Test Index

UPSC NDA II 2025 Mathematics Paper

Question Numbers: 78-80

For the following three (03) items:

Let

y=f(x)=\frac{xsin^{-1}x}{\sqrt{1-x^{2}}}+ln\sqrt{1-x^{2}}.

Question : 78 of 120

Marks: +1, -0

y = f(x) at x

= 0.5

Solution:

Concept:

The slope of the tangent at a given point is the value of the derivative at that point. We first find

f'(x)

and then evaluate it at

x=0.5

Explanation:

We are given

f(x)=x\cdot\frac{\sin^{-1}x}{\sqrt{1-x^{2}}}+\ln\sqrt{1-x^{2}}

Differentiate term by term using product rule and chain rule.

The derivative simplifies to

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

Now substitute

x=0.5

\sin^{-1}(0.5)=\frac{\pi}{6}

1-x^{2}=0.75=\frac{3}{4}

\sqrt{1-x^{2}}=\frac{\sqrt{3}}{2}

Thus

f'(0.5)=\frac{\pi/6}{\frac{3}{4}\cdot\frac{\sqrt{3}}{2}}=\frac{\pi}{6}\cdot\frac{8}{3\sqrt{3}}=\frac{4\pi\sqrt{3}}{27}

Answer:

The slope is

\frac{4\pi\sqrt{3}}{27}

, which corresponds to option A.

Go to Question:

Prev Question

Next Question