The quadratic equation whose roots are m and | vedclass

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(B) First,we calculate $m$: $m = \lim_{x ightarrow 0} \frac{x \log(1+2x)}{x 	an x} = \lim_{x ightarrow 0} \frac{\log(1+2x)}{	an x} = \lim_{x i

Vedclass · Accepted Answer

$x^2-3x+2=0$

Vedclass · Answer

(B) First,we calculate $m$: $m = \lim_{x ightarrow 0} \frac{x \log(1+2x)}{x 	an x} = \lim_{x ightarrow 0} \frac{\log(1+2x)}{	an x} = \lim_{x ightarrow 0} \left( \frac{\log(1+2x)}{2x} 	imes \frac{x}{	an x} 	imes 2 ight) = 1 	imes 1 	imes 2 = 2$. Next,we calculate $n$: $n = \lim_{x ightarrow 0} \frac{\log x + \log(\frac{1+x}{x})}{x} = \lim_{x ightarrow 0} \frac{\log(x 	imes \frac{1+x}{x})}{x} = \lim_{x ightarrow 0} \frac{\log(1+x)}{x} = 1$. The quadratic equation with roots $m=2$ and $n=1$ is given by $x^2 - (m+n)x + mn = 0$. Substituting the values,we get $x^2 - (2+1)x + (2 	imes 1) = 0$,which simplifies to $x^2 - 3x + 2 = 0$.

The quadratic equation whose roots are $m$ and $n$,where $m = \lim_{x \rightarrow 0} \frac{x \log(1+2x)}{x \tan x}$ and $n = \lim_{x \rightarrow 0} \frac{\log x + \log(\frac{1+x}{x})}{x}$,is

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