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Inverse Trigonometric Functions — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSInverse Trigonometric Functions1 Mark
Question
Assertion $(A)$ : The range of the function $f(x)=2 \sin ^{-1} x+\frac{3 \pi}{2}$, where $x \in[-1,1]$, is $\left[\frac{\pi}{2}, \frac{5 \pi}{2}\right]$.
Reason $(R)$ : The range of the principal value branch of $\sin ^{-1}(x)$ is $[0, \pi]$
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Answer

Given, $f(x)=2 \sin ^{-1} x+\frac{3 \pi}{2}, x \in[-1,1]$
As, $-\frac{\pi}{2} \leq \sin ^{-1} x \leq \frac{\pi}{2}$
$\Rightarrow-2 \times \frac{\pi}{2} \leq 2 \sin ^{-1} x \leq 2 \times \frac{\pi}{2} \Rightarrow-\pi \leq 2 \sin ^{-1} x \leq \pi$
$\Rightarrow \quad-\pi+\frac{3 \pi}{2} \leq 2 \sin ^{-1} x+\frac{3 \pi}{2} \leq \pi+\frac{3 \pi}{2}$
$\Rightarrow \frac{\pi}{2} \leq 2 \sin ^{-1} x+\frac{3 \pi}{2} \leq \frac{5 \pi}{2}$
$\therefore \quad$ The range of $f(x)$ is $\left[\frac{\pi}{2}, \frac{5 \pi}{2}\right]$
$\therefore \quad$ Assertion $(A)$ is true.
Now, the range of the principal branch of $\sin ^{-1}(x)$ is $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$
$\therefore \quad$ Reason (R) false.

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