Q66 / 150 · MHT CET 2023 (12 May Shift 2) · Mathematics

Given $\mathrm{f}(x)=\left\{\begin{array}{cc}\frac{1-\cos 4 x}{x^2} & , \text { if } x < 0 \\ \mathrm{a} & , \text { if } x=0 \\ \frac{\sqrt{x}}{\sqrt{16-\sqrt{x}-4}}, & \text { if } x>0\end{array}\right.$ If $\mathrm{f}(x)$ is continuous at $x=0$, then value of $\mathrm{a}$ is

  • A
    $-8$
  • B
    $2$
  • C
    $-2$
  • Correct answer
    D
    $8$

Solution

As $\mathrm{f}(x)$ is continuous at $x=0$, we get $$ \begin{array}{ll} & \lim _{x \rightarrow 0^{-}} \mathrm{f}(x)=\mathrm{a}=\lim _{x \rightarrow 0^{+}} \mathrm{f}(x) \\ \therefore \quad & \lim _{x \rightarrow 0^{-}} \mathrm{f}(x)=\mathrm{a} \\ \therefore \quad & \lim _{x \rightarrow 0} \frac{1-\cos 4 x}{x^2}=\mathrm{a} \\ \therefore \quad & \lim _{x \rightarrow 0} 4 \frac{2 \sin ^2 2 x}{(2 x)^2}=\mathrm{a} \\ \therefore \quad & 8=\mathrm{a} \end{array} $$

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

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