JEE MAIN - Mathematics (2020 - 6th September Morning Slot - No. 9)
If I1 = $$\int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{100}}} dx$$ and
I2 = $$\int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{101}}} dx$$ such
that I2 = $$\alpha $$I1 then $$\alpha $$ equals to :
I2 = $$\int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{101}}} dx$$ such
that I2 = $$\alpha $$I1 then $$\alpha $$ equals to :
$${{5051} \over {5050}}$$
$${{5050} \over {5051}}$$
$${{5050} \over {5049}}$$
$${{5049} \over {5050}}$$
Explanation
I2 = $$\int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{101}}} dx$$
I2 = $$\int\limits_0^1 {\left( {1 - {x^{50}}} \right){{\left( {1 - {x^{50}}} \right)}^{100}}dx} $$
I2 = $$\int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{100}}dx} - \int\limits_0^1 {{x^{50}}{{\left( {1 - {x^{50}}} \right)}^{100}}dx} $$
I2 = I1 - $$\int\limits_0^1 {x.{x^{49}}{{\left( {1 - {x^{50}}} \right)}^{100}}dx} $$
Now apply IBP
I2 = I1 -
$$\left[ {x\int\limits_0^1 {{x^{49}}{{\left( {1 - {x^{50}}} \right)}^{100}}dx} - \int {{{d\left( x \right)} \over {dx}}.\int {{{d\left( x \right)} \over {dx}}\int\limits_0^1 {{x^{49}}{{\left( {1 - {x^{50}}} \right)}^{100}}dx} } } } \right]$$
Let (1 – x50) = t
$$ \Rightarrow $$ -50x49dx = dt
I2 = I1 -
$$\left[ {x.\left( { - {1 \over {50}}} \right){{{{\left( {1 - {x^{50}}} \right)}^{101}}} \over {101}}} \right]_0^1$$ - $$ - \int\limits_0^1 {\left( { - {1 \over {50}}} \right){{{{\left( {1 - {x^{50}}} \right)}^{101}}} \over {101}}dx} $$
= I1 - 0 - $${ - {1 \over {50}}.{1 \over {101}}{I_2}}$$
I2 = I1 - $${1 \over {5050}}{I_2}$$
$$ \Rightarrow $$ I2 + $${1 \over {5050}}{I_2}$$ = I1
$$ \Rightarrow $$ $${{5051} \over {5050}}{I_2}$$ = I1
$$ \Rightarrow $$ I2 = $${{5050} \over {5051}}$$I1
Given I2 = $$\alpha $$I1
$$ \therefore $$ $$\alpha $$ = $${{5050} \over {5051}}$$
I2 = $$\int\limits_0^1 {\left( {1 - {x^{50}}} \right){{\left( {1 - {x^{50}}} \right)}^{100}}dx} $$
I2 = $$\int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{100}}dx} - \int\limits_0^1 {{x^{50}}{{\left( {1 - {x^{50}}} \right)}^{100}}dx} $$
I2 = I1 - $$\int\limits_0^1 {x.{x^{49}}{{\left( {1 - {x^{50}}} \right)}^{100}}dx} $$
Now apply IBP
I2 = I1 -
$$\left[ {x\int\limits_0^1 {{x^{49}}{{\left( {1 - {x^{50}}} \right)}^{100}}dx} - \int {{{d\left( x \right)} \over {dx}}.\int {{{d\left( x \right)} \over {dx}}\int\limits_0^1 {{x^{49}}{{\left( {1 - {x^{50}}} \right)}^{100}}dx} } } } \right]$$
Let (1 – x50) = t
$$ \Rightarrow $$ -50x49dx = dt
I2 = I1 -
$$\left[ {x.\left( { - {1 \over {50}}} \right){{{{\left( {1 - {x^{50}}} \right)}^{101}}} \over {101}}} \right]_0^1$$ - $$ - \int\limits_0^1 {\left( { - {1 \over {50}}} \right){{{{\left( {1 - {x^{50}}} \right)}^{101}}} \over {101}}dx} $$
= I1 - 0 - $${ - {1 \over {50}}.{1 \over {101}}{I_2}}$$
I2 = I1 - $${1 \over {5050}}{I_2}$$
$$ \Rightarrow $$ I2 + $${1 \over {5050}}{I_2}$$ = I1
$$ \Rightarrow $$ $${{5051} \over {5050}}{I_2}$$ = I1
$$ \Rightarrow $$ I2 = $${{5050} \over {5051}}$$I1
Given I2 = $$\alpha $$I1
$$ \therefore $$ $$\alpha $$ = $${{5050} \over {5051}}$$
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