If f(x)={25-x²,} then what is {x→1}{lim}f(x)-f(1)/x-1 equal to?

Correct Answer is : −124-{1}{{24}}−241

Correct Answer is : −124-{1}{{24}}−241

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UPSC NDA II 2018 Mathematics Solved Paper

Question : 69 of 120

Marks: +1, -0

f(x)=\sqrt{25-x^{2},}

\underset{x→1}{lim}\frac{f(x)-f(1)}{x-1}

equal to?

Solution:

Concept:

The given limit represents the derivative of

f(x)

x = 1

, i.e.,

\underset{x \to 1}{\lim} \frac{f(x) - f(1)}{x - 1} = f'(1)

Explanation:

We have

f(x) = \sqrt{25 - x^2}

Differentiate

f(x)

using the chain rule:

f'(x) = \frac{1}{2\sqrt{25 - x^2}} \cdot ( -2x ) = \frac{-x}{\sqrt{25 - x^2}}

Now evaluate at

x = 1

f'(1) = \frac{-1}{\sqrt{25 - 1^2}} = \frac{-1}{\sqrt{24}}

Thus, the limit equals

f'(1) = -\frac{1}{\sqrt{24}}

Answer:

-\frac{1}{\sqrt{24}}

(Option A)

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