Question

If $f\left(x\right)={\cos }^{-1}\left(\frac{2\cos x+3\sin x}{\sqrt{13}}\right)$, then $\left[f^{\prime }\right(x{)}^2]$ is equal to

• A. $\sqrt{1+x}$
• B. $1+2x$
• C. $2$
• D. $1$
• E. $0$

Answer 1

Answer: (D)

$1$

Given, $f\left(x\right)={\cos }^{-1}\left\{\frac{2\cos x+3\sin x}{\sqrt{13}}\right\}$
$={\cos }^{-1}\{\frac{2}{\sqrt{13}}\cos x+\frac{3}{\sqrt{13}}\sin x\}$
$={\cos }^{-1}(\cos \alpha \cos x+\sin \alpha \cdot \sin x)$
$={\cos }^{-1}\cos (\alpha +x)=\alpha +x$
On. differentiating w.r.t. $x$, we get
$f^{\prime }\left(x\right)=1\Rightarrow \left[f^{\prime }\right(x){]}^2=1$