Question

Let $f\left(x\right)=co{s}^{-1}\left[\frac{1}{\sqrt{13}}\right(2cosx-3sinx\left)\right]$. Then $f^{\prime }\left(0.5\right)=$____

• A. $0.5$
• B. $1$
• C. $0$
• D. $-1$

Answer 1

Answer: (B)

$1$

Consider the following
$\frac{2}{\sqrt{13}}cosx-\frac{3}{\sqrt{13}}sinx$
$=cos\theta cosx-sin\theta .sinx$ ....where $\theta =co{s}^{-1}\left(\frac{2}{\sqrt{13}}\right)=si{n}^{-1}\left(\frac{3}{\sqrt{13}}\right)$
$=cos(\theta +x)$
Hence
$co{s}^{-1}[\frac{2}{\sqrt{13}}cosx-\frac{3}{\sqrt{13}}sinx]$
$=co{s}^{-1}\left[cos\right(\theta +x\left)\right]$
$=\theta +x$
$=co{s}^{-1}\left(\frac{2}{\sqrt{13}}\right)+x$
Hence
$f\left(x\right)=co{s}^{-1}\left(\frac{2}{\sqrt{13}}\right)+x$
$f^{\prime }\left(x\right)=1$
Hence $f^{\prime }\left(x\right)$ is a constant function.
Therefore
$f^{\prime }\left(0.5\right)=1$.