Let $f(x)=\int\limits_0^x(\sin t-\cos t)\left(e^t-2\right)(t-1)^3(t-2)^5 d t, 0<x \leq 4$. Then, the number of points where f(x) assumes local maximum value, is

3

We have,

$f(x) =\int\limits_0^x(\sin t-\cos t)\left(e^t-2\right)(t-1)^3(t-2)^5 d t, 0<x \leq 4$

$\Rightarrow f'(x) =(\sin x-\cos x)\left(e^x-2\right)(x-1)^3(x-2)^5, 0<x \leq 4$

For local maximum or minimum, we must have

$f'(x)=0$

$\Rightarrow \sin x-\cos x=0, e^x-2=0,(x-1)^3=0,(x-2)^5=0$

$\Rightarrow \tan x=1, e^x=2, x=1,2$

$\Rightarrow x=\frac{\pi}{4}, \frac{5 \pi}{4}, x=\log _e 2, x=1,2$               $[∵ 0<x \leq 4]$

$\Rightarrow x=0.785,3.925,0.693,1,2$

The changes in sign s of $f'(x)$ in the neighbourhoods of these points are shown in Figure.

Clearly, $x=\log _e 2, \frac{\pi}{4}$ and $\frac{5 \pi}{4}$ are points of local maxima.