JEE MAIN - Mathematics (2021 - 22th July Evening Shift - No. 13)
Let f : R $$\to$$ R be defined as $$f(x) = \left\{ {\matrix{
{{{{x^3}} \over {{{(1 - \cos 2x)}^2}}}{{\log }_e}\left( {{{1 + 2x{e^{ - 2x}}} \over {{{(1 - x{e^{ - x}})}^2}}}} \right),} & {x \ne 0} \cr
{\alpha ,} & {x = 0} \cr
} } \right.$$
If f is continuous at x = 0, then $$\alpha$$ is equal to :
If f is continuous at x = 0, then $$\alpha$$ is equal to :
1
3
0
2
Explanation
For continuity
$$\mathop {\lim }\limits_{x \to 0} {{{x^3}} \over {4{{\sin }^4}x}}(\ln (1 + 2x{e^{ - 2x}}) - 2\ln (1 - x{e^{ - x}})) = \alpha $$
$$\mathop {\lim }\limits_{x \to 0} {1 \over {4x}}[2x{e^{ - 2x}} + 2x{e^{ - x}}] = \alpha $$
$$ = {1 \over 4}(4) = \alpha = 1$$
$$\mathop {\lim }\limits_{x \to 0} {{{x^3}} \over {4{{\sin }^4}x}}(\ln (1 + 2x{e^{ - 2x}}) - 2\ln (1 - x{e^{ - x}})) = \alpha $$
$$\mathop {\lim }\limits_{x \to 0} {1 \over {4x}}[2x{e^{ - 2x}} + 2x{e^{ - x}}] = \alpha $$
$$ = {1 \over 4}(4) = \alpha = 1$$
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