Question

Let $f(x)={\cos }^{-1}\left[\frac{1}{\sqrt{13}}(2\cos x-3\sin x)\right]$ Then, f'(0.5) is equal to

• A. 0.5
• B. 1
• C. 0
• D. -1

Answer 1

Answer: (B)

Given, $f(x)={\cos }^{-1}\left\{\frac{1}{\sqrt{13}}(2\cos x-3\sin x)\right\}$
$\Rightarrow f(x)={\cos }^{-1}\left\{\frac{2}{\sqrt{13}}\cos x-\frac{3}{\sqrt{13}}\sin x\right\}$

$\Rightarrow f(x)={\cos }^{-1}\{\cos \alpha \cdot \cos x-\sin \alpha \cdot \sin x\}$
$={\cos }^{-1}\{\cos (x+\alpha )\}$
$=x+\alpha$
On differentiating w.r.t. $x$, we get
$f^{\prime }(x)=1=$ constant value
$\therefore f^{\prime }(0.5)=1$