Inverse Trigonometric Functions 4 marks Questions, CBSE Class 12 PYQs 2024–2026

1 different 4 marks questions on Inverse Trigonometric Functions from CBSE Class 12 Maths board exams 2024–2026, newest first.

Q384 marksCase StudyInverse Trigonometric Functions CBSE 2024 · 65/1/1

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)
(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)
(iii) Draw the graph of $sin^{- 1} x$ from [– 1, 1] to its principal value branch. (2)
OR (iii) Find the domain and range of $f (x) = 2 sin^{- 1} (1 - x)$ . (2)

Diagram for CBSE 2024 Class 12 Maths question 38

Show answer & solution

Answer: (i) e.g.

[\frac{π}{2}, \frac{3 π}{2}]

(ii)

- \frac{2 π}{3}

(iii) graph of

y = sin^{- 1} x

from

(- 1, - \frac{π}{2})

through O to

(1, \frac{π}{2})

OR (iii) domain [0, 2], range

[- π, π]

(i) Sine is one-one and onto [– 1, 1] on $[\frac{π}{2}, \frac{3 π}{2}]$ (or $[- \frac{3 π}{2}, - \frac{π}{2}]$ ), so A can be such an interval.
(ii) $sin^{- 1} (- \frac{1}{2}) - sin^{- 1} (1) = - \frac{π}{6} - \frac{π}{2} = - \frac{2 π}{3}$ .
(iii) Reflect the part of $y = sin x$ on $[- \frac{π}{2}, \frac{π}{2}]$ in the line $y = x$ : an increasing curve from $(- 1, - \frac{π}{2})$ through (0, 0) to $(1, \frac{π}{2})$ , with domain [– 1, 1] and range $[- \frac{π}{2}, \frac{π}{2}]$ .
OR (iii) Need $- 1 \leq 1 - x \leq 1$ , i.e. $0 \leq x \leq 2$ : domain [0, 2].
$sin^{- 1} (1 - x) \in [- \frac{π}{2}, \frac{π}{2}]$ , so $f (x) \in [- π, π]$ : range $[- π, π]$ .

Also asked in: 2024 65/1/2, 2024 65/1/3

Inverse Trigonometric Functions: 4 marks Questions (CBSE Class 12)