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UPSC NDA I 2018 Mathematics Solved Paper

Question : 109 of 120

Marks: +1, -0

f(x)=\begin{cases} x^{2}ln | x | & x≠0 \\ 0 & x=0 \end{cases}

What is f’(0) equal to?

Solution:

Concept:

The derivative at a point for a piecewise function is computed using the limit definition of the derivative.

Explanation:

We have

f(x) = x^2 \ln |x|

for

x \neq 0

and

f(0)=0

By definition,

f'(0) = \lim_{x \to 0} \frac{f(x)-f(0)}{x} = \lim_{x \to 0} \frac{x^2 \ln |x|}{x} = \lim_{x \to 0} x \ln |x|

This limit is of the form

0 \cdot (-\infty)

Rewrite as

\lim_{x \to 0} \frac{\ln |x|}{1/x}

, which is of the form

\frac{-\infty}{\infty}

Apply L'Hôpital's rule: differentiate numerator and denominator:

\frac{d}{dx}(\ln |x|) = \frac{1}{x}

\frac{d}{dx}(1/x) = -\frac{1}{x^2}

Thus

\lim_{x \to 0} \frac{1/x}{-1/x^2} = \lim_{x \to 0} (-x) = 0

Therefore,

f'(0) = 0

Answer:

0

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