Animated Solution for Mathematics - Definite Integration: Comprehension Passage

Unveiling the Hidden Truth

Welcome, future engineer. Today, we are going to dissect a polynomial that might look like a standard textbook problem, but beneath its surface lies a beautiful lesson in calculus and logical deduction.

We are looking at $f(x)=4x^3+3x^2+2x+1$. It is a cubic, and at first glance, you might feel the urge to hunt for its roots using complex algebraic identities. Resist that urge. In the world of JEE Advanced, we don't just solve; we observe.

The Monotonicity Mystery

Before we find the root $s$, we must understand the 'personality' of this function. Is it dancing up and down, or is it a steady climber? To find out, we look at the slope, the derivative $f^{\prime }(x)$.

Calculating the derivative is straightforward:

Now, look at this quadratic. Does it ever cross the x-axis? Let's check the discriminant, $D=b^2-4ac$.

Substituting our values, we get $D=6^2-4(12)(2)=36-96=-60$. Since the discriminant is negative, the derivative $f^{\prime }(x)$ never touches zero.

Because the leading coefficient is positive, $f^{\prime }(x)$ is always positive. This is a profound realization: our function $f(x)$ is strictly increasing everywhere. It climbs from negative infinity to positive infinity, crossing the x-axis exactly once. That unique crossing point is our root, $s$.

Trapping the Root

Now that we know $s$ exists, how do we find it? We use the Intermediate Value Theorem (IVT). We need to find two points where the function changes sign.

Let's test $x=-\frac{1}{2}$ and $x=-\frac{3}{4}$. Evaluating at $x=-\frac{1}{2}$:

This is positive! Now, let's test $x=-\frac{3}{4}$:

This is negative! The function crossed from negative to positive between $-\frac{3}{4}$ and $-\frac{1}{2}$. Therefore, $s\in \left(-\frac{3}{4},-\frac{1}{2}\right)$.

The Area Under the Curve

We are given $t=|s|$. Since $s$ is negative, $t$ must be positive, specifically $t\in \left(\frac{1}{2},\frac{3}{4}\right)$.

We need the area bounded by $y=f(x)$, $x=0$, $y=0$, and $x=t$. This is the integral:

Integrating term by term, we get:

Since $t$ is positive, this area function is also strictly increasing. To find the interval for the area, we simply evaluate this expression at the boundaries of $t$.

At $t=\frac{1}{2}$, the area is $\frac{15}{16}$. At $t=\frac{3}{4}$, the area is $\frac{525}{256}$. Thus, our area lies strictly within this range.

The Behavior of the Slope

Finally, let's examine $f^{\prime }(x)=12x^2+6x+2$. Is it increasing or decreasing? We need the second derivative, $f^{\prime \prime }(x)=24x+6$.

Setting this to zero gives us the critical point $x=-\frac{1}{4}$. For $x<-\frac{1}{4}$, $f^{\prime \prime }(x)<0$, meaning the slope $f^{\prime }(x)$ is decreasing.

For $x>-\frac{1}{4}$, $f^{\prime \prime }(x)>0$, meaning the slope is increasing. Since our interval of interest is $(-t,t)$, and $-t<-\frac{1}{4}<t$, we see the slope decrease and then increase.

This problem was not just about finding a number; it was about understanding the flow of a function. You have successfully navigated the derivative, the integral, and the second derivative. Keep this analytical mindset, and no problem will ever be too complex for you.