Question

Let $f : R \rightarrow R ^{+}$be a differentiable function satisfying $f ^{\prime}( x )=2 f ( x ) \forall x \in R$. Also $f (0)=1$ and $g(x)=f(x) \cdot \cos ^2 x$. If $n_1$ represent number of points of local maxima of $g(x)$ in $[-\pi, \pi]$ and $n_2$ is the number of points of local minima of $g ( x )$ in $[-\pi, \pi]$ and $n _3$ is the number of points in $[-\pi, \pi]$ where $g ( x )$ attains its global minimum value, then find the value of $\left( n _1+ n _2+ n _3\right)$.

Answer 1

Given $\mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{f}(\mathrm{x})$ $\therefore \frac{\mathrm{f}^{\prime}}{\mathrm{f}}=2 \Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{A}^{2 \mathrm{x}}$ $ \mathrm{f}(0) =1=\mathrm{A} $ $ \mathrm{f}(\mathrm{x}) =\mathrm{e}^{2 \mathrm{x}}$ $\therefore \mathrm{f}(\mathrm{x})=\mathrm{e}^{2 \mathrm{x}}$ Now, $g(x)=e^{2 x} \cdot \cos ^{2} x$ $g^{\prime}(x)=2 \cos x e^{2 x}(\cos x-\sin x)$ $\mathrm{g}^{\prime}(\mathrm{x})=0 \Rightarrow \mathrm{x}=\frac{-3 \pi}{4}, \frac{\pi}{4}, \frac{\pi}{2}, \frac{-\pi}{2}$ $\therefore $ Points of maxima are $\frac{-3 \pi}{4}, \frac{\pi}{4}$ and $\pi$ points of minima are $-\pi, \frac{-\pi}{2}, \frac{\pi}{2}$ and global minimum value occurs at $\frac{ \pm \pi}{2}$ which is zero. Hence $\mathrm{n}_{1}=3, \mathrm{n}_{2}=3, \mathrm{n}_{3}=2 \Rightarrow \mathrm{n}_{1}+\mathrm{n}_{2}+\mathrm{n}_{3}=8$.