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UPSC NDA 2 2016 Mathematics Solved Paper

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Consider the following for the next three (03) items that follow:

Let

f(x)=[|x|-|x-1|]^{2}

[NDA II 2016]

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Question : 79 of 120

Marks: +1, -0

What is

f'(x)

0 < x < 1

0
$2x-1$
$4x-2$
$8x-4$

Solution:

Given

f(x)=(|x|-|x+1|)^{2}

f(x)=\begin{cases}1, & x \le 0 \\ (2x-1)^{2}, & 0 < x < 1 \\ 1, & x \ge 1 \end{cases}

When

0 < x < 1

f(x)=(2x-1)^{2}

f'(x)=2(2x-1)\cdot2=4(2x-1)

f'(x)=8x-4

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