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UPSC NDA 2 2017 Mathematics Solved Paper

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Question : 95 of 120

Marks: +1, -0

If

{\rm{f}}\left( {\rm{x}} \right) = \frac{{4{\rm{x}} + {{\rm{x}}^4}}}{{1 + 4{{\rm{x}}^3}}}

{\rm{g}}\left( {\rm{x}} \right) = {\rm{In\;}}\left( {\frac{{1 + {\rm{x}}}}{{1 - {\rm{x}}}}} \right)

, then what is the value of

{\rm{f\;o\;g\;}}\left( {\frac{{{\rm{e}} - 1}}{{{\rm{e}} + 1}}} \right)

equal to?

2
1
0
$\frac{1}{2}$

Solution:

Concept:

Composite function: evaluate inner function

g

first, then apply

f

.

Explanation:

Given

f(x)=\frac{4x+x^4}{1+4x^3}

and

g(x)=\ln\left(\frac{1+x}{1-x}\right)

.

We need

f\circ g\left(\frac{e-1}{e+1}\right)

.

First, compute

g\left(\frac{e-1}{e+1}\right)

.

g\left(\frac{e-1}{e+1}\right)=\ln\left(\frac{1+\frac{e-1}{e+1}}{1-\frac{e-1}{e+1}}\right)

.

Simplify the numerator:

1+\frac{e-1}{e+1}=\frac{(e+1)+(e-1)}{e+1}=\frac{2e}{e+1}

.

Simplify the denominator:

1-\frac{e-1}{e+1}=\frac{(e+1)-(e-1)}{e+1}=\frac{2}{e+1}

.

The fraction inside log becomes:

\frac{\frac{2e}{e+1}}{\frac{2}{e+1}}=e

.

Thus

g\left(\frac{e-1}{e+1}\right)=\ln(e)=1

.

Now compute

f(1)

:

f(1)=\frac{4\cdot1+1^4}{1+4\cdot1^3}=\frac{4+1}{1+4}=\frac{5}{5}=1

.

Therefore,

f\circ g\left(\frac{e-1}{e+1}\right)=1

.

Answer:

1

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