← 2025 Paper 1

UPSC Maths 2025 Paper 1 Q2b — Solution

15 marks · Section A

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

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