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

4

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8

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π^{2} + 4 π + 1

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2 π

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Solution

The correct option is A $4$
$f (x) = {sin}^{- 1} x + 2 {tan}^{- 1} x + x^{2} + 4 x + 1$
Domain of $f (x)$ is $[- 1, 1]$
$f^{'} (x) = (\frac{1}{\sqrt{1 - x^{2}}}) + (\frac{2}{1 + x^{2}}) + (2 x + 4)$
$f^{'} (x) > 0$ in the given domain.
Hence $f (x)$ is an increasing function in the domain.
$∴ f (x)_{m i n .} = f (- 1) = - \frac{π}{2} + 2 (\frac{- π}{4}) + 1 - 4 + 1$ $= - π - 2$
$∴ f (x)_{m a x .} = f (1) = \frac{π}{2} + 2. \frac{π}{4} + 1 + 4 + 1$ $= π + 6$
$∴$ Range of $f (x)$ is $[- π - 2, π + 6]$
$\Rightarrow p + q = 4$

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