Question

If range of the function $f(x)={\sin }^{-1}x+2{\tan }^{-1}x+x^2+4x+1$ is [p, q] then find the value of $(p+q)$

Answer 1

Answer: 0004

We have $f(x)={\sin }^{-1}x+2{\tan }^{-1}x+x^2+4x+1$
Clearly domain of $f(x)$ is $[-1,1]$.
Also $f(x)$ is increasing function in its domain.
$\therefore p=f_{\min .}(x)=f(-1)=-\frac{\pi }{2}+2\left(\frac{-\pi }{4}\right)+1-4+1=-\pi -2.$
$q=f_{\max .}(x)=f(1)=\frac{\pi }{2}+2\left(\frac{\pi }{4}\right)+1+4+1=\pi +6$
$\therefore \text{ Range of }f(x)\text{ is }[-\pi -2,\pi +6]$
$\text{ Hence }(p+q)=4$
$\text{ Note }:\text{ Vertex of }y=x^2+4x+1\text{ is at }x=-2\text{ and hence in the domain }\left(x^2+4x+1\right)\text{ is increasing.] }$