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^tdt , then the ordered pair (m, n) is equal to

Definite Integration PYQ — JEE Main 2022 · 27 Jun · Shift 2 · Q25 with Solution

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Correct answer: B

Differentiate the given function

Given

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

Using the Fundamental Theorem of Calculus with chain rule:

f'(x)=\frac{(x^2)^2-5x^2+4}{2+e^{x^2}}\cdot \frac{d}{dx}(x^2)

So,

f'(x)=\frac{2x(x^4-5x^2+4)}{2+e^{x^2}}

Factor:

x^4-5x^2+4=(x^2-1)(x^2-4)

Hence

f'(x)=\frac{2x(x^2-1)(x^2-4)}{2+e^{x^2}}

Find critical points

Since $2+e^{x^2}>0$ for all $x$ , the sign of $f'(x)$ depends only on

2x(x^2-1)(x^2-4)=2x(x-1)(x+1)(x-2)(x+2)

Thus critical points are:

x=-2,-1,0,1,2

Sign analysis of $f'(x)$

We check intervals determined by these points:

For $x<-2$ , take $x=-3$ : $x(x^2-1)(x^2-4)<0$ so $f'(x)<0$
For $-2<x<-1$ , take $x=-\tfrac32$ : $x<0,\quad x^2-1>0,\quad x^2-4<0$ product $>0$ , so $f'(x)>0$
For $-1<x<0$ , take $x=-\tfrac12$ : $x<0,\quad x^2-1<0,\quad x^2-4<0$ product $<0$ , so $f'(x)<0$
For $0<x<1$ , take $x=\tfrac12$ : $x>0,\quad x^2-1<0,\quad x^2-4<0$ product $>0$ , so $f'(x)>0$
For $1<x<2$ , take $x=\tfrac32$ : $x>0,\quad x^2-1>0,\quad x^2-4<0$ product $<0$ , so $f'(x)<0$
For $x>2$ , take $x=3$ : $x>0,\quad x^2-1>0,\quad x^2-4>0$ product $>0$ , so $f'(x)>0$

So the sign chart is:

(-\infty,-2):-\,,\quad (-2,-1):+\,,\quad (-1,0):-\,,\quad (0,1):+\,,\quad (1,2):-\,,\quad (2,\infty):+

Identify local maxima and minima

At $x=-2$ : $f'$ changes $- \to +$ , so local minimum
At $x=-1$ : $f'$ changes $+ \to -$ , so local maximum
At $x=0$ : $f'$ changes $- \to +$ , so local minimum
At $x=1$ : $f'$ changes $+ \to -$ , so local maximum
At $x=2$ : $f'$ changes $- \to +$ , so local minimum

Therefore,

m=2,\qquad n=3

Match with options

Thus,

(m,n)=(2,3)

which is Option B.

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View written solutionFree

Correct answer: B

Differentiate the given function

Given

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

Using the Fundamental Theorem of Calculus with chain rule:

f'(x)=\frac{(x^2)^2-5x^2+4}{2+e^{x^2}}\cdot \frac{d}{dx}(x^2)

So,

f'(x)=\frac{2x(x^4-5x^2+4)}{2+e^{x^2}}

Factor:

x^4-5x^2+4=(x^2-1)(x^2-4)

Hence

f'(x)=\frac{2x(x^2-1)(x^2-4)}{2+e^{x^2}}

Find critical points

Since $2+e^{x^2}>0$ for all $x$ , the sign of $f'(x)$ depends only on

2x(x^2-1)(x^2-4)=2x(x-1)(x+1)(x-2)(x+2)

Thus critical points are:

x=-2,-1,0,1,2

Sign analysis of $f'(x)$

We check intervals determined by these points:

For $x<-2$ , take $x=-3$ : $x(x^2-1)(x^2-4)<0$ so $f'(x)<0$
For $-2<x<-1$ , take $x=-\tfrac32$ : $x<0,\quad x^2-1>0,\quad x^2-4<0$ product $>0$ , so $f'(x)>0$
For $-1<x<0$ , take $x=-\tfrac12$ : $x<0,\quad x^2-1<0,\quad x^2-4<0$ product $<0$ , so $f'(x)<0$
For $0<x<1$ , take $x=\tfrac12$ : $x>0,\quad x^2-1<0,\quad x^2-4<0$ product $>0$ , so $f'(x)>0$
For $1<x<2$ , take $x=\tfrac32$ : $x>0,\quad x^2-1>0,\quad x^2-4<0$ product $<0$ , so $f'(x)<0$
For $x>2$ , take $x=3$ : $x>0,\quad x^2-1>0,\quad x^2-4>0$ product $>0$ , so $f'(x)>0$

So the sign chart is:

(-\infty,-2):-\,,\quad (-2,-1):+\,,\quad (-1,0):-\,,\quad (0,1):+\,,\quad (1,2):-\,,\quad (2,\infty):+

Identify local maxima and minima

At $x=-2$ : $f'$ changes $- \to +$ , so local minimum
At $x=-1$ : $f'$ changes $+ \to -$ , so local maximum
At $x=0$ : $f'$ changes $- \to +$ , so local minimum
At $x=1$ : $f'$ changes $+ \to -$ , so local maximum
At $x=2$ : $f'$ changes $- \to +$ , so local minimum

Therefore,

m=2,\qquad n=3

Match with options

Thus,

(m,n)=(2,3)

which is Option B.

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