 0 ⇒ 0 ≤ x ≤ 2. Domain = [0, 2].
−π/2 ≤ sin⁻¹(1 − x) ≤ π/2 ⇒ −π ≤ 2 sin⁻¹(1 − x) ≤ π. Range = [−π, π].
−π/2 ≤ sin⁻¹(1 − x) ≤ π/2 ⇒ −π ≤ 2 sin⁻¹(1 − x) ≤ π. Range = [−π, π].
4
— END —
Questions restated from the CBSE Class XII Board Examination 2024 (set 65/1/1) for practice. Reading passages and map items are described, not reproduced.