Q226 / 270 · NDA 2018 (Phase 2) · Mathematics

If $\mathrm{f}(\mathrm{x})=\sqrt{25-\mathrm{x}^2}$, then what is $\lim _{\mathrm{x} \rightarrow 1} \frac{\mathrm{f}(\mathrm{x})-\mathrm{f}(1)}{\mathrm{x}-1}$ equal to

  • Correct answer
    A
    $-\frac{1}{\sqrt{24}}$
  • B
    $\frac{1}{\sqrt{24}}$
  • C
    $-\frac{1}{4 \sqrt{3}}$
  • D
    $\frac{1}{4 \sqrt{3}}$

Solution

$\lim _{x \rightarrow 1} \frac{f(x)-f(1)}{x-1}=f^{\prime}(x)$ Now, $f^{\prime}(x)=\frac{-2 x}{2 \sqrt{25-x^2}}=\frac{-x}{\sqrt{25-x^2}}$ $f^{\prime}(1)=\frac{-1}{\sqrt{25-1}}=\frac{-1}{\sqrt{24}}$

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

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