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UPSC NDA I 2016 Mathematics Solved Paper

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For the next three (3) items that follow :

Let

f:\mathbb{R}\rightarrow\mathbb{R}

be function such that

f(x)=x^3+x^2f'(1)+xf''(2)+f'''(3)

for

x\in\mathbb{R}

[NDA I 2016]

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Question : 61 of 120

Marks: +1, -0

What is f(1) equal to ?

$-2$
$-1$
0
4

Solution:

f(x)=x^{3}+x^{2}f'(1)+xf''(2)+f'''(3)

...(1)

f'(x)=3x^{2}+2xf'(1)+f'(2)

......(2)

f''(x)=6x+2f'(1)

......(3)

f'''(x)=6

....(4)

f'(1)=3+2f'(1)+f''(2)

.....(5)

f'(2)=12+2f'(1)

......(6)

Using (6) and (5),we get

f'(1)=3+2f'(1)+12+2f''(1)

-3f'(1)=15

f'(1)=-5

Using this value in equation (6) we get

f'(2)=12+2\times(-5)

f''(2)=2

Using

x=3

in eqn (4),

f'''(3)=6

Putting value of

f'(1)+f''(2)

and

f'''(3)

in eq (1)

we get

f(x)=x^{3}+x^{2}(-5)+x(2)+6

=x^{3}-5x^{2}+2x+6

Putting

x=1

f(1)=(1)^{3}-5(1)^{2}+2(1)+6

f(1)-4

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