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NDA 2018
Exam
NDA
Year
2018
Subject
Mathematics
Chapter
Continuity and Differentiability

NDA 2018 · Mathematics · Continuity and Differentiability

If f(x)= 25-x^2 , then what is _ x arrow 1 f(x)-f(1) x-1 equal to

If f(x)=25−x2\mathrm{f}(\mathrm{x})=\sqrt{25-\mathrm{x}^2}, then what is lim⁡x→1f(x)−f(1)x−1\lim _{\mathrm{x} \rightarrow 1} \frac{\mathrm{f}(\mathrm{x})-\mathrm{f}(1)}{\mathrm{x}-1} equal to

Options

  1. A−124-\frac{1}{\sqrt{24}}
  2. B124\frac{1}{\sqrt{24}}
  3. C−143-\frac{1}{4 \sqrt{3}}
  4. D143\frac{1}{4 \sqrt{3}}

Correct answer

A. −124-\frac{1}{\sqrt{24}}

Step-by-step solution

lim⁡x→1f(x)−f(1)x−1=f′(x)\lim _{x \rightarrow 1} \frac{f(x)-f(1)}{x-1}=f^{\prime}(x) Now, f′(x)=−2x225−x2=−x25−x2f′(1)=−125−1=−124f^{\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}}
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