Question

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

• A. $1$
• B. $-1$
• C. $2$
• D. $-2$

Answer 1

The correct option is D $-2$

$f\left(x\right)=\left|\begin{array}{ccc}cosx& x& 1\\ 2sinx& x^2& 2x\\ tanx& x& 1\end{array}\right|$
$R_1\to R_1-R_3$
$f\left(x\right)=\left|\begin{array}{ccc}cosx-tanx& 0& 0\\ 2sinx& x^2& 2x\\ tanx& x& 1\end{array}\right|$
$\Rightarrow f\left(x\right)=(cosx-tanx)(-x^2)$
$\Rightarrow f\left(x\right)=-x^2\cos x+x^2\tan x$
$f^{\prime }\left(x\right)=x^2\sin x-2x\cos x+x{\sec }^{2}x+2\tan x$
$\Rightarrow \underset{x\to 0}{lim}\frac{f^{\prime }\left(x\right)}{x}=\underset{x\to 0}{lim}x\sin x-2\cos x+x{\sec }^{2}x+2\tan x$
$=-2$