MHT CET · Maths · Continuity and Differentiability
If the function \(f(x)=\frac{\log 10+\log (0.1+2 x)}{2 x}\) if \(x \neq 0\)
\(=k \quad\) if \(x=0\)
is continuous at \(x=0\), then \(k+2=\)
- A 2
- B 10
- C 12
- D 11
Answer & Solution
Correct Answer
(C) 12
Step-by-step Solution
Detailed explanation
\(\lim _{x \rightarrow 0} \frac{\log 10+\log (0.1+2 x)}{2 x}=k\)
\(\therefore \lim _{x \rightarrow 0} \frac{\log (1+20 x)}{2 x}=k \Rightarrow \lim _{x \rightarrow 0} \log (1+20 x)^{\frac{1}{2 x}}=k\)
\(\therefore \log \lim _{x \rightarrow 0}\left[(1+20 x)^{\frac{1}{20 x}}\right]^{10}=k \Rightarrow \log e^{10}=k\) \(\Rightarrow k=10\) \(\Rightarrow k+2=10+2=12\)
\(\therefore \lim _{x \rightarrow 0} \frac{\log (1+20 x)}{2 x}=k \Rightarrow \lim _{x \rightarrow 0} \log (1+20 x)^{\frac{1}{2 x}}=k\)
\(\therefore \log \lim _{x \rightarrow 0}\left[(1+20 x)^{\frac{1}{20 x}}\right]^{10}=k \Rightarrow \log e^{10}=k\) \(\Rightarrow k=10\) \(\Rightarrow k+2=10+2=12\)
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