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UPSC NDA I 2024 Mathematics Paper

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Question Numbers: 99-100

Consider the following for the items that follow:

Let

3 f(x)+f\left(\frac{1}{x}\right)=\frac{1}{x}+1

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Question : 100 of 120

Marks: +1, -0

What is

8 \int_1^2 f(x) d x

equal to ?

ln(8√e)
ln(4√e)
ln 2
ln 2 - 1

Solution:

Concept:

Solve a functional equation to find

f(x)

and then evaluate a definite integral.

Explanation:

Given

3f(x)+f\!\left(\frac1x\right)=\frac1x+1

.

Replace

x

by

\frac1x

:

3f\!\left(\frac1x\right)+f(x)=x+1

.

Let

a=f(x)

and

b=f\!\left(\frac1x\right)

. Then we have the system:

3a+b=\frac1x+1

and

3b+a=x+1

.

Multiply the first equation by

3

:

9a+3b=\frac3x+3

.

Subtract the second equation:

(9a+3b)-(3b+a)=\left(\frac3x+3\right)-(x+1)

.

This simplifies to

8a=\frac3x-x+2

, so

a=\frac18\left(\frac3x-x+2\right)

.

Thus

f(x)=\frac{3}{8x}-\frac{x}{8}+\frac14

.

Now compute

8\int_1^2 f(x)\,dx

:

=8\int_1^2\left(\frac{3}{8x}-\frac{x}{8}+\frac14\right)dx

=8\left[\frac38\ln x-\frac{x^2}{16}+\frac{x}{4}\right]_1^2

=\left[3\ln x-\frac{x^2}{2}+2x\right]_1^2

Evaluate:

(3\ln2-2+4)-(3\ln1-\frac12+2)

.

Since

\ln1=0

, we get

3\ln2+2-\frac32=3\ln2+\frac12

.

Write

\frac12=\ln\sqrt{e}

, so

3\ln2+\ln\sqrt{e}=\ln(2^3)+\ln\sqrt{e}=\ln(8\sqrt{e})

.

Answer:

\ln(8\sqrt{e})

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