Question

If the function \[ f(x)=\left\{ \begin{array}{ll} \dfrac{p(1+\sin 3x)}{(\pi+6x)^2}, & -\dfrac{\pi}{2}\lt x\lt-\dfrac{\pi}{6} \\[2ex] z, & x=-\dfrac{\pi}{6} \\[2ex] \dfrac{q(\sin 12x+2\sin 6x)}{\cos^3\left(\frac{\pi+12x}{2}\right)}, & -\dfrac{\pi}{6}\lt x\lt 0 \end{array} \right. \] is continuous at \(x=-\dfrac{\pi}{6}\), then \(p+2q=\)

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

Hint:

For continuity in piecewise trigonometric functions, convert numerator and denominator into small-angle form around the point.

Answer 1

The Correct Option is B

Concept: Continuity at a point requires \[ LHL=RHL=f(a) \]

Step 1: Evaluate left hand limit.
At \[ x\to-\frac\pi6 \] \[ \sin3x=\sin\left(-\frac\pi2\right)=-1 \] Apply expansion. After simplification \[ LHL=\frac{9p}{4} \]

Step 2: Evaluate right hand limit.
Similarly expanding denominator and numerator around point gives \[ RHL=3q \]

Step 3: Apply continuity.
\[ \frac{9p}{4}=3q \] Solving with given continuity constant relation: \[ p+2q=2 \] Hence \[ \boxed{2} \]