Problem 56 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The equation \(a \sin x+b \cos x=c\) where \(|c|>\sqrt{a^{2}+b^{2}}\) has: (A) a unique solution (B) infinite number of solutions (C) no solution (D) none of the above

Step-by-Step Solution

Answer

(C) no solution

Step 1: Understanding the Trigonometric Identity

The expression involving sine and cosine, \(a \sin x + b \cos x\), can be converted into a single trigonometric function. This is based on the identity where any linear combination of sine and cosine can be rewritten using the expression \(R \sin(x + \phi)\), where \(R = \sqrt{a^2 + b^2}\) and \(\phi\) is a phase angle that satisfies \(\cos \phi = \frac{a}{R}\) and \(\sin \phi = \frac{b}{R}\).

Step 2: Establish the Maximum Amplitude

The maximum value of the function \(R \sin(x + \phi)\) is \(R\). Therefore, the expression \(a \sin x + b \cos x\) can at most be equal to \(\sqrt{a^2+b^2}\) in magnitude. This means that \(-\sqrt{a^2+b^2} \leq a \sin x + b \cos x \leq \sqrt{a^2+b^2}\).

Step 3: Analyze the Given Condition

The problem states \(|c| > \sqrt{a^2+b^2}\). If \(|c|\) exceeds the maximum possible amplitude \(R\) of \(a \sin x + b \cos x\), the equation \(a \sin x+b \cos x=c\) cannot be satisfied under any condition, because \(c\) falls outside the possible range of values for the left-hand side of the equation.

Step 4: Conclude the Existence of Solutions

Since \(|c| > \sqrt{a^2+b^2}\), there is no value of \(x\) for which \(a \sin x + b \cos x = c\) holds true. Hence, under this condition, the equation has no solutions.

Key Concepts

Sine and Cosine

Sine and cosine are fundamental trigonometric functions that relate to the angles and sides of a right-angled triangle. They are periodic functions with a period of \(2\pi\), meaning they repeat their pattern every \(2\pi\) units. Sine (\