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Definite Integration question

2020 · 6 Sep · Shift 1 · Q28
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  5. /2020 · 6 Sep · Shift 1 · Q28

Definite Integration question

2020 · 6 Sep · Shift 1 · Q28

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1
If I1 = ∫01(1−x50)100dx\int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{100}}} dx0∫1​(1−x50)100dx and I2 =∫01(1−x50)101dx\int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{101}}} dx0∫1​(1−x50)101dx such that I2 =α\alphaα I1 then α\alphaα equals to :
  1. A
    50515050{{5051} \over {5050}}50505051​
  2. B
    50505051{{5050} \over {5051}}50515050​
  3. C
    50505049{{5050} \over {5049}}50495050​
  4. D
    50495050{{5049} \over {5050}}50505049​
View written solutionFree

Correct answer: B

  1. We need to compare I1=∫01(1−x50)100 dx,I2=∫01(1−x50)101 dxI_1=\int_0^1 (1-x^{50})^{100}\,dx,\qquad I_2=\int_0^1 (1-x^{50})^{101}\,dxI1​=∫01​(1−x50)100dx,I2​=∫01​(1−x50)101dx and find α\alphaα such that I2=αI1.I_2=\alpha I_1.I2​=αI1​.

  2. Use the substitution t=x50  ⟹  x=t1/50,dx=150t−49/50 dt.t=x^{50}\implies x=t^{1/50},\qquad dx=\frac{1}{50}t^{-49/50}\,dt.t=x50⟹x=t1/50,dx=501​t−49/50dt.

Then I1=150∫01t−49/50(1−t)100 dtI_1=\frac1{50}\int_0^1 t^{-49/50}(1-t)^{100}\,dtI1​=501​∫01​t−49/50(1−t)100dt and I2=150∫01t−49/50(1−t)101 dt.I_2=\frac1{50}\int_0^1 t^{-49/50}(1-t)^{101}\,dt.I2​=501​∫01​t−49/50(1−t)101dt.

  1. Recognize these as Beta-function integrals: B(p,q)=∫01tp−1(1−t)q−1 dt.B(p,q)=\int_0^1 t^{p-1}(1-t)^{q-1}\,dt.B(p,q)=∫01​tp−1(1−t)q−1dt.

Here, t−49/50=t150−1.t^{-49/50}=t^{\frac1{50}-1}.t−49/50=t501​−1. So, I1=150B(150,101),I_1=\frac1{50}B\left(\frac1{50},101\right),I1​=501​B(501​,101), I2=150B(150,102).I_2=\frac1{50}B\left(\frac1{50},102\right).I2​=501​B(501​,102).

  1. Now use B(p,q)=Γ(p)Γ(q)Γ(p+q).B(p,q)=\frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}.B(p,q)=Γ(p+q)Γ(p)Γ(q)​. Thus,
=\frac{\Gamma\left(\frac1{50}\right)\Gamma(102)}{\Gamma\left(102+\frac1{50}\right)}\cdot \frac{\Gamma\left(101+\frac1{50}\right)}{\Gamma\left(\frac1{50}\right)\Gamma(101)}.$$ Canceling common terms, $$\frac{I_2}{I_1}=\frac{\Gamma(102)}{\Gamma(101)}\cdot \frac{\Gamma\left(101+\frac1{50}\right)}{\Gamma\left(102+\frac1{50}\right)}.$$ Using $\Gamma(z+1)=z\Gamma(z)$, $$\frac{\Gamma(102)}{\Gamma(101)}=101,$$ and $$\Gamma\left(102+\frac1{50}\right)=\left(101+\frac1{50}\right)\Gamma\left(101+\frac1{50}\right).$$ So, $$\frac{\Gamma\left(101+\frac1{50}\right)}{\Gamma\left(102+\frac1{50}\right)}=\frac{1}{101+\frac1{50}}.$$ Therefore, $$\frac{I_2}{I_1}=\frac{101}{101+\frac1{50}}.$$ 5. Simplify: $$101+\frac1{50}=\frac{5050+1}{50}=\frac{5051}{50}.$$ Hence, $$\frac{I_2}{I_1}=101\cdot \frac{50}{5051}=\frac{5050}{5051}.$$ Thus, $$\alpha=\frac{5050}{5051}.$$ 6. Checking options: - A: $\frac{5051}{5050}$ - B: $\frac{5050}{5051}$ - C: $\frac{5050}{5049}$ - D: $\frac{5049}{5050}$ Correct option is **B**.
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