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NDA 2015
Exam
NDA
Year
2015
Subject
Mathematics
Chapter
Limits

NDA 2015 · Mathematics · Limits

If G(x)= 25-x^2 , then what is _ x arrow 1 (G(x)-G(1))/(x-1) equal...

If G(x)=25−x2G(x)=\sqrt{25-x^2}, then what is lim⁡x→1G(x)−G(1)x−1\lim _{x \rightarrow 1} \frac{G(x)-G(1)}{x-1} equal to?

Options

  1. A−126-\frac{1}{2 \sqrt{6}}
  2. B15\frac{1}{5}
  3. C−16-\frac{1}{\sqrt{6}}
  4. D16\frac{1}{\sqrt{6}}

Correct answer

A. −126-\frac{1}{2 \sqrt{6}}

Step-by-step solution

G(x)=25−x2\mathrm{G}(\mathrm{x})=\sqrt{25-\mathrm{x}^2} Now, lim⁡x→1G(x)−G(1)x−1lim⁡x→125−x2−24x−1(÷\lim _{x \rightarrow 1} \frac{G(x)-G(1)}{x-1}\lim _{x \rightarrow 1} \frac{\sqrt{25-x^2}-\sqrt{24}}{x-1} \quad(\div form )) Applying L'Hospital rule, lim⁡x→1(−2x)225−x2=−2225−x2=−2224=−124=−126\begin{aligned} & \lim _{x \rightarrow 1} \frac{(-2 x)}{2 \sqrt{25-x^2}}=\frac{-2}{2 \sqrt{25-x^2}} \\ & =\frac{-2}{2 \sqrt{24}}=\frac{-1}{\sqrt{24}}=\frac{-1}{2 \sqrt{6}} \end{aligned}\therefore \quad$ Option (A) is correct.
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