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UPSC NDA I 2016 Mathematics Solved Paper

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Question Numbers: 61-64

Let

f: \mathbb{R} \to \mathbb{R}

be a function such that

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

for

x \in \mathbb{R}

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Question : 62 of 120

Marks: +1, -0

What is f’(1) equal to?

$-6$
$-5$
1
0

Solution:

Concept:

The derivative of a constant is

0

and the power rule states

\frac{d}{dx}x^n = nx^{n-1}

.

Explanation:

Given

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

.

Differentiate with respect to

x

:

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

. (Equation 1)

Differentiate again:

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

. (Equation 2)

Differentiate again:

f'''(x) = 6

. (Equation 3)

Put

x = 1

in equation 1:

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

. (Equation A)

Put

x = 2

in equation 2:

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

. (Equation B)

Substitute equation B into equation A:

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

.

Simplify:

f'(1) = 15 + 4f'(1)

.

Rearrange:

f'(1) - 4f'(1) = 15

→

-3f'(1) = 15

.

Thus

f'(1) = -5

.

Answer:

-5

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