Animated Solution for Mathematics - Definite Integration: Let $f:\mathbb{R}\to \mathbb{R}$ be given by $f(x)=(x-1)(x-2)(x-5)$. Define $F(x)={\int }_0^{x}f(t)dt,x>0$. Then which of the following options is/are correct ?

My dear student, welcome to the arena. Today, we are not just solving a problem; we are peeling back the layers of a function to reveal its true nature.

We are given $f(x)=(x-1)(x-2)(x-5)$ and a function $F(x)$ defined as the integral of $f(t)$ from $0$ to $x$. This is a classic JEE Advanced setup—it looks like a simple calculus problem, but it demands a deep, intuitive grasp of the Fundamental Theorem of Calculus.

Many students see an integral and immediately panic, reaching for the integration formulas. But pause. Breathe. Look at the relationship: $F(x)={\int }_0^{x}f(t)dt$.

By the Fundamental Theorem of Calculus, we know that the derivative of this integral is simply the integrand itself. That is, $F^{\prime }(x)=f(x)$.

This is our golden key. We don't need to integrate $f(t)$ to understand the shape of $F(x)$. We only need to understand the behavior of $f(x)$.

By setting $F^{\prime }(x)=0$, we find our critical points: $x=1$, $x=2$, and $x=5$. These are the points where the slope of our function $F(x)$ is zero—the peaks and valleys of our landscape.

Now, let us visualize the landscape. We plot our critical points on the number line: $1,2,5$. We use the Wavy Curve method to determine the sign of $F^{\prime }(x)$.

For $x>5$, all factors $(x-1),(x-2),(x-5)$ are positive, so $F^{\prime }(x)>0$. As we move left across each root, the sign flips.

- In $(5,\infty )$, $F^{\prime }(x)>0$ (Function is increasing). - In $(2,5)$, $F^{\prime }(x)<0$ (Function is decreasing). - In $(1,2)$, $F^{\prime }(x)>0$ (Function is increasing). - In $(0,1)$, $F^{\prime }(x)<0$ (Function is decreasing).

This tells us everything! At $x=1$, the slope changes from negative to positive—a local minimum. At $x=2$, it changes from positive to negative—a local maximum. At $x=5$, it changes from negative to positive—another local minimum.

This confirms that Options A and B are correct, while Option D is incorrect because we have two minima and one maximum, not two maxima.

Now, we face the final challenge: Option C. Does $F(x) eq 0$ for all $x\in (0,5)$? This is where many students stumble.

They assume that because $F(x)$ has a local maximum at $x=2$, it must cross the x-axis. But we must check the value of that maximum. To do this, we expand $f(t)$:

Now, we integrate to find $F(x)$:

Let's evaluate this at our local maximum, $x=2$:

Look at that result! Even at its highest point in the interval $(0,5)$, the function is negative. If the peak of the mountain is below sea level, the entire mountain is underwater. Thus, $F(x)$ never touches the x-axis in this interval. Option C is correct.

We have navigated the derivative, the geometry of the curve, and the arithmetic of the integral. We found that $F(x)$ has a local minimum at $x=1$, a local maximum at $x=2$, and remains strictly negative in the interval $(0,5)$.

The final answer is A, B, and C. Remember, in JEE Advanced, the math is the tool, but the visualization is the master. Keep practicing, keep visualizing, and you will conquer any problem they throw at you.