If {{f}}≤ft( {{x}} ) = {25 - {{{x}}²} {{\;}},{{\;}} then what is {{{Lim}}}_{{{x}} → 1} {{{{f}}≤ft( {{x}} ) - {{f}}≤ft( 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 2015 Mathematics Paper

Question : 104 of 120

Marks: +1, -0

{\rm{f}}\left( {\rm{x}} \right) = \sqrt {25 - {{\rm{x}}^2}} {\rm{\;}},{\rm{\;}}

\mathop {{\rm{Lim}}}\limits_{{\rm{x}} \to 1} \frac{{{\rm{f}}\left( {\rm{x}} \right) - {\rm{f}}\left( 1 \right)}}{{{\rm{x}} - 1}}

equal to?

Solution:

Concept:

The given limit is in the

0/0

form and represents the derivative of

f(x)

x=1

Explanation:

We have

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

The limit

\displaystyle \lim_{x \to 1} \frac{f(x) - f(1)}{x - 1}

is exactly

f'(1)

by definition.

So we first differentiate

f(x)

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

Now substitute

x = 1

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

Therefore the required limit is

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

Answer:

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

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