The number of integral value(s) of $k$ for which ${sin}^{- 1} (x^{2} + 4 x + 5) + {tan}^{- 1} (x^{2} - 2 x + 1 - k^{2}) > \frac{π}{2}$ is

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Solution

Given: ${sin}^{- 1} (x^{2} + 4 x + 5) + {tan}^{- 1} (x^{2} - 2 x + 1 - k^{2}) > \frac{π}{2}$
Now, $x^{2} + 4 x + 5 = (x + 2)^{2} + 1$
${sin}^{- 1} ((x + 2)^{2} + 1)$ is defined only when $(x + 2)^{2} = 0 \Rightarrow - 2$
So, inequality becomes : ${sin}^{- 1} (1) + {tan}^{- 1} (4 + 4 + 1 - k^{2}) > \frac{π}{2}$
$\Rightarrow \frac{π}{2} + {tan}^{- 1} (9 - k^{2}) > \frac{π}{2}$
$\Rightarrow {tan}^{- 1} (9 - k^{2}) > 0$
$\Rightarrow 9 - k^{2} > 0 \Rightarrow - 3 < k < 3$
$∴$ The integral values are $- 2, - 1, 0, 1, 2$
So, number of intergral values of $k$ is $5.$

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