Animated Solution for Mathematics - Differentiation: Let $f$ be a twice differentiable function on $(1,6)$. If $f(2)=8$, $f^{\prime }(2)=5$, $f^{\prime }(x)\ge 1$ and $f^{\prime \prime }(x)\ge 4$ for all $x\in (1,6)$, then :

Imagine you are standing on a mountain path, defined by a function $f(x)$. You know exactly where you are at $x=2$, and you know how steep the path is at that moment.

But you need to predict your altitude and your steepness at $x=5$. This is the essence of the problem we are solving today.

We are using the powerful, elegant machinery of the Mean Value Theorem (LMVT) to predict the future of a function based on its constraints.

The Mean Value Theorem is the bridge between the average and the instantaneous. It tells us that for a smooth, continuous function, there is always a point where the instantaneous slope (the derivative) is exactly equal to the average slope (the secant line) over an interval.

In our case, we have two layers of reality: the function $f(x)$ and its derivative $f^{\prime }(x)$.

To solve this, we must apply the theorem twice. We are looking for a lower bound for the sum $f(5)+f^{\prime }(5)$. This means we need to find the minimum possible value for $f(5)$ and the minimum possible value for $f^{\prime }(5)$ separately, and then combine them.

Let us start with the derivative. We know $f^{\prime }(2)=5$ and we are given that $f^{\prime \prime }(x)\ge 4$.

This second derivative is the rate of change of the slope itself. If we apply LMVT to the function $f^{\prime }(x)$ on the interval $[2,5]$, the theorem guarantees there exists some $c_2\in (2,5)$ such that:

Since we know $f^{\prime \prime }(x)\ge 4$ for all $x$, it must be true that $f^{\prime \prime }(c_2)\ge 4$. Substituting our known values, we get:

With a simple algebraic step, multiplying by $3$ and adding $5$, we find that $f^{\prime }(5)\ge 17$. This is our first major victory; we have successfully bounded the steepness of the path at $x=5$.

Now, we turn our attention to the function $f(x)$ itself. We apply LMVT again, this time to $f(x)$ on the interval $[2,5]$.

There exists some $c_1\in (2,5)$ such that:

We are given the constraint $f^{\prime }(x)\ge 1$ for all $x$. Therefore, $f^{\prime }(c_1)\ge 1$. Substituting $f(2)=8$, we get:

Solving this inequality, we multiply by $3$ to get $f(5)-8\ge 3$, which leads us to $f(5)\ge 11$.

We have arrived at the summit. We have two independent lower bounds: $f^{\prime }(5)\ge 17$ and $f(5)\ge 11$.

By the properties of inequalities, if $A\ge a$ and $B\ge b$, then $A+B\ge a+b$. Thus:

This gives us the final result:

We have navigated the constraints, applied the theorems, and arrived at the truth. This is the beauty of JEE Advanced mathematics—it is not about memorizing formulas, but about understanding how to use them to construct a logical path through the unknown.