AP EAMCET · Maths · Limits
\(l=\lim _{\theta \rightarrow 0}\left(\frac{3 \sin \theta-4 \sin ^3 \theta}{\theta}\right)\) and \(m=\lim _{\theta \rightarrow 0}\left(\frac{2 \tan \theta}{\theta\left(1-\tan ^2 \theta\right)}\right)\) is
- A \(x^2+5 x+6=0\)
- B \(x^2-5 x+6=0\)
- C \(x^2-5 x-6=0\)
- D \(x^2+5 x-6=0\)
Answer & Solution
Correct Answer
(B) \(x^2-5 x+6=0\)
Step-by-step Solution
Detailed explanation
\(l=\lim _{\theta \rightarrow 0}\left(\frac{\sin 3 \theta}{\theta}\right)=3\) \(m=\lim _{\theta \rightarrow 0}\left(\frac{\tan \theta}{\theta} \cdot \frac{2}{1-\tan ^2 \theta}\right)=1 \cdot \frac{2}{1-0}=2\) \(S=l+m=3+2=5\) \(P=l \cdot m=3 \cdot 2=6\) \(x^2-Sx+P=0\)…
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