JEE MAIN - Mathematics (2021 - 17th March Morning Shift - No. 20)
If the function $$f(x) = {{\cos (\sin x) - \cos x} \over {{x^4}}}$$ is continuous at each point in its domain and $$f(0) = {1 \over k}$$, then k is ____________.
Answer
6
Explanation
$$\mathop {\lim }\limits_{x \to 0} f\left( x \right) = \mathop {\lim }\limits_{x \to 0} {{\cos \left( {\sin x} \right) - \cos x} \over {{x^4}}}$$
$$ \Rightarrow $$ $${1 \over k} = \mathop {\lim }\limits_{x \to 0} {{2\sin \left( {{{\sin x + x} \over 2}} \right)\sin \left( {{{x - \sin x} \over 2}} \right)} \over {{x^4}}}$$
= $$\mathop {\lim }\limits_{x \to 0} {{2\sin \left( {{{x + \sin x} \over 2}} \right)} \over {\left( {{{x + \sin x} \over 2}} \right)}} \times {{\sin \left( {{{x - \sin x} \over 2}} \right)} \over {\left( {{{x - \sin x} \over 2}} \right)}} \times {{{x^2} - {{\sin }^2}x} \over {4{x^4}}}$$
= $$\mathop {\lim }\limits_{x \to 0} 2 \times 1 \times \left( {{{x + \sin x} \over x}} \right)\left( {{{x - \sin x} \over {{x^3}}}} \right) \times {1 \over 4}$$
= $$\mathop {\lim }\limits_{x \to 0} 2 \times 1 \times \left( {{{1 + \cos x} \over 1}} \right)\left( {{{1 - \cos x} \over {3{x^2}}}} \right) \times {1 \over 4}$$
= $$\mathop {\lim }\limits_{x \to 0} 2 \times 1 \times \left( {{{1 + 1} \over 1}} \right)\left( {{{1 - \cos x} \over {3{x^2}}}} \right) \times {1 \over 4}$$
= $$\mathop {\lim }\limits_{x \to 0} 2 \times 1 \times \left( {{{1 + 1} \over 1}} \right)\left( {{{1 + \sin x} \over {6x}}} \right) \times {1 \over 4}$$
= $$2 \times 2 \times {1 \over 6} \times {1 \over 4}$$ = $${1 \over 6}$$
$$ \Rightarrow $$ $${1 \over k} = \mathop {\lim }\limits_{x \to 0} {{2\sin \left( {{{\sin x + x} \over 2}} \right)\sin \left( {{{x - \sin x} \over 2}} \right)} \over {{x^4}}}$$
= $$\mathop {\lim }\limits_{x \to 0} {{2\sin \left( {{{x + \sin x} \over 2}} \right)} \over {\left( {{{x + \sin x} \over 2}} \right)}} \times {{\sin \left( {{{x - \sin x} \over 2}} \right)} \over {\left( {{{x - \sin x} \over 2}} \right)}} \times {{{x^2} - {{\sin }^2}x} \over {4{x^4}}}$$
= $$\mathop {\lim }\limits_{x \to 0} 2 \times 1 \times \left( {{{x + \sin x} \over x}} \right)\left( {{{x - \sin x} \over {{x^3}}}} \right) \times {1 \over 4}$$
= $$\mathop {\lim }\limits_{x \to 0} 2 \times 1 \times \left( {{{1 + \cos x} \over 1}} \right)\left( {{{1 - \cos x} \over {3{x^2}}}} \right) \times {1 \over 4}$$
= $$\mathop {\lim }\limits_{x \to 0} 2 \times 1 \times \left( {{{1 + 1} \over 1}} \right)\left( {{{1 - \cos x} \over {3{x^2}}}} \right) \times {1 \over 4}$$
= $$\mathop {\lim }\limits_{x \to 0} 2 \times 1 \times \left( {{{1 + 1} \over 1}} \right)\left( {{{1 + \sin x} \over {6x}}} \right) \times {1 \over 4}$$
= $$2 \times 2 \times {1 \over 6} \times {1 \over 4}$$ = $${1 \over 6}$$
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