Question

JEE MAINS MATHS QUESTION SEPTEMBER 2020

Answer 1

Answer:

Bhai ye to bhaot high level wala question mil gaya abb kya karen???

Step-by-step explanation:

Answer 2

we have to find the value of λ where it has given that $I_1=\int\limits^1_0{(1-x^{50})^{100}}\,dx$ and $I_2=\int\limits^1_0{(1-x^{50})^{101}}\,dx$ and $I_1=\lambda I_2$

solution : let's solve I₂

$I_2=\int\limits^1_0{(1-x^{50})^{101}}\,dx$

$=\int\limits^1_0{(1-x^{50})(1-x^{50})^{100}}\,dx$

$=\int\limits^1_0{(1-x^{50})^{100}}\,dx-\int\limits^1_0{x^{50}(1-x^{50})^{100}}\,dx$[/tex]

⇒I₂ = I₁ - $\int\limits^1_0{x \{x^{49}(1-x^{50})^{100}\}}\,dx$........(1)

now using integration by part for $\int\limits^1_0{x \{x^{49}(1-x^{50})^{100}\}}\,dx$

here 1st function = x and 2nd function = x^49(1 - x^50)

so, I = $\int\limits^1_0{x \{x^{49}(1-x^{50})^{100}\}}\,dx$

= $x\int\limits^1_0{x^{49}(1-x^{50})^{100}}\,dx-\int\limits^1_0{\frac{dx}{dx}\int{x^{49}(1-x^{50})^{100}}}\,dx$

let (1 - x^50) = p

differentiating both sides we get,

-50x^49 dx = dp

so, x^49 dx = -dp/50

so, I = $\left[x\frac{-1}{50}\frac{(1-x^{50})^{101}}{101}\right]^1_0-\int\limits^1_0{\left(\frac{-1}{50}\right)\frac{(1-x^{50})^{101}}{101}}\,dx$

I = 0 + (1/50) × (1/101) I₂ = 1/5050 I₂

now putting I in equation (1) we get,

I₂ = I₁ - 1/5050 I₂ =

so, I₁ = 5051/5050I₂

now on comparing with I₁ = λI₂

we get, λ = 5051/5050

Therefore the value of λ is 5051/5050