Animated Solution for Mathematics - Definite Integration: Comprehension Passage

Unraveling the Mystery of $F(x)$

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving a problem; we are peeling back the layers of a function to reveal the elegant geometry hidden beneath.

We are given a thrice differentiable function $F(x)$ defined on $\mathbb{R}$, with specific boundary conditions: $F(1)=0$ and $F(3)=-4$. We are also told that $F(x)<0$ for all $x\in (1,3)$. Our companion function is $f(x)=xF(x)$.

Let us embark on this journey.

The Geometry of $f(x)$

Imagine standing at $x=1$. We know $F(1)=0$, so $f(1)=1\cdot F(1)=0$.

Now, move to $x=3$. We know $F(3)=-4$, so $f(3)=3\cdot (-4)=-12$.

Between these two points, we are told $F(x)$ is strictly negative. Since $x$ is positive in the interval $(1,3)$, the product $xF(x)$ must also be strictly negative.

This is our first realization: $f(x)$ is a function that starts at zero, drops into the negative realm, and ends at $-12$. This simple sign analysis tells us immediately that $f(2)<0$. We have successfully oriented ourselves in the coordinate plane.

The Derivative Dance

Now, we must understand the slope. We define $f(x)=xF(x)$.

To find the rate of change, we apply the product rule:

This simplifies to:

Let us look at the left boundary, $x=1$. We have $f^{\prime }(1)=F(1)+1\cdot F^{\prime }(1)$. Since $F(1)=0$, this becomes $f^{\prime }(1)=F^{\prime }(1)$.

Because $F(x)$ starts at zero and immediately becomes negative for $x>1$, the function must be decreasing. Therefore, $F^{\prime }(1)$ must be negative, which forces $f^{\prime }(1)<0$. This is not just algebra; it is the physical reality of a curve diving below the x-axis.

The Art of Integration by Parts

This is where the problem transforms from a simple analysis into a beautiful puzzle. We are given two integrals:

These look intimidating, but they are invitations to use Integration by Parts (IBP). Recall the formula: $\int udv=uv-\int vdu$.

For the first integral, let $u=x^2$ and $dv=F^{\prime }(x)dx$. Then $du=2xdx$ and $v=F(x)$. The integral becomes:

Notice the magic here! The term $\int 2xF(x)dx$ is simply $2{\int }_1^{3}f(x)dx$. We have successfully linked the integral of $F^{\prime }(x)$ to the integral of $f(x)$.

Substituting our boundary values:

With $F(3)=-4$ and $F(1)=0$, we get $-12=-36-2{\int }_1^{3}f(x)dx$, which simplifies to:

The complexity has vanished, leaving behind a clean, elegant result.

The Grand Finale

Finally, we tackle the second integral: ${\int }_1^3{x}^3{F}^{\prime \prime }(x)dx=40$. Again, we use IBP.

Let $u=x^3$ and $dv=F^{\prime \prime }(x)dx$. Then $du=3x^{2}dx$ and $v=F^{\prime }(x)$. This yields:

We already know the value of ${\int }_1^3{x}^2{F}^{\prime }(x)dx$ is $-12$. Substituting this, we get:

To relate this to $f^{\prime }(x)$, we recall $f^{\prime }(x)=F(x)+x{F}^{\prime }(x)$. At $x=3$, $f^{\prime }(3)=F(3)+3F^{\prime }(3)=-4+3F^{\prime }(3)$. Thus, $3F^{\prime }(3)=f^{\prime }(3)+4$.

Multiplying by $9$, we get $27F^{\prime }(3)=9f^{\prime }(3)+36$. Substituting this back into our equation, we arrive at:

We have traversed the entire landscape of this function. We started with simple points, moved through the dynamics of derivatives, and used the power of integration to uncover a deep relationship between the boundary slopes. This is the essence of JEE Advanced mathematics—not just calculation, but the art of connecting disparate concepts into a single, harmonious truth.