Question

Between any two roots of equation $\sin x{e}^{-x^2}-1=0,$ there lies at least one root of equation $\cos x{e}^{-x^2}-2x=0$. This is because of

• A. Lagrange's Mean Value Theorem
• B. Rolle's Theorem
• C. Intermediate value theorem
• D. none of these

Answer 1

Answer: (B)

Let $f(x)=\sin x-e^{x^2}$
Let $\alpha$ and $\beta$ be roots of equation $\sin x-e^{x^2}=0$
$\Rightarrow f(\alpha )=f(\beta )$
As function is continuous in $[\alpha ,\beta ]$ and differentiable in $(\alpha ,\beta )$ there exists a number ${}^{\prime }{\gamma }^{\prime }\in (\alpha ,\beta )$ such that $f^{\prime }(\gamma )=0$
$\Rightarrow \cos \gamma -2\gamma e^{{\gamma }^2}=0$
$\Rightarrow \cos \gamma e^{-{\gamma }^2}-2\gamma =0$