Animated Solution for Mathematics - Functions: If $f(x)=\sin x+\cos x,g(x)=x^2-1$, then $g(f(x))$ is invertible in the domain

Welcome, JEE aspirants! Today, we embark on a journey to decode the mystery of composite functions. Imagine you are standing on the edge of a mathematical cliff, looking at two functions: $f(x)=\sin x+\cos x$ and $g(x)=x^2-1$.

The question asks us to find the domain where $g(f(x))$ is invertible. This is not just about solving an equation; it is about understanding the soul of a function.

First, we must construct the composite function $g(f(x))$. This is like a machine within a machine. We take the output of $f(x)$ and feed it into $g(x)$.

So, $g(f(x))=(f(x){)}^2-1$. Substituting $f(x)=\sin x+\cos x$, we get:

This is the raw material we need to work with.

Now, let's expand this. Using the identity $(a+b{)}^2=a^2+b^2+2ab$, we get:

We know that ${\sin }^{2}x+{\cos }^{2}x=1$. So, the expression simplifies to $1+2\sin x\cos x-1$.

The $1$ and $-1$ cancel out, leaving us with $2\sin x\cos x$. This is the double angle identity for sine: $\sin (2x)$. It is truly elegant how a seemingly complex expression collapses into something so simple.

Now, we have $y=\sin (2x)$. To be invertible, a function must be one-to-one. This means it must pass the horizontal line test.

The sine function is periodic, so it is not one-to-one over its entire domain. We need to restrict it to a region where it is monotonic. The principal domain for $\sin \theta$ is $[-\frac{\pi }{2},\frac{\pi }{2}]$.

In this interval, the sine function travels from its minimum to its maximum exactly once, without turning back.

Since our function is $\sin (2x)$, we set the argument $2x$ within the principal domain:

Dividing by $2$, we get:

This is our domain! It is the interval where the function is strictly increasing. I hope this journey has made the concept clear. Keep practicing, and you will master these concepts!