QJEEM20Mace45: JEE Main 2020 Mathematics - Class 12 JEEM20Mace45 Solution: If = 2 -k then the value of k is _______ .

Question

Verified Solution • 2025–26 Session moderate

If

\mathop {\lim }\limits_{x \to 0} \left\{ {{1 \over {{x^8}}}\left( {1 - \cos {{{x^2}} \over 2} - \cos {{{x^2}} \over 4} + \cos {{{x^2}} \over 2}\cos {{{x^2}} \over 4}} \right)} \right\}

= 2^-k

then the value of k is _______ .

Accepted Answer

\mathop {\lim }\limits_{x \to 0} \left\{ {{1 \over {{x^8}}}\left( {1 - \cos {{{x^2}} \over 2} - \cos {{{x^2}} \over 4} + \cos {{{x^2}} \over 2}\cos {{{x^2}} \over 4}} \right)} \right\} = {2^{ - k}}

\Rightarrow

\mathop {\lim }\limits_{x \to 0} {{\left( {1 - \cos {{{x^2}} \over 2}} \right)} \over {4{{\left( {{{{x^2}} \over 2}} \right)}^2}}}{{\left( {1 - \cos {{{x^2}} \over 4}} \right)} \over {16{{\left( {{{{x^2}} \over 4}} \right)}^2}}}

=

{2^{ - k}}

\Rightarrow

\mathop {\lim }\limits_{x \to 0} {{2{{\sin }^2}{{{x^2}} \over 4}} \over {16{{\left( {{{{x^2}} \over 4}} \right)}^2}}} \times {{2{{\sin }^2}{{{x^2}} \over 8}} \over {64{{\left( {{{{x^2}} \over 8}} \right)}^2}}}

= 2^-k

\Rightarrow

{1 \over 8} \times {1 \over {32}} = {2^{ - k}}

\Rightarrow

2^-8 = 2^-k

\Rightarrow

k = 8

Step-by-Step Verified Solution