Animated Solution for Mathematics - Differentiation: There exist a function $f(x)$, satisfying $f(0)=1,f^{\prime }(0)=-1,f(x)>0$ for all $x$, and

Welcome, future engineers! Today, we are going to explore a problem that isn't just about crunching numbers; it's about understanding the soul of a function.

We are given a function $f(x)$ with three simple, yet powerful constraints: $f(0)=1$, $f^{\prime }(0)=-1$, and $f(x)>0$ for all $x$. Our mission is to determine the nature of its second derivative, $f^{\prime \prime }(x)$.

Imagine you are standing on a coordinate plane. The first condition, $f(0)=1$, tells us exactly where our journey begins: at the point $(0,1)$ on the $y$-axis.

Now, look at the second condition, $f^{\prime }(0)=-1$. This is the slope of the tangent line at our starting point. A slope of $-1$ means that for every step we take to the right, we must take a step down.

If we were to draw this tangent line, its equation would be $y=1-x$. This is our initial trajectory.

Here is where the drama begins. If our function were to simply follow that straight tangent line $y=1-x$, it would hit the $x$-axis at $x=1$.

But wait! Our third condition, $f(x)>0$ for all $x$, acts as an impenetrable barrier. The function is strictly positive; it can never, ever touch or cross the $x$-axis.

If it were to follow the straight line, it would dive into the negative territory, violating our constraint. This is the 'cliff' our function must avoid.

So, how does the function survive? To avoid the $x$-axis, the curve must 'bend' upwards, pulling away from the downward tangent line.

In the language of calculus, a curve that bends upwards like a bowl is called concave upwards. Mathematically, this bending is governed by the second derivative, $f^{\prime \prime }(x)$.

When a function is concave upwards, its slope is increasing—meaning it is becoming less negative. This direct relationship tells us that $f^{\prime \prime }(x)$ must be strictly positive.

Let's make this concrete. We need a function that starts at $(0,1)$, has a slope of $-1$ at $x=0$, and stays positive forever. A classic candidate is the exponential function $f(x)=e^{-x}$.

Let's test it:

1. $f(0)=e^{-0}=1$. (Check!)

2. $f^{\prime }(x)=-e^{-x}$, so $f^{\prime }(0)=-e^0=-1$. (Check!)

3. $f(x)=e^{-x}$ is always positive for all real $x$. (Check!)

Now, let's find the second derivative of our candidate:

Since $e^{-x}$ is always positive, we have confirmed that $f^{\prime \prime }(x)>0$. Our geometric intuition was spot on!

We have successfully navigated the constraints. A positive function starting with a negative slope must be concave upwards to avoid becoming negative.

This forces $f^{\prime \prime }(x)>0$ for all $x$. This elegant result perfectly matches Option A.

Remember, in JEE Advanced, always look for the geometric reality behind the algebraic constraints. It will turn a terrifying problem into a beautiful story.