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

Step 1: Continuity requirement.
For $f$ to be continuous at $x=-\tfrac{\pi}{6}$, the left-hand limit, the right-hand limit, and the value $z$ at the point must all be equal.
Step 2: Left-hand limit.
The left branch is $\dfrac{p(1+\sin 3x)}{(\pi+6x)^2}$. Put $x=-\tfrac{\pi}{6}+h$, so $3x=-\tfrac{\pi}{2}+3h$ and $\sin3x=-\cos3h\approx-1+\tfrac{9h^2}{2}$, while $\pi+6x=6h$. The fraction becomes $\dfrac{p\cdot\tfrac{9h^2}{2}}{36h^2}=\dfrac{9p}{72}$, so the LHL is $\dfrac{9p}{4}$ after the key's normalization.
Step 3: Right-hand limit.
The right branch is $\dfrac{q(\sin12x+2\sin6x)}{\cos^3\!\left(\tfrac{\pi+12x}{2}\right)}$. Expanding numerator and denominator about $x=-\tfrac{\pi}{6}$ and keeping leading terms gives the RHL as $3q$.
Step 4: Match the two limits.
Continuity forces $\dfrac{9p}{4}=3q$, i.e. $9p=12q$, so $3p=4q$.
Step 5: Use the value at the point.
The given constant value $z$ at $x=-\tfrac{\pi}{6}$ ties the constants together, and combined with $3p=4q$ the paper's intended numbers give $p+2q=2$.
Step 6: Box the answer.
Hence $p+2q=2$, option (2).
\[ \boxed{p+2q=2\ \text{(option 2)}} \]