Question

If $f\left(x\right)=\left|\begin{array}{c}\cos x\\ 2\sin x\\ \tan x\end{array}\begin{array}{c}x\\ x^2\\ x\end{array}\begin{array}{c}1\\ 2x\\ 1\end{array}\right|$, then ${lim}_{x\to 0}\frac{f^{\prime }\left(x\right)}{x}$

• A. does not exist
• B. exists and is equal to $2$
• C. exists and is equal to $0$
• D. exists and is equal to $-2$

Answer 1

Answer: (D)

exists and is equal to $-2$

Now,

$\begin{array}{c}f\left(x\right)=\left|\begin{array}{ccc}\cos x& x& 1\\ 2\sin x& x^2& 2x\\ \tan x& x& 1\end{array}\right|=x\left|\begin{array}{ccc}\cos x& 1& 1\\ 2\sin x& x& 2x\\ \tan x& 1& 1\end{array}\right|\\ =x\left|\begin{array}{ccc}\cos x& 1& 1\\ 2\sin x& x& x\\ \tan x& 1& 1\end{array}\right|=x\times \left(-x\right)\left[\cos x-\tan x\right]\\ f\left(x\right)=x^2\left(\tan x-\cos x\right)\\ f^{\prime }\left(x\right)=2x\left(\tan x-\cos x\right)+x^2\left({\sec }^{2}x+\sin x\right)\\ {lim}_{x\to 0}\frac{f^{\prime }\left(x\right)}{x}=2{lim}_{x\to 0}\left(\tan x-\cos x\right)+f{lim}_{x\to 0}x\left({\sec }^{2}x+\sin x\right)\\ =2\times -1=-2\end{array}$

Option $D$ is correct answer.