Question

If the equation ${\sin }^{-1}(x-4)+{\cos }^{-1}(x-6)+{\tan }^{-1}(\frac{40}{15+x^2})=m$ holds, then value of ${}^{\prime }{m}^{\prime }$ is

• A. $\frac{\pi }{4}$
• B. $\frac{3\pi }{4}$
• C. $\frac{7\pi }{4}$
• D. $2\pi$

Answer 1

The correct option is C $\frac{7\pi }{4}$

$-1\le x-4\le 1,-1\le x-6\le 13\le x\le 5\cap 5\le x\le 7$
so, $x=5$ is only point in domain.
Therefore $m={\sin }^{-1}\left(1\right)+{\cos }^{-1}(-1)+{\tan }^{-1}(\frac{40}{40})$
$\Rightarrow m=\frac{\pi }{2}+\pi +\frac{\pi }{4}=\frac{7\pi }{4}$