Animated Solution for Mathematics - Functions: If $f:R\to S$, defined by $f(x)=\sin x-\sqrt{3}\cos x+1$, is onto, then the interval of $S$ is

Welcome, aspiring physicist and mathematician. Today, we are going to peel back the layers of a problem that might seem like a simple trigonometric exercise, but is actually a beautiful lesson in the nature of functions.

We are looking at the function $f(x)=\sin x-\sqrt{3}\cos x+1$ and we are told it is 'onto'.

In the world of functions, the term 'onto'—or surjective—is a powerful constraint. It is a promise that the function is not leaving any part of its codomain $S$ untouched.

If $f:R\to S$ is onto, it means that for every element in $S$, there exists at least one $x$ in the domain such that $f(x)$ maps to it. Mathematically, this is the most elegant way of saying: the codomain $S$ is exactly the range of the function.

So, our task is not to solve for $x$, but to map the territory of the function's output. We need to find the range of $f(x)=\sin x-\sqrt{3}\cos x+1$.

Now, look at the expression $\sin x-\sqrt{3}\cos x$. This is a classic superposition of two waves. In physics, when you combine two oscillations of the same frequency, they interfere to form a new wave.

We use the identity $a\sin x+b\cos x$. The range of this expression is always bounded by $[-\sqrt{a^2+b^2},\sqrt{a^2+b^2}]$.

Why does this work? Imagine a vector in the Cartesian plane with components $a$ and $b$. The magnitude of this vector is $\sqrt{a^2+b^2}$.

When we combine the sine and cosine terms, we are essentially projecting this vector onto a rotating axis. The maximum value it can reach is the length of the vector itself, and the minimum is the negative of that length.

For our function, $a=1$ and $b=-\sqrt{3}$. Let us calculate the amplitude $R$:

This tells us that the oscillating part, $\sin x-\sqrt{3}\cos x$, is trapped forever between $-2$ and $2$. Mathematically, we write:

We have the oscillating core, but our function isn't just the wave; it has a $+1$ at the end. This is a vertical shift.

Think of the wave as a physical object—a string vibrating between $-2$ and $2$. Adding $1$ is like taking that entire string and lifting it up by one unit. The new equilibrium position is no longer $0$, but $1$.

To find the new range, we simply add $1$ to every part of our inequality:

This simplifies beautifully to:

We have arrived at the destination. The function $f(x)$ takes all values in the closed interval $[-1,3]$.

Since the function is onto, the codomain $S$ must be exactly this range. Thus, $S=[-1,3]$.

This problem is a perfect example of how complex-looking expressions are often just simple waves in disguise. Whenever you see $a\sin x+b\cos x+c$, do not be intimidated.

Identify your $a$, $b$, and $c$, calculate the amplitude $\sqrt{a^2+b^2}$, and shift it by $c$. You have mastered the wave. Keep this intuition with you, and you will find that even the most daunting JEE problems start to look like familiar friends.