← 2014 Paper 1

UPSC 2014 Maths Optional Paper 1 Q1c — Step-by-Step Solution

10 marks · Section A

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

Question

Prove that between two real roots of $e^x\cos x+1=0$, a real root of $e^x\sin x+1=0$ lies.

Technique

Rolle’s theorem applied to $\phi(x)=e^{-x}\cdot(\text{original LHS})$.

Solution

Strategy. Apply Rolle’s theorem to a cleverly chosen auxiliary function $\phi(x)$ such that:

1. $\phi$ vanishes at the roots of $e^x\cos x+1=0$.
2. $\phi'$ vanishing gives $e^x\sin x+1=0$.

The trick is to multiply the original $e^x\cos x+1$ by $e^{-x}$ to extract a clean derivative.

Step 1 — Define auxiliary function

Let $\alpha,\beta$ be two real roots of $e^x\cos x+1=0$ with $\alpha<\beta$. Define

Since $e^{-x}>0$, $\phi(x)=0\iff e^x\cos x+1=0$. In particular, $\phi(\alpha)=\phi(\beta)=0$.

Step 2 — Apply Rolle’s theorem to $\phi$

$\phi$ is continuous on $[\alpha,\beta]$ (sum of $\cos x$ and $e^{-x}$, both $C^\infty$) and differentiable on $(\alpha,\beta)$, with $\phi(\alpha)=\phi(\beta)=0$. By Rolle’s theorem, there exists $\xi\in(\alpha,\beta)$ with

Step 3 — Compute $\phi'$ and translate back

So $\phi'(\xi)=0$ becomes

So $\xi$ is a root of $e^x\sin x+1=0$, and $\alpha<\xi<\beta$.

Answer