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

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Solution

Now the domain of $s i n^{- 1} (x)$ is $[- 1, 1]$
Hence the domain of the above function is $[- 1, 1]$ since the domain of $t a n (x)$ is $(- \infty, \infty)$ and the rest is a linear function.
Hence
Domain is $[- 1, 1]$
Now
$f (1) = \frac{π}{2} + 2 \frac{π}{4} + 1 + 4 + 1$
$= π + 6$ .
$f (- 1) = \frac{- π}{2} - 2 \frac{π}{4} + 1 - 4 + 1$
$= - π - 2$
Hence
$p + q$
$= π + 6 + (- π - 2)$
$= 6 - 2$
$= 4$

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