Question

Assertion (A) : The maximum value of $f\left(x\right)={\sin }^{-1}x+{\cos }^{-1}x+{\tan }^{-1}x$ is $\frac{3\pi }{4}$
Reason (R) : ${\sin }^{-1}x>{\cos }^{-1}x$ for all $x$ in $R$

• A. Both A and R are true and R is the correct explanation of A
• B. Both A and R are true and R is not correct explanation of A
• C. A is true but R is false
• D. A is false but R is true

Answer 1

Answer: (C)

A is true but R is false

$f\left(x\right)=si{n}^{-1}\left(x\right)+co{s}^{-1}\left(x\right)+ta{n}^{-1}\left(x\right)$
Now
$si{n}^{-1}\left(x\right)+co{s}^{-1}\left(x\right)=\frac{\pi }{2}$
Hence
$f\left(x\right)=ta{n}^{-1}\left(x\right)+\frac{\pi }{2}$
Substituting x=1, we get the maximum as
$\frac{\pi }{4}+\frac{\pi }{2}$
$=\frac{3\pi }{4}$
Reason.
The intersection point of sin(x) and cos(x) is $x=\frac{\pi }{4}$ between $[0,\frac{\pi }{2}]$.
Now cos(x) is a decreasing function in the above interval while sin(x) is an increasing function in the above interval.
Also.
$cos\left(x\right)>sin\left(x\right)$ between $[0,\frac{\pi }{4}]$ while
$sin\left(x\right)>cos\left(x\right)$ between $[\frac{\pi }{4},\frac{\pi }{2}]$
Hence
$si{n}^{-1}\left(x\right)>co{s}^{-1}\left(x\right)$ between $[\frac{1}{\sqrt{2}},1]$.
Hence reason is false.