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Q.
Between any two roots of equation $\sin\, x e^{-x^{2}}-1=0,$ there lies at least one root of equation $\cos x e^{-x^{2}}-2 x=0$. This is because of
Application of Derivatives
Solution:
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 $'\gamma' \in(\alpha, \beta)$ such that $f'(\gamma)=0$
$\Rightarrow \cos \gamma-2 \gamma e^{\gamma^{2}}=0$
$\Rightarrow \cos \gamma e^{-\gamma^{2}}-2 \gamma=0$