If a function f : X → Y defined as f(x) = y is one-one and onto, then we can define a unique function g : Y → X such that g(y) = x, where x X and y = f(x), y Y. Function g is called the inverse of function f. The domain of sine function is R and function sine : R → R is neither one-one nor onto. The following graph shows the sine function. Let sine function be defined from set A to [-1, 1] such that inverse of sine function exists, i.e., ^-1x is defined from [-1, 1] to A. On the basis of the above information, answer the following questions : (i) If A is the interval other than principal value branch, give an example of one such interval. (ii) If ^-1(x) is defined from [-1, 1] to its principal value branch, find the value of ^-1 (-1/2 ) - ^-1(1). (iii) (a) Draw the graph of ^-1x from [-1, 1] to its principal value branch. (iii) (b) Find the domain and range of f(x) = 2^-1(1-x).

(i) [(π)/(2),(3π)/(2) ] (or any similar interval); (ii) -(2π)/(3); (iii)(a) increasing curve from (-1,-π/2) through (0,0) to (1,π/2); (iii)(b) domain [0,2], range [-π,π]

MathsCBSE Class 12 · Board exam 2024, set 65/1/1 · 4 marks · Case-based

CBSE Class 12 Mathematics 2024: question 38

Set in the CBSE Class 12 board exam of 2024, paper 65/1/1, section E, question 38, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q38 · 4 marks · Case-basedCompetency-basedInverse Trigonometric Functions

If a function $f : X \to Y$ defined as $f (x) = y$ is one-one and onto, then we can define a unique function $g : Y \to X$ such that $g (y) = x$ , where $x \in X$ and $y = f (x), y \in Y$ . Function g is called the inverse of function f.

The domain of sine function is R and function sine : $R \to R$ is neither one-one nor onto. The following graph shows the sine function.

Let sine function be defined from set A to $[- 1, 1]$ such that inverse of sine function exists, i.e., $sin^{- 1} x$ is defined from $[- 1, 1]$ to A.

On the basis of the above information, answer the following questions :

(i)
If A is the interval other than principal value branch, give an example of one such interval.
1 mark
(ii)
If $sin^{- 1} (x)$ is defined from $[- 1, 1]$ to its principal value branch, find the value of $sin^{- 1} (- \frac{1}{2}) - sin^{- 1} (1)$ .
1 mark
(iii)
(a) Draw the graph of $sin^{- 1} x$ from $[- 1, 1]$ to its principal value branch.
2 marks
OR
(iii)
(b) Find the domain and range of $f (x) = 2 sin^{- 1} (1 - x)$ .
2 marks

Figure in the paper: Graph of y = sin x on X'X and Y'Y axes, a wave passing through the origin, with marked x-values at -5π/2, -2π, -π, -π/2, π/2, π, 3π/2, 2π, 5π/2 and y-values 1 and -1. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 65/1/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

Look at the sine graph for an interval of length $π$ on which sine is one-one and covers all of $[- 1, 1]$ .

Hint 2

Recall that the principal value branch of $sin^{- 1}$ is $[- π /2, π /2]$ ; find the angles in it whose sines are $- 1/2$ and 1.

Hint 3

For the graph, reflect the principal part of $y = sin x$ in the line $y = x$ . For the domain, the input $1 - x$ of $sin^{- 1}$ must lie in $[- 1, 1]$ .

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

(i) On $[\frac{π}{2}, \frac{3 π}{2}]$ , sine is one-one and takes every value in $[- 1, 1]$ (it falls from 1 to −1). So $A = [\frac{π}{2}, \frac{3 π}{2}]$ works. Any such interval, e.g. $[- \frac{3 π}{2}, - \frac{π}{2}]$ , is also correct.
1 mark
(ii) With principal values in $[- \frac{π}{2}, \frac{π}{2}]$ :
$sin^{- 1} (- \frac{1}{2}) - sin^{- 1} (1) = - \frac{π}{6} - \frac{π}{2} = - \frac{2 π}{3}$
1 mark
(iii)(a) **Graph of $y = sin^{- 1} x$ , domain $[- 1, 1]$ , range $[- \frac{π}{2}, \frac{π}{2}]$ :**
- Plot the points $(- 1, - \frac{π}{2})$ , $(- \frac{1}{2}, - \frac{π}{6})$ , $(0, 0)$ , $(\frac{1}{2}, \frac{π}{6})$ , $(1, \frac{π}{2})$ .
- Join them with a smooth increasing curve through the origin. It is symmetric about the origin and becomes vertical at the end-points $x = \pm 1$ .
- This curve is the mirror image of $y = sin x$ , $x \in [- \frac{π}{2}, \frac{π}{2}]$ , in the line $y = x$ .
- Mark $\pm \frac{π}{2}$ on the y-axis and $\pm 1$ on the x-axis.
2 marks
(iii)(b) Domain: $sin^{- 1} u$ is defined only for $- 1 \leq u \leq 1$ , so $- 1 \leq 1 - x \leq 1 \Rightarrow - 2 \leq - x \leq 0 \Rightarrow 0 \leq x \leq 2$ .
½ mark
(iii)(b) Domain $= [0, 2]$ .
½ mark
(iii)(b) Range: $- \frac{π}{2} \leq sin^{- 1} (1 - x) \leq \frac{π}{2}$ ; multiplying by 2, $- π \leq 2 sin^{- 1} (1 - x) \leq π$ .
½ mark
(iii)(b) Range $= [- π, π]$ .
½ mark

Answer: (i) $[\frac{π}{2}, \frac{3 π}{2}]$ (or any similar interval); (ii) $- \frac{2 π}{3}$ ; (iii)(a) increasing curve from $(- 1, - π /2)$ through $(0, 0)$ to $(1, π /2)$ ; (iii)(b) domain $[0, 2]$ , range $[- π, π]$

Where the marks are

(i) Gives a valid interval other than the principal branch, e.g. $[π /2, 3 π /2]$ (1 mark)
(ii) Uses the principal values $- π /6$ and $π /2$ to get $- 2 π /3$ (1 mark)
(iii)(a) Correct graph of $sin^{- 1} x$ on $[- 1, 1]$ with range $[- π /2, π /2]$ marked (2 marks)
(iii)(b) Writes the condition $- 1 \leq 1 - x \leq 1$ (½ mark)
(iii)(b) Gets domain $[0, 2]$ (½ mark)
(iii)(b) Uses the bounds $- π /2 \leq sin^{- 1} (1 - x) \leq π /2$ (½ mark)
(iii)(b) Gets range $[- π, π]$ (½ mark)

Revise the chapter

Chapter: Inverse Trigonometric Functions (notes, formulas, where marks are lost and every past question)
Topic: Principal Values of Inverse Trigonometric Functions

Practise this chapterBoard-style questions from Inverse Trigonometric Functions, one at a time, marked against the answer key and marking points.Back to the full paper2024 board paper, set 65/1/1: every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.

CBSE Class 12 Mathematics 2024: question 38

Hints

Step-by-step solution

Revise the chapter

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