"If m and n respectively are the number of local maximum and local minimum points of the function $f(x) = \int\limits_0^{{x^2}} {{{{t^2} - 5t + 4} \over {2 + {e^t}}}dt}$, then the ordered pair (m, n) is equal to"

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$f(x) = \int_0^{{x^2}} {{{{t^2} - 5t + 4} \over {2 + {e^t}}}dt}$

$f'(x) = 2x\left( {{{{x^4} - 5{x^2} + 4} \over {2 + {e^{{x^2}}}}}} \right) = 0$

$x = 0$, or $({x^2} - 4)({x^2} - 1) = 0$

$x = 0,$ $x = \pm 2,\, \pm 1$

Now, $$f'(x) = {{2x(x + 1)(x - 1)(x + 2)(x - 2)} \over {\left( {{e^{{x^2}}} + 2} \right)}}$$

f'(x) changes sign from positive to negative at x = $-$1, 1 So, number of local maximum points = 2

f'(x) changes sign from negative to positive at x = $-$2, 0, 2 So, number of local minimum points = 3

$\therefore$ m = 2, n = 3

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If m and n respectively are the number of local maximum and local minimum points of the function $f(x) = \int\limits_0^{{x^2}} {{{{t^2} - 5t + 4} \over {2 + {e^t}}}dt}$, then the ordered pair (m, n) is equal to

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If m and n respectively are the number of local maximum and local minimum points of the function $f(x) = \int\limits_0^{{x^2}} {{{{t^2} - 5t + 4} \over {2 + {e^t}}}dt}$, then the ordered pair (m, n) is equal to

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