Question

Let $f:R\to R^{+}$be a differentiable function satisfying $f^{\prime }(x)=2f(x)\forall x\in R$. 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

Answer: 0008

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, $g(x)=e^{2x}\cdot {\cos }^{2}x$ $g^{\prime }(x)=2\cos x{e}^{2x}(\cos x-\sin x)$ ${\mathrm{g}}^{\prime }(\mathrm{x})=0\Rightarrow \mathrm{x}=\frac{-3\pi }{4},\frac{\pi }{4},\frac{\pi }{2},\frac{-\pi }{2}$ $\therefore$ 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 ${\mathrm{n}}_1=3,{\mathrm{n}}_2=3,{\mathrm{n}}_3=2\Rightarrow {\mathrm{n}}_1+{\mathrm{n}}_2+{\mathrm{n}}_3=8$.