Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f:\mathbb{R}\to (0,1)$ be a continuous function. Then, which of the following function(s) has(have) the value zero at some point in the interval $(0,1)$ ?

Welcome, fellow traveler on the road to JEE Advanced. Today, we are not just solving a problem; we are exploring the beautiful, logical landscape of the Intermediate Value Theorem (IVT).

Often, students look at a problem involving integrals and functions and immediately reach for complex calculus machinery. But sometimes, the most powerful tool is the simplest one—the one that tells us that to get from the basement to the attic, you must pass through the ground floor.

Understanding Our Constraints

We are given a continuous function $f:\mathbb{R}\to (0,1)$. This is a goldmine of information.

It tells us two things: first, the function is well-behaved (continuous), and second, it is trapped in a narrow corridor: $0<f(x)<1$. This constraint is our secret weapon.

Whenever you see an integral involving $f(t)$, remember that $f(t)$ is never zero and never reaches one. It is always a positive fraction.

The IVT Test

Our goal is to find which of the given functions $g(x)$ hits zero somewhere in the interval $(0,1)$.

The IVT states that if a continuous function $g(x)$ satisfies $g(0)<0$ and $g(1)>0$ (or vice versa), then there must exist some $c\in (0,1)$ such that $g(c)=0$. We will test each option by checking the signs at the boundaries.

Imagine you are tracking the growth of $e^x$ against the accumulation of the integral. Since $f(t)<1$ and $\sin t<1$ for $t\in (0,1)$, the integrand $f(t)\sin t$ is strictly less than 1.

Thus, the integral is strictly less than ${\int }_0^{x}1dt=x$. Because $e^x>1+x$ for all $x>0$, it is clear that $e^x$ will always stay above the integral.

This function never touches zero. It stays positive, floating safely above the x-axis.

This is where the magic happens. Let $h(x)=x-f(x)$.

At $x=0$, $h(0)=0-f(0)$. Since $f(0)>0$, $h(0)$ is negative.

Now, look at $x=1$: $h(1)=1-f(1)$. Because $f(1)<1$, the value $1-f(1)$ must be positive. We started below the axis and ended above it. By the grace of the IVT, $h(x)$ must cross zero. Option B is a winner!

This one is a trap for the unwary. Both $f(x)$ and the integral are strictly positive.

Adding two positive numbers will never yield zero. This function is strictly positive for all $x$. No root here.

Finally, let $m(x)=x-{\int }_0^{\pi /2-x}f(t)\cos tdt$.

At $x=0$, $m(0)=0-{\int }_0^{\pi /2}f(t)\cos tdt$. Since $f(t)$ and $\cos t$ are positive on $(0,\pi /2)$, the integral is positive, so $m(0)<0$.

At $x=1$, $m(1)=1-{\int }_0^{\pi /2-1}f(t)\cos tdt$. We know:

Since $\frac{\pi }{2}-1\approx 0.57$, the integral is less than $0.57$. Thus, $m(1)=1-(\text{something}<0.57)$, which is clearly positive. Again, we have a sign change. Option D is also correct!

Mathematics is not about memorizing formulas; it is about observing the behavior of functions. By using the IVT, we turned a potentially daunting calculus problem into a simple game of signs.

Keep this perspective, stay curious, and remember: even the most complex problems are just a collection of simple, logical steps waiting to be discovered.