MHT CET · Maths · Continuity and Differentiability
If \(f(x)=\frac{(81)^{x}-(9)^{x}}{(k)^{x}-1}\) if \(x \neq 0\)
\(=2 \quad\) if \(\quad x=0\)
is continuous at \(x=0\), then the value of \(k\) is
- A 3
- B 9
- C 2
- D 4
Answer & Solution
Correct Answer
(A) 3
Step-by-step Solution
Detailed explanation
\(\lim _{x \rightarrow 0} \frac{81^{x}-9^{x}}{k^{x}-1}=2\)
\(\therefore \lim _{x \rightarrow 0} \frac{9^{x}\left(9^{x}-1\right)}{k^{x}-1}=2 \Rightarrow \frac{\left(\lim _{x \rightarrow 0} 9^{x}\right)\left(\lim _{x \rightarrow 0} \frac{9^{x}-1}{x}\right)}{\lim _{x \rightarrow 0} \frac{k^{x}-1}{x}}=2\)
\(\frac{\log 9}{\log k}=2 \Rightarrow \log 9=2 \log k \Rightarrow \log k^{2}=\log 9 \Rightarrow k^{2}\) \(=9 \Rightarrow k=3\)
\(\therefore \lim _{x \rightarrow 0} \frac{9^{x}\left(9^{x}-1\right)}{k^{x}-1}=2 \Rightarrow \frac{\left(\lim _{x \rightarrow 0} 9^{x}\right)\left(\lim _{x \rightarrow 0} \frac{9^{x}-1}{x}\right)}{\lim _{x \rightarrow 0} \frac{k^{x}-1}{x}}=2\)
\(\frac{\log 9}{\log k}=2 \Rightarrow \log 9=2 \log k \Rightarrow \log k^{2}=\log 9 \Rightarrow k^{2}\) \(=9 \Rightarrow k=3\)
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