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.

Answer 2

The given equation is:

\[ p\cos\theta + q\sin\theta = r \]

We can rewrite this equation using the auxiliary angle method. Let \( p = R\cos\phi \) and \( q = R\sin\phi \), where \( R = \sqrt{p^2 + q^2} \) and \( \phi = \arctan(q/p) \).

Then the equation becomes:

\[ R\cos\phi\cos\theta + R\sin\phi\sin\theta = r \] \[ R\cos(\theta - \phi) = r \] \[ \cos(\theta - \phi) = \frac{r}{R} = \frac{r}{\sqrt{p^2 + q^2}} \]

Since \( |r| > \sqrt{p^2 + q^2} \), we have \( \left|\frac{r}{R}\right| > 1 \). 

However, the range of the cosine function is \([-1, 1]\). 

Therefore, there are no real solutions for \( \theta \) in this case. 

The equation has no solution.