Question

If the real valued function \( f(x) = \begin{cases} \frac{\cos 3x - \cos x}{x \sin x}, & \text{if } x < 0 \\ p, & \text{if } x = 0 \\ \frac{\log(1 + q \sin x)}{x}, & \text{if } x > 0 \end{cases} \) is continuous at \( x = 0 \), then \( p + q = \) 

• A. 4
• B. -4
• C. 8
• D. -8

Hint:

For continuity problems, the core task is to evaluate the left-hand limit, the right-hand limit, and the function value at the point. Utilize standard limits like $\lim_{x\to 0} \frac{\sin x}{x}=1$ and $\lim_{x\to 0} \frac{\log(1+x)}{x}=1$ to simplify calculations.

Answer 1

The Correct Option is D

For the function $f(x)$ to be continuous at $x=0$, we must have:
Left-Hand Limit (LHL) = Right-Hand Limit (RHL) = $f(0)$.
Step 1: Calculate the LHL.
LHL = $\lim_{x\to 0^-} \frac{\cos 3x - \cos x}{x \sin x}$.
Using the trigonometric identity $\cos C - \cos D = -2\sin(\frac{C+D}{2})\sin(\frac{C-D}{2})$:
LHL = $\lim_{x\to 0^-} \frac{-2\sin(\frac{3x+x}{2})\sin(\frac{3x-x}{2})}{x \sin x} = \lim_{x\to 0^-} \frac{-2\sin(2x)\sin(x)}{x \sin x}$.
LHL = $\lim_{x\to 0^-} \frac{-2\sin(2x)}{x}$.
Using the standard limit $\lim_{\theta\to 0} \frac{\sin \theta}{\theta} = 1$:
LHL = $\lim_{x\to 0^-} -2 \cdot \frac{\sin(2x)}{2x} \cdot 2 = -2 \cdot (1) \cdot 2 = -4$.
Step 2: Calculate the RHL.
RHL = $\lim_{x\to 0^+} \frac{\log(1+q \sin x)}{x}$.
Using the standard limit $\lim_{y\to 0} \frac{\log(1+y)}{y}=1$:
RHL = $\lim_{x\to 0^+} \frac{\log(1+q \sin x)}{q \sin x} \cdot \frac{q \sin x}{x} = (1) \cdot q \cdot \lim_{x\to 0^+} \frac{\sin x}{x} = q \cdot (1) = q$.
Step 3: Apply the continuity condition.
We have $f(0)=p$.
So, LHL = RHL = $f(0) \implies -4 = q = p$.
This gives us $p=-4$ and $q=-4$.
The required value is $p+q = (-4) + (-4) = -8$.