The function f(x) = _ - 1 ^x t( e^t - 1)(t - 1) (t - 2) ^3 (t - 3… - Vidyadip

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
$2$
D
None of These

✓

Answer

Correct option: B.

$1$

b
(b) $f(x) = \int_{ - 1}^x {t({e^t} - 1)(t - 1){{(t - 2)}^3}{{(t - 3)}^5}} dt$

$\therefore f'(x) = x({e^x} - 1)(x - 1){(x - 2)^3}{(x - 3)^5}$

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.

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