UPSC NDA II 2015 Mathematics Paper

Question Numbers: 74-75

For the next two (02) items that follow:

Consider

{\rm{f'}}\left( {\rm{x}} \right) = \frac{{{{\rm{x}}^2}}}{2} - {\rm{kx}} + 1

such that f(0) = 0 and f(3) = 15

Question : 74 of 120

Marks: +1, -0

The value of k is

Solution:

Concept:

Integration of a derivative gives back the original function plus a constant of integration:

\int f'(x)\,dx = f(x) + C

Explanation:

Given

f'(x) = \frac{x^2}{2} - kx + 1

Integrate both sides:

\int f'(x)\,dx = \int \left( \frac{x^2}{2} - kx + 1 \right) dx

This yields

f(x) = \frac{x^3}{6} - \frac{kx^2}{2} + x + C

Apply the condition

f(0) = 0

\frac{0^3}{6} - \frac{k \cdot 0^2}{2} + 0 + C = 0

, so

C = 0

Now use

f(3) = 15

\frac{3^3}{6} - \frac{k \cdot 3^2}{2} + 3 = 15

Simplify:

\frac{27}{6} - \frac{9k}{2} + 3 = 15

\frac{9}{2} - \frac{9k}{2} + 3 = 15

Combine constants:

\frac{9}{2} + 3 = \frac{9}{2} + \frac{6}{2} = \frac{15}{2}

Thus

\frac{15}{2} - \frac{9k}{2} = 15

Multiply both sides by 2:

15 - 9k = 30

-9k = 15

, hence

k = -\frac{15}{9} = -\frac{5}{3}

Answer:

k = -\frac{5}{3}

(Option C).

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