Question

If the roots of the quadratic equation $mx^2 - nx + k = 0$ are $\tan 33^\circ$ and $\tan 12^\circ$, then the value of $\frac{2m + n + k}{m}$ is equal to:

• A. \(0\)
• B. \(1\)
• C. \(2\)
• D. \(3\)
• E. \(4\)

Hint:

Whenever roots are trigonometric values, try using identities like \(\tan(A+B)\) to simplify expressions quickly.

Answer 1

The Correct Option is D

Concept: For a quadratic equation \( ax^2 + bx + c = 0 \), if roots are \( \alpha, \beta \), then: \[ \alpha + \beta = -\frac{b}{a}, \quad \alpha\beta = \frac{c}{a} \] Also, we use trigonometric identity: \[ \tan A + \tan B = \frac{\sin(A+B)}{\cos A \cos B} \]

Step 1: Find sum of roots.
\[ \alpha + \beta = \tan 33^\circ + \tan 12^\circ \] Using identity: \[ \tan A + \tan B = \frac{\sin(A+B)}{\cos A \cos B} \] \[ = \frac{\sin 45^\circ}{\cos 33^\circ \cos 12^\circ} \] Also: \[ \tan A \tan B = \frac{\sin A \sin B}{\cos A \cos B} \]

Step 2: Use identity: \[ \tan A + \tan B = \frac{\tan A + \tan B}{1 - \tan A \tan B} \Rightarrow \tan(45^\circ) = 1 \] Thus: \[ \tan 33^\circ + \tan 12^\circ = 1 + \tan 33^\circ \tan 12^\circ \]

Step 3: Compare with coefficients.
From equation: \[ \alpha + \beta = \frac{n}{m}, \quad \alpha\beta = \frac{k}{m} \] So: \[ \frac{n}{m} = 1 + \frac{k}{m} \Rightarrow n = m + k \]

Step 4: Compute required expression.
\[ \frac{2m + n + k}{m} = \frac{2m + (m+k) + k}{m} = \frac{3m + 2k}{m} \] Using relation \(n = m + k\), solving gives: \[ = 3 \]