The quadratic equation whose roots are l and | vedclass

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(B) First,we calculate $l$:$l = \lim_{	heta ightarrow 0} \left( \frac{3 \sin 	heta - 4 \sin^2 	heta}{	heta} ight) = \lim_{	heta ightarrow 0

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$x^2-5x+6=0$

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(B) First,we calculate $l$:$l = \lim_{	heta ightarrow 0} \left( \frac{3 \sin 	heta - 4 \sin^2 	heta}{	heta} ight) = \lim_{	heta ightarrow 0} \left( 3 \frac{\sin 	heta}{	heta} - 4 \sin 	heta \cdot \frac{\sin 	heta}{	heta} ight)$Since $\lim_{	heta ightarrow 0} \frac{\sin 	heta}{	heta} = 1$,we have $l = 3(1) - 4(0)(1) = 3$.Next,we calculate $m$:$m = \lim_{	heta ightarrow 0} \frac{2 	an 	heta}{	heta(1 - 	an^2 	heta)} = \lim_{	heta ightarrow 0} \frac{	an(2	heta)}{	heta}$Using the limit $\lim_{x ightarrow 0} \frac{	an(kx)}{x} = k$,we get $m = 2$.The quadratic equation with roots $l=3$ and $m=2$ is given by $x^2 - (l+m)x + lm = 0$.Substituting the values,we get $x^2 - (3+2)x + (3 	imes 2) = 0$,which simplifies to $x^2 - 5x + 6 = 0$.

The quadratic equation whose roots are $l$ and $m$,where
$\begin{aligned}
& l=\lim _{\theta \rightarrow 0}\left(\frac{3 \sin \theta-4 \sin ^2 \theta}{\theta}\right), \\
& m=\lim _{\theta \rightarrow 0} \frac{2 \tan \theta}{\theta\left(1-\tan ^2 \theta\right)}, \text{ is}
\end{aligned}$

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The quadratic equation whose roots are $l$ and $m$,where$\begin{aligned}& l=\lim _{\theta \rightarrow 0}\left(\frac{3 \sin \theta-4 \sin ^2 \theta}{\theta}\right), \\& m=\lim _{\theta \rightarrow 0} \frac{2 \tan \theta}{\theta\left(1-\tan ^2 \theta\right)}, \text{ is}\end{aligned}$