Q66 / 180 · KCET 2012 · Mathematics

If $f(x)=\left|\begin{array}{ccc}\sin x & \cos x & \tan x \\ x^{3} & x^{2} & x \\ 2 x & 1 & x\end{array}\right|$, then $\lim _{n -> \infty} \frac{f(x)}{x^{2}}$ is equal to

  • A
    0
  • B
    3
  • C
    2
  • Correct answer
    D
    1

Solution

Given, $\mathrm{f}(\mathrm{x})=\left|\begin{array}{ccc}\sin \mathrm{x} & \cos x & \tan \mathrm{x} \\ \mathrm{x}^{3} & \mathrm{x}^{2} & \mathrm{x} \\ 2 \mathrm{x} & 1 & \mathrm{x}\end{array}\right|$ Expanding the determinant along the first row, we get $$ \begin{aligned} &=\sin x\left(x^{3}-x\right)-\cos x\left(x^{4}-2 x^{2}\right) \\ &=\sin x\left(x^{3}-x\right)-\cos x \cdot x^{2}\left(x^{2}-2\right) \\ &=\sin x\left(x^{3}-2 x^{3}\right) \\ &\left.\therefore \quad \lim _{x \rightarrow 0} \frac{f(x)}{x^{2}}\right) \tan x\left(-x^{3}\right) \\ &=\lim _{x \rightarrow 0} \frac{\left[x \sin x\left(x^{2}-1\right)-\cos x \cdot x^{2}\left(x^{2}-2\right)\right.}{x^{2}} \cdot x^{2}\left(x^{2}-2\right)-\tan x\left(x^{3}\right) \\ &=\lim _{x \rightarrow 0}\left\{\frac{\sin x}{x}\left(x^{2}-1\right)-\cos x\left(x^{2}-2\right)-\tan x \cdot x\right\} \\ &=[1(0-1)-\cos 0(0-2)-\tan 0 \cdot 0]-[-1+2] \\ &=1 \end{aligned} $$

Hence, the correct answer is option: 'D'.

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