Question

If the function \(f(x)\) is defined as below
\(f(x)=\begin{cases} \frac{(3^{x}+3^{-x}-2)\log (1+4x)}{(1-\cos 4x)\tan (2x)}, & x\neq 0 \\ \frac{(\log p)^{r}}{q}, & x=0 \end{cases}\)
is continuous at \(x=0\) then \(p+q-r=\ldots\)

• A. \(0\)
• B. \(3\)
• C. \(5\)
• D. \(7\)

Hint:

Evaluate the limit at 0 using standard limits and compare with (log p)^r / q.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Continuity at \(x=0\) needs \(\lim_{x\rightarrow 0}f(x)=f(0)\).

Step 2: Simplify numerator
\(3^{x}+3^{-x}-2=\left(3^{x/2}-3^{-x/2}\right)^{2}\approx (x\log 3)^{2}\), and \(\log(1+4x)\approx 4x\).

Step 3: Simplify denominator
\(1-\cos 4x=2\sin^{2}2x\approx 8x^{2}\) and \(\tan 2x\approx 2x\).

Step 4: Limit
\(\lim f(x)=\dfrac{x^{2}(\log 3)^{2}\cdot 4x}{8x^{2}\cdot 2x}=\dfrac{(\log 3)^{2}}{4}\).

Step 5: Match
Comparing with \(\dfrac{(\log p)^{r}}{q}\): \(p=3\), \(r=2\), \(q=4\).
\(p+q-r=3+4-2=5\).

Final Answer:
The value of p + q - r is 5. \[ \boxed{\text{(C) }5} \]