Question

JEE Mains Maths question is in the image

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

TO DETERMINE

k when

$\displaystyle \lim_{x \to 0} \frac{(1- \cos \frac{ {x}^{2} }{2} ) (1- \cos \frac{ {x}^{2} }{4} ) }{ {x}^{8} } = {2}^{ - k} \: \: ...(1)$

EVALUATION

$\displaystyle \lim_{x \to 0} \frac{(1- \cos \frac{ {x}^{2} }{2} ) (1- \cos \frac{ {x}^{2} }{4} ) }{ {x}^{8} }$

$= \displaystyle \lim_{x \to 0} \frac{(2 { \sin}^{2} \frac{ {x}^{2} }{2} ) (2 { \sin}^{2} \frac{ {x}^{2} }{4} )}{ {x}^{8} }$

$= 4 \times \displaystyle \lim_{x \to 0} \frac{( { \sin}^{2} \frac{ {x}^{2} }{2} ) ( { \sin}^{2} \frac{ {x}^{2} }{4} )}{ {x}^{8} }$

First Limit

$= \displaystyle \lim_{x \to 0} \frac{({ \sin}^{2} \frac{ {x}^{2} }{2} ) }{ {x}^{4} }$

Let

$\displaystyle \: u = \frac{ {x}^{2} }{2}$

Then

$u \to \: 0 \: \: as \: \: x \to \: 0$

So

$\displaystyle \lim_{x \to 0} \frac{{ \sin}^{2} \frac{ {x}^{2} }{2} }{ {x}^{4} }$

$= \displaystyle \lim_{u \to 0} \frac{{ \sin}^{2} u }{ 4{u}^{2} }$

$= \frac{1}{4} \times \displaystyle \lim_{u \to 0} \frac{{ \sin} u }{ {u}} \times \lim_{u \to 0} \frac{{ \sin} u }{ {u}}$

$= \displaystyle \: \frac{1}{4} \times 1 \times 1$

$= \displaystyle \: \frac{1}{4}$

Second Limit

$= \displaystyle \lim_{x \to 0} \frac{ { \sin}^{2} \frac{ {x}^{2} }{4} }{ {x}^{4} }$

Let

$\displaystyle \: u = \frac{ {x}^{2} }{4}$

Then

$u \to \: 0 \: \: as \: \: x \to \: 0$

So

$\displaystyle \lim_{x \to 0} \frac{ { \sin}^{2} \frac{ {x}^{2} }{4} }{ {x}^{4} }$

$= \displaystyle \lim_{u \to 0} \frac{{ \sin}^{2} u }{ 16{u}^{2} }$

$= \displaystyle\frac{1}{16} \times \lim_{u \to 0} \frac{{ \sin} u }{ {u}} \times \lim_{u \to 0} \frac{{ \sin} u }{ {u}}$

$= \displaystyle\frac{1}{16} \times1 \times 1$

$= \displaystyle\frac{1}{16}$

Hence from above

$4 \times \displaystyle \frac{1}{4} \times \frac{1}{16} = {2}^{ - k}$

$\implies \: \displaystyle \frac{1}{16} = {2}^{ - k}$

$\implies \: \displaystyle {2}^{ - k} = {2}^{ - 4}$

Hence

$k = 4$