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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 : 80 of 120

Marks: +1, -0

Which of the following equations is/are correct?

f(-2)=f(5)

f''(-2)+f''(0.5)+f''(3)=4

Select the correct answer using the code given below:

1 only
2 only
Both 1 and 2
Neither 1 nor 2

Solution:

Given

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

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

For

x=-2

f(x)=1

so

f(-2)=1

For

x=5

f(x)=1 \Rightarrow f(5)=1

Hence

f(-2)=f(5)

Now, for

x=-1

f''(x)=0

f''(-2)=0

For

x=0.5

f''(x)=8 \Rightarrow f''(0.5)=8

For

x=3

f''(x)=0 \Rightarrow f''(3)=0

\Rightarrow f''(-2)+f''(0.5)+f''(3)=8 \neq 4

Only statement 1 is correct.

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