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KCET 2026 Mathematics Solved Paper

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Question : 34 of 60

Marks: +1, -0

If

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

f^{\prime}\left(\frac{1}{2}\right)=

$\frac{8}{5}$
$\frac{5}{8}$
$\frac{4}{5}$
0

Solution:

Concept:

Differentiate the inverse sine function using the chain rule and then evaluate at the given point.

Explanation:

Let

u = \frac{2x}{1+x^2}

. Then

f(x) = \sin^{-1} u

.

Using the chain rule,

f'(x) = \frac{u'}{\sqrt{1-u^2}}

.

First, differentiate

u

with respect to

x

:

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

u' = \frac{2+2x^2 - 4x^2}{(1+x^2)^2} = \frac{2(1-x^2)}{(1+x^2)^2}

.

Now compute

1-u^2

:

1-u^2 = 1 - \left(\frac{2x}{1+x^2}\right)^2

1-u^2 = \frac{(1+x^2)^2 - 4x^2}{(1+x^2)^2} = \frac{(1-x^2)^2}{(1+x^2)^2}

.

Thus,

\sqrt{1-u^2} = \frac{|1-x^2|}{1+x^2}

.

At

x = \frac{1}{2}

, we have

1-x^2 > 0

, so

\sqrt{1-u^2} = \frac{1-x^2}{1+x^2}

.

Substitute into the derivative formula:

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

.

Evaluate at

x = \frac{1}{2}

:

f'\left(\frac{1}{2}\right) = \frac{2}{1+\left(\frac{1}{2}\right)^2} = \frac{2}{1+\frac{1}{4}} = \frac{2}{\frac{5}{4}} = \frac{8}{5}

.

Answer:

\frac{8}{5}

, which is option A.

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