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Limits — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question

Let

L=\displaystyle\lim_{x\to 1}\dfrac{\sin(6\cos^{-1}x)}{\sqrt{1-x^{2}}}

and

M=\displaystyle\lim_{x\to 1}\dfrac{1-\cos(6\cos^{-1}x)}{1-x^{2}}

. Which of the following is/are correct?

L+M=24

correct

\dfrac{M}{L}=3

correct

L-M=-12

correct

LM=108

correct

Solution

Step 1: Substitute

\cos^{-1}x=\theta

, so

x=\cos\theta

. As

x\to 1

\theta\to 0

. Step 2: Compute

L

L=\lim_{\theta\to 0}\dfrac{\sin 6\theta}{\sqrt{1-\cos^{2}\theta}}=\lim_{\theta\to 0}\dfrac{\sin 6\theta}{|\sin\theta|}=\lim_{\theta\to 0}\dfrac{\sin 6\theta}{\sin\theta}

Step 3: Use

\dfrac{\sin 6\theta}{\sin\theta}=\dfrac{\sin 6\theta}{6\theta}\cdot\dfrac{\theta}{\sin\theta}\cdot 6

L=1\cdot 1\cdot 6=6

Step 4: Compute

M

M=\lim_{\theta\to 0}\dfrac{1-\cos 6\theta}{1-\cos^{2}\theta}=\lim_{\theta\to 0}\dfrac{1-\cos 6\theta}{\sin^{2}\theta}

Step 5: Use

1-\cos 6\theta=2\sin^{2}3\theta

M=\lim_{\theta\to 0}\dfrac{2\sin^{2}3\theta}{\sin^{2}\theta}=2\lim_{\theta\to 0}\left(\dfrac{\sin 3\theta}{\sin\theta}\right)^{2}=2\cdot 9=18

Step 6: So

L=6

and

M=18

. Check options: (1)

L+M=6+18=24

. True. (2)

\dfrac{M}{L}=\dfrac{18}{6}=3

. True. (3)

L-M=6-18=-12

. True. (4)

LM=6\cdot 18=108

. True. Correct answers: (1), (2), (3) and (4)

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