Question

If $f\left(x\right)=\left|\begin{array}{ccc}x& x^2& 2x\\ \sin x& \tan x& \sin 2x\\ \cos x& -x^3& 5x\end{array}\right|$ then $\underset{x\to 0}{lim}\frac{f^{\prime }\left(x\right)}{x}$ equals

• A. $2$
• B. $-3$
• C. $-2$
• D. $-4$

Answer 1

Answer: (B)

$-3$

$f\left(x\right))=\left|\begin{array}{ccc}x& x^2& 2x\\ \sin x& \tan x& \sin 2x\\ \cos x& -x^3& 5x\end{array}\right|$
$f^{\prime }\left(x\right)=\underset{\left(i\right)}{\left|\begin{array}{ccc}1& x^2& 2x\\ \cos x& \tan x& \sin 2x\\ -\sin x& -x^3& 5x\end{array}\right|}$+$\underset{\left(ii\right)}{\left|\begin{array}{ccc}x& 2x& 2x\\ \sin x& {\sec }^{2}x& \sin 2x\\ \cos x& -3x^2& 5x\end{array}\right|}$+$\underset{\left(iii\right)}{\left|\begin{array}{ccc}x& x^2& 2\\ \sin x& \tan x& 2\cos 2x\\ \cos x& -x^3& 5\end{array}\right|}$
Using $\underset{i}{\underset{}{C_2\to \frac{1}{x}{C}_2}}$, $\underset{ii}{\underset{}{C_3\to \frac{1}{x}{C}_3}}$, $\underset{iii}{\underset{}{C_2\to \frac{1}{x}{C}_2}}$
$\therefore \frac{f^{\prime }\left(x\right)}{x}=\left|\begin{array}{ccc}1& x& 2x\\ \cos x& \frac{\tan x}{x}& \sin 2x\\ -\sin x& -x^2& 5x\end{array}\right|$+$\left|\begin{array}{ccc}x& 2x& 2\\ \sin x& {\sec }^{2}x& \frac{\sin 2x}{x}\\ \cos x& -3x^2& 5\end{array}\right|$ +$\left|\begin{array}{ccc}x& x& \cos x\\ \sin x& \frac{\tan x}{x}& -x^2\\ \cos x& 2\cos 2x& 5\end{array}\right|$

$\therefore \underset{x\to 0}{lim}\frac{f^{\prime }\left(x\right)}{x}$ $=\left|\begin{array}{ccc}1& 0& 0\\ 1& 1& 0\\ 0& 0& 0\end{array}\right|+\left|\begin{array}{ccc}0& 0& 2\\ 0& 1& 2\\ 1& 0& 5\end{array}\right|+\left|\begin{array}{ccc}0& 0& 1\\ 0& 1& 0\\ 1& 2& 5\end{array}\right|$

$=0+1(-2)+1(-1)=-3$