JEE MAIN - Mathematics (2020 - 3rd September Morning Slot - No. 20)
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 _______ .
then the value of k is _______ .
Answer
8
Explanation
$$\mathop {\lim }\limits_{x \to 0} \left\{ {{1 \over {{x^8}}}\left( {1 - \cos {{{x^2}} \over 2} - \cos {{{x^2}} \over 4} + \cos {{{x^2}} \over 2}\cos {{{x^2}} \over 4}} \right)} \right\} = {2^{ - k}}$$
$$ \Rightarrow $$ $$\mathop {\lim }\limits_{x \to 0} {{\left( {1 - \cos {{{x^2}} \over 2}} \right)} \over {4{{\left( {{{{x^2}} \over 2}} \right)}^2}}}{{\left( {1 - \cos {{{x^2}} \over 4}} \right)} \over {16{{\left( {{{{x^2}} \over 4}} \right)}^2}}} $$ = $${2^{ - k}}$$
$$ \Rightarrow $$ $$\mathop {\lim }\limits_{x \to 0} {{2{{\sin }^2}{{{x^2}} \over 4}} \over {16{{\left( {{{{x^2}} \over 4}} \right)}^2}}} \times {{2{{\sin }^2}{{{x^2}} \over 8}} \over {64{{\left( {{{{x^2}} \over 8}} \right)}^2}}}$$ = 2-k
$$ \Rightarrow $$ $$ {1 \over 8} \times {1 \over {32}} = {2^{ - k}}$$
$$ \Rightarrow $$ 2-8 = 2-k
$$ \Rightarrow $$ k = 8
$$ \Rightarrow $$ $$\mathop {\lim }\limits_{x \to 0} {{\left( {1 - \cos {{{x^2}} \over 2}} \right)} \over {4{{\left( {{{{x^2}} \over 2}} \right)}^2}}}{{\left( {1 - \cos {{{x^2}} \over 4}} \right)} \over {16{{\left( {{{{x^2}} \over 4}} \right)}^2}}} $$ = $${2^{ - k}}$$
$$ \Rightarrow $$ $$\mathop {\lim }\limits_{x \to 0} {{2{{\sin }^2}{{{x^2}} \over 4}} \over {16{{\left( {{{{x^2}} \over 4}} \right)}^2}}} \times {{2{{\sin }^2}{{{x^2}} \over 8}} \over {64{{\left( {{{{x^2}} \over 8}} \right)}^2}}}$$ = 2-k
$$ \Rightarrow $$ $$ {1 \over 8} \times {1 \over {32}} = {2^{ - k}}$$
$$ \Rightarrow $$ 2-8 = 2-k
$$ \Rightarrow $$ k = 8
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