Let $f : R \to R^{+}$ be a differentiable function satisfying $f^{'} (x) = 2 f (x) \forall x \in R$ . Also $f (0) = 1 a n d g (x) = f (x) . \cos^{2} x$ . If $n_{1}$ represent number points of local maxima of $g (x)$ in $[- π, π]$ and $n_{2}$ is the number of points of local minima of $g (x)$ in $[- π, π]$ and $n_{3}$ is the number of points in $[- π, π]$ where $g (x)$ attains its global minimum value, then find the value of $(n_{1} + n_{2} + n_{3})$

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

answer is 6.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Given $f^{'} (x) = 2 f (x)$
$∴ \frac{f^{'}}{f} = 2 \Rightarrow f (x) = A e^{2 x}$
$\begin{array}{l} f (0) = 1 = A \\ ∴ f (x) = e^{2 x} \end{array}$
Question Image
NOW, $\begin{array}{l} g (x) = e^{2 x} . \cos^{2} x \\ g^{'} (x) = e^{2 x} (- 2 \cos x \sin x + 2 \cos^{2} x) \end{array}$
$g^{'} (x) = 2 \cos x e^{2 x} (\cos x - \sin x)$
$g^{'} (x) = 0 \Rightarrow x = \frac{- 3 π}{4}, \frac{π}{4}, \frac{π}{2}, \frac{- π}{2}$
$∴ p o int s o f \max i m a a r e \frac{- 3 π}{4}, \frac{π}{4}$ and points of minima are $\frac{- π}{2}, \frac{π}{2}$
And global minimum value occurs at $\frac{\pm π}{2}$ which is zero
Hence, $n_{1} = 2, n_{2} = 2, n_{3} = 2 \Rightarrow n_{1} + n_{2} + n_{3} = 6$