Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f:R\to R$ be a function defined by $f(x)=\min \{x+1,|x|+1\}$. Then which of the following is true?

A Journey into Differentiability

Welcome, future engineer! Today, we are going to dissect a problem that seems to be about calculus, but is actually a beautiful exercise in visualization and logical deduction.

We are looking at the function $f(x)=\min \{x+1,|x|+1\}$. At first glance, the presence of the modulus function $|x|$ might make you nervous. You might be expecting a sharp corner, a cusp, or a point of non-differentiability.

But let us slow down and peel back the layers of this function together.

Meet the Competitors

Imagine you are a referee in a competition. We have two contestants: $g(x)=x+1$ and $h(x)=|x|+1$. Our function $f(x)$ is the referee that always picks the smaller value of the two.

First, $g(x)=x+1$. This is a classic, well-behaved straight line. It has a slope of $1$ and crosses the y-axis at $1$. It is predictable, smooth, and differentiable everywhere.

Second, $h(x)=|x|+1$. This is the 'challenger.' Because of the modulus, it behaves differently depending on where you stand on the x-axis.

For $x\ge 0$, it behaves exactly like $x+1$. For $x<0$, it behaves like $-x+1$. If you were to graph this, you would see a classic V-shape, with the vertex at $(0,1)$.

The Arena of Comparison

Now, let us place them in the arena. We need to find $f(x)=\min \{g(x),h(x)\}$. Let us split the number line into two regions: the positive side ($x\ge 0$) and the negative side ($x<0$).

For $x\ge 0$: Here, $g(x)=x+1$ and $h(x)=x+1$. They are identical! Since they are the same, the minimum is simply $x+1$.

For $x<0$: This is where it gets interesting. We are comparing $g(x)=x+1$ and $h(x)=-x+1$.

Since $x$ is a negative number, let us test a value, say $x=-2$. Then $g(-2)=-2+1=-1$, while $h(-2)=-(-2)+1=3$.

Clearly, $g(x)<h(x)$. Algebraically, since $x<-x$ for all negative $x$, it follows that $x+1<-x+1$. Therefore, in this region, the line $g(x)$ is strictly below the V-shape $h(x)$.

The Final Verdict

Look at what we have discovered! For $x\ge 0$, $f(x)=x+1$. For $x<0$, $f(x)=x+1$.

Putting it all together, we realize that for all real numbers $x$:

The entire function $f(x)$ is just a single, continuous, straight line. The 'challenger' $h(x)$ was never actually the minimum; it was always 'above' or 'equal to' our steady player $g(x)$.

The Differentiability Check

Now, the question asks: is $f(x)$ differentiable? We have reduced the entire problem to $f(x)=x+1$.

This is a simple linear polynomial. Its derivative is:

There are no sharp corners, no cusps, and no breaks. The function is smooth, continuous, and differentiable everywhere.

We have successfully navigated the trap of the modulus function by relying on rigorous algebraic comparison. Remember, in JEE Advanced, always trust your algebra over your initial intuition. Sometimes, the most complex-looking functions simplify into the most elegant solutions.