Animated Solution for Mathematics - Differentiation: If m and n respectively are the number of local maximum and local minimum points of the function $f(x)={\int }_0^{x^2}\frac{t^2-5t+4}{2+e^t}dt$, then the ordered pair (m, n) is equal to

Unlocking the Integral

Welcome, future engineers. Today, we are going to dismantle a problem that, at first glance, looks like a nightmare of integration. You see an integral, you see an exponential in the denominator, and your instinct might be to panic.

But here is the secret of JEE Advanced: the problem is not about integration; it is about the behavior of the function. We are going to use the Leibniz Rule to peel back the layers of this function and reveal its hidden geometry.

The Magic of Leibniz

We are given $f(x)={\int }_0^{x^2}\frac{t^2-5t+4}{2+e^t}dt$. The question asks for local maxima and minima. To find these, we need the derivative $f^{\prime }(x)$.

We invoke the Leibniz Rule to differentiate the integral. The rule states that the rate of change of this integral depends on the boundaries of the integration.

Applying the rule, we get:

Notice how the lower limit, being zero, vanishes because its derivative is zero. The derivative of $x^2$ is $2x$, and suddenly, the complexity of the integral collapses into a simple algebraic expression.

The Algebraic Landscape

Now, let us simplify our derivative:

Look at the denominator: $2+e^{x^2}$. Since $e^{x^2}$ is always positive, this denominator is always greater than $2$. It is a 'safe' zone that will never be zero and will never flip the sign of our derivative.

We can focus entirely on the numerator: $2x(x^4-5x^2+4)$. Factorizing the quadratic in $x^2$, we see $x^4-5x^2+4$ factors neatly into $(x^2-1)(x^2-4)$.

Breaking this down further using the difference of squares, we get:

This is the heart of the problem. Our critical points are where $f^{\prime }(x)=0$, which are $x=0,\pm 1,\pm 2$.

The Wavy Curve Dance

Now, we arrange these points on a number line: $-2,-1,0,1,2$. These points divide the real line into six intervals. We use the 'wavy curve' method to determine the sign of $f^{\prime }(x)$ in each interval.

Starting from the rightmost interval ($x>2$), all factors are positive, so $f^{\prime }(x)>0$. Because each root has an odd multiplicity, the sign alternates as we cross each point:

For $x>2$: Positive ($+$) For $1<x<2$: Negative ($-$) For $0<x<1$: Positive ($+$) For $-1<x<0$: Negative ($-$) For $-2<x<-1$: Positive ($+$) For $x<-2$: Negative ($-$)

Now, we simply read the map we have created. A local maximum occurs when the slope changes from positive to negative. Looking at our signs, this happens at $x=1$ and $x=-1$. That gives us $m=2$ local maxima.

A local minimum occurs when the slope changes from negative to positive. This happens at $x=2,x=0,$ and $x=-2$. That gives us $n=3$ local minima.

We have navigated the complexity and arrived at the truth. The ordered pair $(m,n)$ is $(2,3)$. With the right tools and a calm mind, even the most intimidating integrals become simple stories of slopes and signs.