Question

Let $f(x)=\sin ^{-1} x+\cos ^{-1} x+\pi(|x|-2)$. The true set of values of $k$ for which the equation $f(x)=k \pi$ possesses real solution is $[a, b]$ then the value of $2|a|+4|b|$ is equal to

• A. 2
• B. 3
• C. 5
• D. 7

Answer 1

Answer: (C)

$ f(x)=\frac{\pi}{2}+\pi(|x|-2) x \in[-1,1]$
Equation $f ( x )= k \pi \Rightarrow \frac{1}{2}+| x |-2= k$
$\text { or } | x |= k +\frac{3}{2} \in[0,1]$
$\Rightarrow 0 \leq k +\frac{3}{2} \leq 1 \Rightarrow-\frac{3}{2} \leq k \leq-\frac{1}{2}$
$\Rightarrow a =\frac{-3}{2} \text { and } b =\frac{-1}{2} $
$\therefore 2| a |+4| b |=5$