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UPSC NDA II 2023 Mathematics Solved Paper

Question Numbers: 53-54

Consider the following for the next items that follow:

Let

f(x)=\frac{a^{x-1}+b^{x-1}}{2}

and g(x) = x - 1.

Question : 54 of 120

Marks: +1, -0

What is

\underset{x→1}{lim}f(x)^{\frac{1}{g(x)}}

equal to?

Solution:

Concept:

Use the limit of

f(x)^{\frac{1}{g(x)}}

x\to1

by rewriting it as

e^{\lim_{x\to1}\frac{f(x)-1}{g(x)}}

, applying L'Hopital's rule to the exponent.

Explanation:

Given

f(x)=\frac{a^{x-1}+b^{x-1}}{2}

and

g(x)=x-1

We need

\displaystyle L=\lim_{x\to1} f(x)^{\frac{1}{g(x)}}

Assume

\displaystyle L=e^{\lim_{x\to1}\frac{f(x)-1}{g(x)}}

Let

\displaystyle M=\lim_{x\to1}\frac{f(x)-1}{g(x)}

Substitute

f(x)

and

g(x)

\displaystyle M=\lim_{x\to1}\frac{\frac{a^{x-1}+b^{x-1}}{2}-1}{x-1}

Set

y=x-1

, so

y\to0

. Then:

\displaystyle M=\lim_{y\to0}\frac{\frac{a^{y}+b^{y}}{2}-1}{y}=\lim_{y\to0}\frac{a^{y}+b^{y}-2}{2y}

Split the limit:

\displaystyle M=\frac12\lim_{y\to0}\frac{(a^{y}-1)+(b^{y}-1)}{y}=\frac12\left(\lim_{y\to0}\frac{a^{y}-1}{y}+\lim_{y\to0}\frac{b^{y}-1}{y}\right)

Using standard limits:

\displaystyle\lim_{y\to0}\frac{a^{y}-1}{y}=\ln a

and

\displaystyle\lim_{y\to0}\frac{b^{y}-1}{y}=\ln b

Thus

M=\frac12(\ln a+\ln b)=\frac12\ln(ab)=\ln\sqrt{ab}

Therefore

L=e^{\ln\sqrt{ab}}=\sqrt{ab}

Answer:

\sqrt{ab}

(Option A).

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