Question

Let $f\left(x\right)=\sqrt{({\sin }^{-1}x-{\cos }^{-1}x)},$ $g\left(x\right)=\sqrt{({\tan }^{-1}x-{\cot }^{-1}x)}$ and $h\left(x\right)=\sqrt{({\sec }^{-1}x-{\text{cosec}}^{-1}x)}$. Then CORRECT statement(s) is(are)

• A. Domain of $f\left(x\right)+g\left(x\right)$ is $\left\{1\right\}$
• B. Domain of $g\left(x\right)+h\left(x\right)$ is $[\sqrt{2},\infty )$
• C. Domain of $h\left(x\right)+f\left(x\right)$ is $[\frac{1}{\sqrt{2}},1]$
• D. Domain of $f\left(x\right)+g\left(x\right)+h\left(x\right)$ is $\varphi$

Answer 1

Answer: (A), (B), (D)

Domain of $f\left(x\right)+g\left(x\right)+h\left(x\right)$ is $\varphi$

${\sin }^{-1}-{\cos }^{-1}x=2{\sin }^{-1}x-\frac{\pi }{2}$

$f\left(x\right)=\sqrt{(2{\sin }^{-1}x-\pi /2)}$
For $x$ being in domain, $2{\sin }^{-1}x-\frac{\pi }{2}\ge 0$
$\Rightarrow {\sin }^{-1}x\ge \frac{\pi }{4}$
So, Domain of, $f\left(x\right):\frac{1}{\sqrt{2}}\le x\le 1$

Similarly $g\left(x\right)=\sqrt{(2{\tan }^{-1}x-\frac{\pi }{2})}$
For $x$ being in domain: ${\tan }^{-1}x\ge \frac{\pi }{4}$
So domain of $g\left(x\right):1\le x<\infty$

Similarly $h\left(x\right)=\sqrt{(2{\sec }^{-1}x-\frac{\pi }{2})}$
For $x$ being in domain: ${\sec }^{-1}x\ge \frac{\pi }{4}$
So, Domain of $h\left(x\right):(-\infty ,-1]\cup [\sqrt{2},\infty )$

Now, domain of $f\left(x\right)+g\left(x\right)$ is $\left\{1\right\}$ only.
for $g\left(x\right)+h\left(x\right),$ domain is $[\sqrt{2},\infty )$
for $h\left(x\right)+f\left(x\right),$ domain is null set
for $g\left(x\right)+h\left(x\right)+f\left(x\right),$ domain is null set.