UPSC 2025 Maths Optional Paper 1 Q2b — Step-by-Step Solution

15 marks · Section A

Mean-value theorems (Rolle, Lagrange, Cauchy) · Calculus · asked 3× in 13 yrs · Read the full method →

Question

Using Mean Value Theorem, prove that $\dfrac{\pi}{6} + \dfrac{\sqrt{3}}{15} < \sin^{-1}\left(\dfrac{3}{5}\right) < \dfrac{\pi}{6} + \dfrac{1}{8}$ .

Technique

Apply Lagrange’s Mean Value Theorem to $f(x)=\sin^{-1}x$ on $[1/2,\,3/5]$ , then bound the unknown intermediate value using the monotonicity of $f'$ .

Solution

Step 1 — Choose the function and interval.

Let $f(x)=\sin^{-1}x$ , continuous on $[1/2,3/5]$ and differentiable on $(1/2,3/5)$ , with

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

Note $f\!\left(\tfrac12\right)=\sin^{-1}\tfrac12=\dfrac{\pi}{6}$ , and the interval length is $\dfrac35-\dfrac12=\dfrac{1}{10}$ .

Step 2 — Apply MVT.

There exists $c\in\left(\tfrac12,\tfrac35\right)$ with

f\!\left(\tfrac35\right)-f\!\left(\tfrac12\right)=f'(c)\left(\tfrac35-\tfrac12\right),

i.e.

\sin^{-1}\!\left(\tfrac35\right)=\frac{\pi}{6}+\frac{1}{10}\cdot\frac{1}{\sqrt{1-c^2}}.\qquad(\star)

Step 3 — Bound $f'(c)$ by monotonicity.

For $0<x<1$ , $f'(x)=\dfrac{1}{\sqrt{1-x^2}}$ is strictly increasing. Since $\tfrac12<c<\tfrac35$ ,

f'\!\left(\tfrac12\right)<f'(c)<f'\!\left(\tfrac35\right).

Compute the endpoints:

f'\!\left(\tfrac12\right)=\frac{1}{\sqrt{1-\tfrac14}}=\frac{1}{\sqrt{3}/2}=\frac{2}{\sqrt3},\qquad f'\!\left(\tfrac35\right)=\frac{1}{\sqrt{1-\tfrac{9}{25}}}=\frac{1}{\sqrt{16/25}}=\frac{1}{4/5}=\frac54.

Hence

\frac{2}{\sqrt3}<\frac{1}{\sqrt{1-c^2}}<\frac54.

Step 4 — Insert into $(\star)$ .

Multiply through by $\tfrac{1}{10}$ and add $\tfrac{\pi}{6}$ :

\frac{\pi}{6}+\frac{1}{10}\cdot\frac{2}{\sqrt3}<\sin^{-1}\!\left(\tfrac35\right)<\frac{\pi}{6}+\frac{1}{10}\cdot\frac54.

Simplify the additive terms:

\frac{1}{10}\cdot\frac{2}{\sqrt3}=\frac{2}{10\sqrt3}=\frac{1}{5\sqrt3}=\frac{\sqrt3}{15},\qquad \frac{1}{10}\cdot\frac54=\frac{5}{40}=\frac18.

Therefore

\frac{\pi}{6}+\frac{\sqrt3}{15}<\sin^{-1}\!\left(\frac35\right)<\frac{\pi}{6}+\frac18.\qquad\blacksquare

Answer

\boxed{\;\dfrac{\pi}{6}+\dfrac{\sqrt3}{15}<\sin^{-1}\!\left(\dfrac35\right)<\dfrac{\pi}{6}+\dfrac18\quad\text{(proved via MVT on }[1/2,3/5]).\;}