JEE MAIN - Mathematics (2020 - 7th January Evening Slot - No. 4)
The value of $$\alpha $$ for which
$$4\alpha \int\limits_{ - 1}^2 {{e^{ - \alpha \left| x \right|}}dx} = 5$$, is:
$$4\alpha \int\limits_{ - 1}^2 {{e^{ - \alpha \left| x \right|}}dx} = 5$$, is:
$${\log _e}2$$
$${\log _e}\sqrt 2 $$
$${\log _e}\left( {{4 \over 3}} \right)$$
$${\log _e}\left( {{3 \over 2}} \right)$$
Explanation
$$4\alpha \int\limits_{ - 1}^2 {{e^{ - \alpha \left| x \right|}}dx} = 5$$
$$ \Rightarrow $$ $$4\alpha \int\limits_{ - 1}^0 {{e^{\alpha x}}dx} + 4\alpha \int\limits_0^2 {{e^{ - \alpha x}}dx} $$ = 5
$$ \Rightarrow $$ $$4\alpha \left[ {{{{e^{\alpha x}}} \over \alpha }} \right]_{ - 1}^0 + 4\alpha \left[ {{{{e^{ - \alpha x}}} \over { - \alpha }}} \right]_0^2$$ = 5
$$ \Rightarrow $$ $$4{e^{ - 2\alpha }} + 4{e^{ - \alpha }} - 3$$ = 0
Let e-$$\alpha $$ = t
$$ \therefore $$ 4t2 + 4t - 3 = 0
$$ \Rightarrow $$ t = $${1 \over 2}$$
$$ \therefore $$ e-$$\alpha $$ = $${1 \over 2}$$
$$ \Rightarrow $$ $$\alpha $$ = $${\log _e}2$$
$$ \Rightarrow $$ $$4\alpha \int\limits_{ - 1}^0 {{e^{\alpha x}}dx} + 4\alpha \int\limits_0^2 {{e^{ - \alpha x}}dx} $$ = 5
$$ \Rightarrow $$ $$4\alpha \left[ {{{{e^{\alpha x}}} \over \alpha }} \right]_{ - 1}^0 + 4\alpha \left[ {{{{e^{ - \alpha x}}} \over { - \alpha }}} \right]_0^2$$ = 5
$$ \Rightarrow $$ $$4{e^{ - 2\alpha }} + 4{e^{ - \alpha }} - 3$$ = 0
Let e-$$\alpha $$ = t
$$ \therefore $$ 4t2 + 4t - 3 = 0
$$ \Rightarrow $$ t = $${1 \over 2}$$
$$ \therefore $$ e-$$\alpha $$ = $${1 \over 2}$$
$$ \Rightarrow $$ $$\alpha $$ = $${\log _e}2$$
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