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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class12  >>  Determinants

If $f(x)=\begin{vmatrix}\sin x&\cos x&\tan x\\x^3&x^2&x\\2x&1&1\end{vmatrix}$ then $\lim\limits_{x\to 0}\large\frac{f(x)}{x^2}$ is

$\begin{array}{1 1}(A)\;3&(B)\;-1\\(C)\;0&(D)\;1\end{array}$

1 Answer

Given :
$f(x)=\begin{vmatrix}\sin x&\cos x&\tan x\\x^3&x^2&x\\2x&1&1\end{vmatrix}$
On expanding we get
$\sin x(x^2-x)-\cos x(x^3-2x^2)+\tan x(x^3-2x^3)$
$x(x-1)\sin x-(x^3-2x^2)\cos x-x^3\tan x$
$x^2\sin x-x^3\cos x-x^3\tan x+2x^2\cos x-x\sin x$
Hence $\lim\limits_{x\to 0}\large\frac{f(x)}{x^2}$
$\Rightarrow \lim\limits_{x\to 0}(\sin x-x\cos x-x\tan x+2\cos x-\large\frac{\sin x}{x})$
$\Rightarrow 0-0-0+2\times 1-1$
$\Rightarrow 2-1=1$
Hence (D) is the correct answer.
answered Apr 23, 2014 by sreemathi.v
 

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