UPSC 2021 Maths Optional Paper 1 Q1d — 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

Show that between any two roots of $e^x\cos x=1$ , there exists at least one root of $e^x\sin x-1=0$ .

Rolle’s theorem on $\phi(x)=e^{-x}-\cos x$ ; $\phi'(x)=\sin x-e^{-x}$ , vanishes where $e^x\sin x=1$ .

Strategy. Apply Rolle’s theorem to a suitable function.

Let $\phi(x)=e^{-x}-\cos x$ . Then $e^x\cos x=1\Leftrightarrow\cos x=e^{-x}\Leftrightarrow\phi(x)=0$ .

So roots of $e^x\cos x=1$ are roots of $\phi$ .

Let $a,b$ ( $a<b$ ) be two roots of $e^x\cos x=1$ . Then $\phi(a)=\phi(b)=0$ .

$\phi$ is continuous and differentiable everywhere. By Rolle’s theorem, there exists $c\in(a,b)$ with $\phi'(c)=0$ .

$\phi'(x)=-e^{-x}+\sin x=\sin x-e^{-x}$ .

So $\phi'(c)=0\Rightarrow\sin c=e^{-c}\Rightarrow e^c\sin c=1\Rightarrow e^c\sin c-1=0$ .

So $c$ is a root of $e^x\sin x-1=0$ lying in $(a,b)$ . $\blacksquare$

\boxed{\;\text{Between any two roots of }e^x\cos x=1,\text{ a root of }e^x\sin x-1=0\text{ exists.}\;}