Question

For \(\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\), if \(2\cos \theta + \sin \theta = 1\), and \(7\cos \theta + 6 \sin \theta = k\), then the possible values of \(k\) are:

• A. \(8, -2\)
• B. \(6, 2\)
• C. \(12, 4\)
• D. \(7, 6\)

Hint:

Use substitution and the Pythagorean identity to find possible values of expressions with trigonometric constraints.

Answer 1

The Correct Option is B

Step 1: Express \(2\cos \theta + \sin \theta = 1\)
Rewrite as: \[ a = 2, \quad b = 1, \quad a \cos \theta + b \sin \theta = 1 \] Step 2: Maximum and minimum of \(a \cos \theta + b \sin \theta\)
\[ R = \sqrt{a^2 + b^2} = \sqrt{4 + 1} = \sqrt{5} \] The equation holds for some \(\theta\) if \(|1| \le R\), which is true. Step 3: Write \(7 \cos \theta + 6 \sin \theta\) in terms of \(a \cos \theta + b \sin \theta\)
Let: \[ X = \cos \theta, \quad Y = \sin \theta \] Given constraint: \[ 2X + Y = 1 \] We want: \[ k = 7X + 6Y \] Step 4: Express \(X\) and \(Y\)
From constraint: \[ Y = 1 - 2X \] Substitute into \(k\): \[ k = 7X + 6(1 - 2X) = 7X + 6 - 12X = 6 - 5X \] Step 5: Use \(X^2 + Y^2 = 1\)
\[ X^2 + (1 - 2X)^2 = 1 \] \[ X^2 + 1 - 4X + 4X^2 = 1 \] \[ 5X^2 - 4X = 0 \] \[ X(5X - 4) = 0 \] Step 6: Solve for \(X\)
\[ X = 0 \quad \text{or} \quad X = \frac{4}{5} \] Step 7: Calculate \(k\)
If \(X=0\), then: \[ k = 6 - 5 \times 0 = 6 \] If \(X = \frac{4}{5}\), then: \[ k = 6 - 5 \times \frac{4}{5} = 6 - 4 = 2 \]