MCQ
The function $f(x) = \int\limits_{ - 1}^x {t({e^t} - 1)(t - 1){{(t - 2)}^3}{{(t - 3)}^5}} dt$ has a local minimum at $ x =$ ..........
- A$0$
- ✓$1$
- C$2$
- DNone of These
$\therefore f'(x) = x({e^x} - 1)(x - 1){(x - 2)^3}{(x - 3)^5}$
For local minima, slope $i.e.$ $f'(x)$ should change sign from $ - ve$ to $ + ve$.
$f'(x) = 0 \Rightarrow x = 0,\,1,\,2,\,3$
If $x = 0 - h,$ where h is a very small number, then
$f'(x) = ( - )( - )( - 1)( - 1)( - 1) = - ve$
If $x = 0 + h$; $f'(x) = ( + )( + )( - )( - 1)( - 1) = - ve$
Hence at $x = 0$neither maxima nor minima.
If $x = 1 - h$
$f'(x) = ( + )( + )( - )( - 1)( - 1) = - ve$
If $x = 1 + h$; $f'(x) = ( + )( + )( + )( - 1)( - 1) = + ve$
Hence at $x = 1$there is a local minima.
If $x = 2 - h$; $f'(x) = ( + )( + 1)( + )( - )( - ) = + ve$
If $x = 2 + h$; $f'(x) = ( + )( + )( + )( + )( - 1) = - ve$
Hence at $x = 2$there is a local maxima.
If $x = 3 - h$; $f'(x) = ( + )( + )( + )( + )( - ) = - ve$
If $x = 3 + h$; $f'(x) = ( + )( + )( + )( + )( + ) = + ve$
Hence at $x = 3$there is a local minima.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.