Question

Let p,q and r be real numbers such that $|r|\gt\sqrt{p^2+q^2}$.Then the equation $p\cos\theta+q\sin\theta=r$ has

• A. exactly one real solution.
• B. exactly two real solutions.
• C. infinite number of real solutions.
• D. no real solution.
• E. integer solution.

Answer 1

The Correct Option is D

Given:

• Real numbers $p, q, r$ with $|r| > \sqrt{p^2 + q^2}$
• Equation: $p \cos \theta + q \sin \theta = r$

Step 1: Rewrite the equation using trigonometric identity: $$p \cos \theta + q \sin \theta = \sqrt{p^2 + q^2} \cos(\theta - \alpha)$$ where $\alpha$ is an angle such that $\cos \alpha = \frac{p}{\sqrt{p^2 + q^2}}$ and $\sin \alpha = \frac{q}{\sqrt{p^2 + q^2}}$.

Step 2: Analyze the range of the left side: $$-\sqrt{p^2 + q^2} \leq p \cos \theta + q \sin \theta \leq \sqrt{p^2 + q^2}$$

Step 3: Compare with the given condition $|r| > \sqrt{p^2 + q^2}$:

• If $r > \sqrt{p^2 + q^2}$, the equation $\sqrt{p^2 + q^2} \cos(\theta - \alpha) = r$ has no solution since the maximum value of cosine is 1.
• Similarly, if $r < -\sqrt{p^2 + q^2}$, there's no solution since the minimum value of cosine is -1.

 

Conclusion: The equation $p \cos \theta + q \sin \theta = r$ has no real solution under the given condition.

The correct answer is (D) no real solution.