MHT CET · Maths · Limits
The quadratic equation whose roots are \(m\) and \(n\), where \(m=\lim _{x \rightarrow 0} \frac{x \log (1+2 x)}{x \tan x}\) and \(n=\lim _{x \rightarrow 0} \frac{\log x+\log \left(\frac{(1+x)}{x}\right)}{x}\), is
- A \(x^2-x+2=0\)
- B \(x^2-3 x+2=0\)
- C \(x^2+x+2=0\)
- D \(x^2+3 x+2=0\)
Answer & Solution
Correct Answer
(B) \(x^2-3 x+2=0\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & m=\lim _{x \rightarrow 0} \frac{x \log (1+2 x)}{x \cdot \tan x}=\lim _{x \rightarrow 0} \frac{\frac{\log (1+2 x)}{2 x} \cdot 2 x}{\frac{\tan x}{x} \cdot x}=2 \\ & n=\lim _{x \rightarrow 0} \frac{\log x+\log \left(\frac{1+x}{x}\right)}{x}=\lim _{x \rightarrow 0} \frac{\log \left(x \times \frac{1+x}{x}\right)}{x} \\ & =\lim _{x \rightarrow 0} \frac{\log (1+x)}{x}=1\end{aligned}\)
now, required quadratic equation is
\(x^2-(m+n) x+m n=0\)
\(\Rightarrow x^2-3 x+2=0\)
now, required quadratic equation is
\(x^2-(m+n) x+m n=0\)
\(\Rightarrow x^2-3 x+2=0\)
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