Problem 156 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(f: R \rightarrow S\), defined by \(f(x)=\sin x-\sqrt{3} \cos x+1\), is onto, then the interval of \(S\) is (A) \([0,3]\) (B) \([-1,1]\) (C) \([0,1]\) (D) \([-1,3]\)

Step-by-Step Solution

Answer

The interval of \(S\) is \([-1, 3]\) (option D).

Step 1: Recall the definition of an onto function

An onto function (also known as a surjective function) requires that every element in the codomain is the image of at least one element from the domain. Thus, for the function to be onto, its range must cover the entire codomain.

Step 2: Express the function in terms of trigonometric identities

Given function is \(f(x) = \sin x - \sqrt{3} \cos x + 1\). Identify this as a linear combination of sine and cosine, which can be rewritten using the trigonometric identity \(a \sin x + b \cos x = R \sin(x + \alpha)\), where \(R = \sqrt{a^2 + b^2}\). Here, \(a = 1\) and \(b = -\sqrt{3}\).

Step 3: Calculate R and rewrite the function using sine

Calculate \(R\) where \(R = \sqrt{1^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = 2\). Therefore, \(f(x) = 2 \sin(x + \alpha) + 1\). To find \(\alpha\), use \(\tan \alpha = \frac{b}{a} = \frac{-\sqrt{3}}{1} = -\sqrt{3}\), which gives \(\alpha = -\frac{\pi}{3}\). The function now is \(f(x) = 2 \sin(x - \frac{\pi}{3}) + 1\).

Step 4: Determine the range of the sine function

The sine function \(\sin(x + \alpha)\) has a range of \([-1, 1]\). Thus, \(2 \sin(x - \frac{\pi}{3})\) will have a range of \([-2, 2]\).

Step 5: Adjust for the +1 constant

Since the function is \(f(x) = 2 \sin(x - \frac{\pi}{3}) + 1\), add the constant to the sine range: \([-2+1, 2+1] = [-1, 3]\).

Step 6: Confirm the function is onto

Since the computed range of \(f(x)\) is \([-1, 3]\), which matches one of the options given for the codomain \(S\), the function is indeed onto when \(S = [-1, 3]\).

Key Concepts

Trigonometric Identities

Trigonometric identities are valuable tools in mathematics, especially when dealing with expressions involving sine and cosine. One such identity is the linear combination equation \( a \sin x + b \cos x = R \sin(x + \alpha) \). This identity can transform complex trigonometric expressions into simpler forms.
In our problem, we use this identity to rewrite the given function \( f(x) = \sin x - \sqrt{3} \cos x + 1 \). Here, the coefficients \( a = 1 \) and \( b = -\sqrt{3} \) are identified. By applying the formula, we can find \( R = \sqrt{a^2 + b^2} = \sqrt{1^2 + (-\sqrt{3})^2} = 2 \). This simplifies the expression to \( f(x) = 2 \sin(x - \frac{\pi}{3}) + 1 \).
Mastering these identities helps solve a variety of problems by transforming them into more manageable forms.

Function Range

The function range is crucial when studying the behavior of functions. It tells us the possible outputs or values that a function can produce. Understanding the range is especially essential when determining if a function is onto, as it should match the entire codomain.
In the exercise, the sine function has the basic range of \([-1, 1]\). When modified by a constant, it accommodates a new range. For \( 2 \sin(x - \frac{\pi}{3}) \), the range extends to \([-2, 2]\). Adding an additional constant (like +1 in our example) shifts this range further, giving us \([-1, 3]\).
Comprehending how these changes affect the function's output is critical, as it ensures that we correctly identify and confirm onto functions.

Sine and Cosine

Sine and cosine are fundamental trigonometric functions, often featured in various mathematical problems, especially those involving periodic behavior or angle-related computations. The sine and cosine functions respectively represent the y and x coordinates on the unit circle.
In our function \( f(x) = \sin x - \sqrt{3} \cos x + 1 \), sine and cosine are combined linearly. By converting this combination into the form \( R \sin(x + \alpha) \), we can use these functions' properties more effectively. For instance, determining the angle \( \alpha \) involves understanding that \( \tan \alpha = \frac{b}{a} \) links \( b = -\sqrt{3} \) and \( a = 1 \) to find \( \alpha = -\frac{\pi}{3} \).
These details are not only fundamental in transforming and simplifying the functions but also in exploring their applications through problems like proving surjectivity.