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UPSC NDA 2 2016 Mathematics Solved Paper

Question Numbers: 78-80

Let

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

Question : 80 of 120

Marks: +1, -0

1. f(-2) = f(5)

f^{\prime\prime}(-2) + f^{\prime\prime}(0.5) + f^{\prime\prime}(3) = 4

Select the correct answer using the code given below:

Solution:

Concept:

f(x) = (|x| - |x-1|)^2

is a piecewise function, and we evaluate functional values and second derivatives in different intervals to check the given statements.

Explanation:

For

x > 1

|x| = x

|x-1| = x-1

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

. Thus

f(5)=1

For

x < 0

|x| = -x

|x-1| = -(x-1)

f(x) = (-x - (-(x-1)))^2 = (-x + x - 1)^2 = 1

. Thus

f(-2)=1

Hence

f(-2) = f(5)

, so statement 1 is correct.

Now for derivatives:

For

x > 1

f(x)=1 \Rightarrow f'(x)=0 \Rightarrow f''(x)=0

, so

f''(3)=0

For

x < 0

f(x)=1 \Rightarrow f''(x)=0

, so

f''(-2)=0

For

0 < x < 1

f(x) = (x - (1-x))^2 = (2x-1)^2 = 4x^2 -4x +1

Differentiate:

f'(x) = 8x -4

f''(x) = 8

. So

f''(0.5)=8

Thus

f''(-2)+f''(0.5)+f''(3)=0+8+0=8 \neq 4

. Statement 2 is false.

Answer:

Option A. 1 only.

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