Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Relations and Functions

Question:

Find the domain of $f(x)=\sqrt{\cos^{-1}x-\sin^{-1}x}$

Options:

$[-1,\frac{1}{\sqrt{2}}]$

$[-1,-\frac{1}{\sqrt{2}}]$

$[-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}]$

$[-\frac{1}{\sqrt{2}},1]$

Correct Answer:

$[-1,\frac{1}{\sqrt{2}}]$

Explanation:

We must have $\cos^{-1}x≥\sin^{-1}x$

or $\frac{π}{2}-\sin^{-1}x≥\sin^{-1}x$

or $\frac{π}{2}≥2\sin^{-1}x$

or $\sin^{-1}x≤\frac{π}{4}$, but $-\frac{π}{2}≤\sin^{-1}x$

or $-\frac{π}{2}≤\sin^{-1}x≤\frac{π}{4}$

or $\sin(-\frac{π}{2})≤x≤\sin\frac{π}{4}$

(∵ $\sin x$ is increasing function in $[-\frac{π}{2},\frac{π}{2}]$)

or $x∈[-1,\frac{1}{\sqrt{2}}]$