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Question:
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The quadratic equation whose roots are $l$ and $m$, where
$$
\begin{aligned}
& l=\lim _{\theta \rightarrow 0}\left(\frac{3 \sin \theta-4 \sin ^2 \theta}{\theta}\right), \\
& m=\lim _{\theta \rightarrow 0} \frac{2 \tan \theta}{\theta\left(1-\tan ^2 \theta\right)}, \text { is }
\end{aligned}
$$
Options:
$$
\begin{aligned}
& l=\lim _{\theta \rightarrow 0}\left(\frac{3 \sin \theta-4 \sin ^2 \theta}{\theta}\right), \\
& m=\lim _{\theta \rightarrow 0} \frac{2 \tan \theta}{\theta\left(1-\tan ^2 \theta\right)}, \text { is }
\end{aligned}
$$
Solution:
1112 Upvotes
Verified Answer
The correct answer is:
$x^2-5 x+6=0$
We have,
$$
l=\lim _{\theta \rightarrow 0}\left[\frac{3 \sin \theta-4 \sin ^2 \theta}{\theta}\right]
$$
Using $L$-hospital's rule
$$
=\lim _{\theta \rightarrow 0} \frac{3 \cos \theta-8 \sin \theta \cos \theta}{1}=3
$$
and
$$
\begin{aligned}
m & =\lim _{\theta \rightarrow 0} \frac{2 \tan \theta}{\theta\left(1-\tan ^2 \theta\right)} \\
& =\lim _{\theta \rightarrow 0} \frac{\tan 2 \theta}{\theta} \\
& =\lim _{\theta \rightarrow 0} \frac{\tan 2 \theta}{2 \theta} \cdot 2=2
\end{aligned}
$$
The required quadratic equation is
$$
\begin{aligned}
& x^2-(l+m) x+l m=0 \\
& \Rightarrow \quad x^2-(3+2) x+3 \cdot 2=0 \\
& x^2-5 x+6=0 \\
&
\end{aligned}
$$
$$
l=\lim _{\theta \rightarrow 0}\left[\frac{3 \sin \theta-4 \sin ^2 \theta}{\theta}\right]
$$
Using $L$-hospital's rule
$$
=\lim _{\theta \rightarrow 0} \frac{3 \cos \theta-8 \sin \theta \cos \theta}{1}=3
$$
and
$$
\begin{aligned}
m & =\lim _{\theta \rightarrow 0} \frac{2 \tan \theta}{\theta\left(1-\tan ^2 \theta\right)} \\
& =\lim _{\theta \rightarrow 0} \frac{\tan 2 \theta}{\theta} \\
& =\lim _{\theta \rightarrow 0} \frac{\tan 2 \theta}{2 \theta} \cdot 2=2
\end{aligned}
$$
The required quadratic equation is
$$
\begin{aligned}
& x^2-(l+m) x+l m=0 \\
& \Rightarrow \quad x^2-(3+2) x+3 \cdot 2=0 \\
& x^2-5 x+6=0 \\
&
\end{aligned}
$$
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