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

$f (x) = {sin}^{- 1} x + 2 {tan}^{- 1} x + (x + 2)^{2} - 3$
Domain of $f (x)$ is $[- 1, 1] .$
Also $f (x)$ is an increasing function in the domain. Therefore,
$p = f_{m i n} (x)$
$= f (- 1) = - \frac{π}{2} + 2 (\frac{- π}{4}) + 1 - 3 = - π - 2$
and $q = f_{m a x} (x)$
$= f (1) = \frac{π}{2} + 2 (\frac{π}{4}) + 9 - 3 = π + 6$
Therefore, the range of $f (x)$ is $[- π - 2, π + 6]$
Hence, $(p + q) = 4$

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