Let `f(x)=|[cosx , x , x] , [2sinx , x , 2x] , [sinx , x , x]|,` then `lim_(x->0)(f(x))/(x^2)` is equal to (a)` 0 `(b) `-1` (c) 2 (d) 3

Explanation

$$f(x)=\left|\begin{array}{ccc}\cos x & x & 1 \\ 2 \sin x & x & 2 x \\ \sin x & x & x\end{array}\right|$$
$$=-x[x \sin x-\sin x-x \sin x+x \cos x]$$
$$=-x(x \cos x-\sin x)$$
$$\therefore \lim _{x \rightarrow 0} \frac{f(x)}{x^{2}}=\lim _{x \rightarrow 0} \frac{x(\sin x-x \cos x)}{x^{2}}$$
$$=\lim _{x \rightarrow 0} \frac{\sin x}{x}-\lim _{x \rightarrow 0} \cos x$$
$$=1-1=0$$