If m and n respectively are the number of local maximum and local minimum points of the function f ( x )=∫ limits0x2 ( t 2-5 t +4/2+ c t ) dt, then the ordered pair ( m , n ) is equal to

Question

If m and n respectively are the number of local maximum and local minimum points of the function f ( x )=∫ limits0x2 ( t 2-5 t +4/2+ c t ) dt, then the ordered pair ( m , n ) is equal to (A) (3,2) (B) (2,3) (C) (2,2) (D) (3,4)

Tardigrade Admin · Accepted Answer

m=L ⋅ max N=L ⋅ min f(x)=∫ limits0x2 (t2-5 t+4/2+et) d t f prime(x)=((x4-5 x2+4) 2 x/2+ex2)=(2 x(x2-1)(x2-4)/2+ex2) =(2 x(x-1)(x+1)(x-2)(x+2)/2+ex2) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/0c9d596a5e857f9cb6bfe91de025e79f-.png" />

Q. If $m$ and $n$ respectively are the number of local maximum and local minimum points of the function $f (x) = 0 \int x^{2} \frac{t ^{2} - 5 t + 4}{2 + c ^{t}} d t$ , then the ordered pair $(m, n)$ is equal to

$(3, 2)$

$(2, 3)$

$(2, 2)$

$(3, 4)$

$m = L \cdot max$
$N = L \cdot min$
$f (x) = 0 \int x^{2} \frac{t ^{2} - 5 t + 4}{2 + e ^{t}} d t$
$f^{'} (x) = \frac{( x ^{4} - 5 x ^{2} + 4 ) 2 x}{2 + e ^{x^{2}}} = \frac{2 x ( x ^{2} - 1 ) ( x ^{2} - 4 )}{2 + e ^{x^{2}}}$
$= \frac{2 x ( x - 1 ) ( x + 1 ) ( x - 2 ) ( x + 2 )}{2 + e ^{x^{2}}}$

Q. If m and n respectively are the number of local maximum and local minimum points of the function f(x)=0∫x2​2+ctt2−5t+4​dt, then the ordered pair (m,n) is equal to

(3,2)

(2,3)

(2,2)

(3,4)