Question

The equation $a\sin x + b\cos x = c$, where $\left| c \right| > \sqrt {{a^2} + {b^2}}$ has

1. A unique solution
2. Infinite number of solutions
3. No solution
4. None of the above

Answer 1

Hint: We will first consider the given equation and as we have to determine the category of the solution so, we will start by finding the value of $c$. Now compare this value with the given value that is, $\left| c \right| > \sqrt {{a^2} + {b^2}}$ and determine whether the solution has no solution or infinite solution or unique solution.

Complete step by step solution:
Consider the given equation, $a\sin x + b\cos x = c$.
Here, we have to check that the given equation has a unique solution or the infinite number of solutions or no solution.
We know that the maximum value of $a\sin x + b\cos x$ is $\sqrt {{a^2} + {b^2}}$ and minimum value of $a\sin x + b\cos x$ is $- \sqrt {{a^2} + {b^2}}$.
So, we can say that

$$\- \sqrt {{a^2} + {b^2}} < a\sin x + b\cos x < \sqrt {{a^2} + {b^2}} \\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left| {a\sin x + b\cos x} \right| \leqslant \sqrt {{a^2} + {b^2}} \\\$$

In this question $a\sin x + b\cos x = c$.
So, we have $\left| {a\sin x + b\cos x} \right| = \left| c \right|$
Therefore, $\left| c \right| \leqslant \sqrt {{a^2} + {b^2}}$.
But according to the given condition in question, $\left| c \right| > \sqrt {{a^2} + {b^2}}$ which is different from what we have obtained above.
Hence, $\left| c \right| \leqslant \sqrt {{a^2} + {b^2}}$ and $\left| c \right| > \sqrt {{a^2} + {b^2}}$ are two different scenarios.
And both of these conditions cannot be true simultaneously. So, we can say that the equation $a\sin x + b\cos x = c$, where $\left| c \right| > \sqrt {{a^2} + {b^2}}$ does not have a solution.
Therefore, the correct option is C.

Note: In this question, first of all, note that the range of $c$ is given. We can find another range of $c$ by evaluating the value of $c$ using the given equation. If the calculated range is same then we can say that there is a solution. If that calculated range is different from the given range, then it will be a contradiction to our hypothesis. Hence, there will be no solution. Also, the maximum value of $a\sin x + b\cos x$ is $\sqrt {{a^2} + {b^2}}$ and minimum value of $a\sin x + b\cos x$ is $- \sqrt {{a^2} + {b^2}}$.