Problem 59 - 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) \([-1,3]\) (B) \([-1,1]\) (C) \([0,1]\) (D) \([0,3]\)

Step-by-Step Solution

Answer

The interval of S is (A) [-1, 3].

Step 1: Identify Components

We are given the function \(f(x) = \sin x - \sqrt{3} \cos x + 1\). This is a trigonometric function involving sine and cosine, and it's shifted by 1.

Step 2: Simplify the Expression

Notice that the function \(f(x) = \sin x - \sqrt{3} \cos x\) can be rewritten in terms of a single trigonometric function. This can be achieved by expressing it as \(R \sin(x + \alpha)\), where \(R\) and \(\alpha\) are constants. Here, \(R = \sqrt{1^2 + (-\sqrt{3})^2} = 2\) and \(\alpha = \tan^{-1}\left(\frac{-\sqrt{3}}{1}\right) = -\frac{\pi}{3}\). Thus, the function becomes \(f(x) = 2\sin(x - \frac{\pi}{3}) + 1\).

Step 3: Determine the Range of the Transformed Function

Since \(\sin(x)\) ranges from -1 to 1, \(2\sin(x - \frac{\pi}{3})\) will range from \(-2\) to \(2\). Therefore, \(f(x) = 2\sin(x - \frac{\pi}{3}) + 1\) will range from \(-2 + 1\) to \(2 + 1\), which is \([-1, 3]\).

Step 4: Determine Onto Interval for S

For a function to be onto, its range must cover the entire interval of the codomain \(S\). Since the range we found is \([-1, 3]\), and given the options, the function covers the interval (A) \([-1, 3]\).

Key Concepts

Range of a Function

Understanding the range of a function is essential to various areas of mathematics. The range is all possible output values of a function based on its defined inputs. In simpler terms, it's the set of all possible results you can get from a function by plugging in different values.

For the function given in the exercise, the range is determined by the transformation made on the standard sine function, transforming it into a linear combination of sine and cosine. Here, the trigonometric transformation results in a new function which affects the traditional range of \([-1, 1]\) (the usual range of a pure sine or cosine wave). This function's range shifts due to the constants and transformations involved.

• First, note the trigonometric function described: \(f(x) = \sin x - \sqrt{3} \cos x + 1\)
• This results in a transformation of the sine wave's normal range \([-1, 1]\).
• By expressing it in the standardized \(R \sin(x + \alpha)\) form, we arrive at the adjusted range of \([-1, 3]\).

Calculating the range precisely depends on understanding these alterations to the original trigonometric function. With the provided transformations, the original limited range expands or shifts to account for the entire output seen in the function.

Sine and Cosine Transformation

Transforming a function that involves sine and cosine can initially seem challenging, but it's easier when you break it down. Combining sine and cosine in an expression like \(f(x) = \sin x - \sqrt{3} \cos x\) involves a process called trigonometric transformation.

The goal is to represent the combination of sine and cosine as a single trigonometric function:

• Use the identity \(R \sin(x + \alpha)\), where \(R\) is the amplitude or resultant amplitude and \(\alpha\) is the phase shift.
• Calculate \(R\) using the formula \(R = \sqrt{a^2 + b^2}\). In this case, with \(a = 1\) and \(b = -\sqrt{3}\), \(R = \sqrt{1 + 3} = 2\).
• Determine the angle \(\alpha\) as \(\alpha = \tan^{-1}\left(\frac{b}{a}\right)\), resulting in \(-\frac{\pi}{3}\) for the exercise.

This transformation simplifies the dual trigonometric components into one neat sine function, facilitating the calculation of other properties such as the range or critical points of the function itself.

Onto Function

An onto function, also known as a surjective function, is a function where every element in the codomain is mapped to by at least one element in the domain. In other words, for a function to be onto, every possible output value within the specified range must be achievable by plugging some input value into the function.

For the exercise function \(f(x) = 2 \sin(x - \frac{\pi}{3}) + 1\), it was required to determine if it was onto by identifying if the full range \([-1, 3]\) could be produced. Since:

• The transformation expands the range of \([-2, 2]\) to \([-1, 3]\) due to the \(+1\) shift.
• All numbers between and including \(-1\) and \(+3\) are within reach of the transformed function, \(f(x)\).
• The codomain \(S\) in this exercise matches this range, confirming it as an onto function.

Thus, every potential y-value in the interval \([-1, 3]\) is obtained, satisfying the condition for the function being onto.