QJEEM20Mae0da: JEE Main 2020 Mathematics - Class 12 JEEM20Mae0da Solution: If I 1 = and I 2 = such that I 2 = I 1 then equals to :

Question

Verified Solution • 2025–26 Session moderate

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 :

Accepted Answer

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}}

Step-by-Step Verified Solution