Question

$f\left(x\right)=\left|\begin{array}{ccc}cosx& x& 1\\ 2sinx& x^2& 2x\\ tanx& x& 1\end{array}\right|$ then $\underset{x⟶0}{lim}\frac{f^{{}^{\prime }}\left(x\right)}{x}=$

• A. 1
• B. -1
• C. 2
• D. -2

Answer 1

Answer: (D)

-2

We have,

$f\left(x\right)=\left|\begin{array}{ccc}cosx& x& 1\\ 2sinx& x^2& 2x\\ tanx& x& 1\end{array}\right|$

Expansion from Row (1)

$f\left(x\right)=\cos x(x^2-2x^2)-x(2\sin x-2x\tan x)+1(2x\sin x-x^2\tan x)$

$\Rightarrow f\left(x\right)=-x^2\cos x-2x\sin x+2x^2\tan x+2x\sin x-x^2\tan x$

$\Rightarrow f\left(x\right)=-x^2\cos x+x^2\tan x$

$\Rightarrow f\left(x\right)=x^2\tan x-x^2\cos x$

$\Rightarrow f\left(x\right)=x^2(\tan x-\cos x)$

On differentiating and we get,

$f^{\prime }\left(x\right)=x^2({\sec }^{2}x+\sin x)+(\tan x-\cos x)2x$

$f^{\prime }\left(x\right)=x[x({\sec }^{2}x+\sin x)+(\tan x-\cos x)2]$

Then,

${lim}_{x\to 0}\frac{f^{\prime }\left(x\right)}{x}$

$={lim}_{x\to 0}\frac{x[x({\sec }^{2}x+\sin x)+(\tan x-\cos x)2]}{x}$

$={lim}_{x\to 0}[x({\sec }^{2}x+\sin x)+(\tan x-\cos x)2]$

Taking limit and we get,

$-2$

Hence, this is the answer.