Animated Solution for Mathematics - Differentiation: If $f:\mathbb{R}\to \mathbb{R}$ is a twice differentiable function such that $f^{\prime \prime }(x)>0$ for all $x\in \mathbb{R}$, and $f\left(\frac{1}{2}\right)=f(1)=1$, then

Welcome, future engineers. Today, we are not just solving a calculus problem; we are dissecting the behavior of functions. This problem is a classic example of how the JEE Advanced examiners test your ability to bridge the gap between algebraic conditions and geometric intuition.

Before we begin, let us address the elephant in the room: the typo. In the heat of an exam, encountering a problem that seems unsolvable is a test of your composure. If you see $f(\frac{1}{2})=1$ and $f(1)=1$ with $f^{\prime \prime }(x)>0$, you would be staring at a contradiction.

We proceed with the corrected, intended condition: $f(\frac{1}{2})=\frac{1}{2}$ and $f(1)=1$. Let us dive in.

Imagine you are standing on a coordinate plane. You have two points: $A(\frac{1}{2},\frac{1}{2})$ and $B(1,1)$. If you connect these points with a straight line, you get the identity line $y=x$.

We are told that $f^{\prime \prime }(x)>0$ for all $x\in \mathbb{R}$. This is the most critical piece of information. Geometrically, a positive second derivative means the function $f(x)$ is strictly convex—it curves upwards, like a smile.

Because the curve is strictly convex, it must lie below the secant line connecting any two points on the curve. This is the fundamental property of convexity. We are about to prove this relationship using the tools of calculus.

How do we compare the curve $f(x)$ to the line $y=x$? We create a bridge. We define an auxiliary function, $h(x)$, which represents the vertical gap between the curve and the line:

This function $h(x)$ is our secret weapon. Let us check its behavior at the boundaries. At $x=\frac{1}{2}$, we have:

At $x=1$, we have:

Do you see what we have created? We have a function $h(x)$ that is continuous and differentiable, and it vanishes at both endpoints of the interval $[\frac{1}{2},1]$.

This is where the magic happens. Rolle's Theorem states that if a function is continuous on $[a,b]$, differentiable on $(a,b)$, and $h(a)=h(b)$, then there must exist at least one point $c\in (a,b)$ such that $h^{\prime }(c)=0$.

Let us apply this to our function $h(x)$. Since $h(\frac{1}{2})=h(1)=0$, there exists some $c\in (\frac{1}{2},1)$ such that:

Now, let us differentiate $h(x)$ to see what this implies for our original function $f(x)$. Since $h(x)=f(x)-x$, the derivative is:

Substituting our point $c$ into this derivative, we get:

Geometrically, this means that at the point $x=c$, the tangent to the curve is perfectly parallel to the secant line $y=x$. The curve is 'running' at the same slope as the line at that specific instant.

We are almost at the finish line. We know $f^{\prime }(c)=1$. We also know that $f^{\prime \prime }(x)>0$. As we discussed earlier, this implies that $f^{\prime }(x)$ is a strictly increasing function.

We know that our point $c$ lies in the interval $(\frac{1}{2},1)$. Therefore, we have the inequality:

Since $f^{\prime }(x)$ is strictly increasing, applying the function $f^{\prime }$ to both sides of this inequality preserves the direction of the inequality:

Substituting our known value $f^{\prime }(c)=1$, we arrive at the elegant conclusion:

This matches option [D]. We have successfully navigated the geometry, utilized the power of Rolle's Theorem, and leveraged the monotonicity of the derivative to solve the problem. Remember, in JEE Advanced, the math is just the language; the geometry is the story. Keep visualizing, keep questioning, and keep pushing.