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Definite Integration question

2014 · Shift 2 · Q38
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  5. /2014 · Shift 2 · Q38

Definite Integration question

2014 · Shift 2 · Q38

JEE AdvancedMathematicsDefinite IntegrationMCQ+3 / −1
Given that for each a∈(0,1),   lim⁡h→0+ ∫h1−ht−a(1−t)a−1dta \in \left( {0,1} \right),\,\,\,\mathop {\lim }\limits_{h \to {0^ + }} \,\int\limits_h^{1 - h} {{t^{ - a}}{{\left( {1 - t} \right)}^{a - 1}}dt}a∈(0,1),h→0+lim​h∫1−h​t−a(1−t)a−1dt exists. Let this limit be g(a).g(a).g(a). In addition, it is given that the function g(a)g(a)g(a) is differentiable on (0,1).(0,1).(0,1). The value of g′(12)g'\left( {{1 \over 2}} \right)g′(21​) is
  1. A
    π2{\pi \over 2}2π​
  2. B
    π\piπ
  3. C
    −π2-{\pi \over 2}−2π​
  4. D
    000
View written solutionFree

Correct answer: D

  1. Identify the integral

We are given

g(a)=lim⁡h→0+∫h1−ht−a(1−t)a−1 dt,a∈(0,1).g(a)=\lim_{h\to 0^+}\int_h^{1-h} t^{-a}(1-t)^{a-1}\,dt, \qquad a\in(0,1).g(a)=h→0+lim​∫h1−h​t−a(1−t)a−1dt,a∈(0,1).

So effectively,

g(a)=∫01t−a(1−t)a−1 dt,g(a)=\int_0^1 t^{-a}(1-t)^{a-1}\,dt,g(a)=∫01​t−a(1−t)a−1dt,

provided the improper integral converges.

Since 0<a<10<a<10<a<1:

  • near t=0t=0t=0, the integrand behaves like t−at^{-a}t−a, which is integrable because a<1a<1a<1;
  • near t=1t=1t=1, the integrand behaves like (1−t)a−1(1-t)^{a-1}(1−t)a−1, which is integrable because a>0a>0a>0.

Hence the integral exists.

  1. Recognize Beta function form

Recall the Beta function:

B(x,y)=∫01tx−1(1−t)y−1 dt,x,y>0.B(x,y)=\int_0^1 t^{x-1}(1-t)^{y-1}\,dt, \qquad x,y>0.B(x,y)=∫01​tx−1(1−t)y−1dt,x,y>0.

Compare with

t−a(1−t)a−1=t(1−a)−1(1−t)a−1.t^{-a}(1-t)^{a-1}=t^{(1-a)-1}(1-t)^{a-1}.t−a(1−t)a−1=t(1−a)−1(1−t)a−1.

So,

g(a)=B(1−a,a).g(a)=B(1-a,a).g(a)=B(1−a,a).

Using the identity

B(x,y)=Γ(x)Γ(y)Γ(x+y),B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)},B(x,y)=Γ(x+y)Γ(x)Γ(y)​,

and here x=1−ax=1-ax=1−a, y=ay=ay=a, so x+y=1x+y=1x+y=1, hence

g(a)=Γ(1−a)Γ(a).g(a)=\Gamma(1-a)\Gamma(a).g(a)=Γ(1−a)Γ(a).

Now use Euler’s reflection formula:

Γ(a)Γ(1−a)=πsin⁡(πa).\Gamma(a)\Gamma(1-a)=\frac{\pi}{\sin(\pi a)}.Γ(a)Γ(1−a)=sin(πa)π​.

Therefore,

g(a)=πsin⁡(πa)=πcsc⁡(πa).g(a)=\frac{\pi}{\sin(\pi a)}=\pi\csc(\pi a).g(a)=sin(πa)π​=πcsc(πa).
  1. Differentiate g(a)g(a)g(a)

We have

g(a)=πcsc⁡(πa).g(a)=\pi\csc(\pi a).g(a)=πcsc(πa).

So

g′(a)=π⋅dda[csc⁡(πa)]=π[−csc⁡(πa)cot⁡(πa)](π).g'(a)=\pi\cdot \frac{d}{da}[\csc(\pi a)] =\pi\bigl[-\csc(\pi a)\cot(\pi a)\bigr](\pi).g′(a)=π⋅dad​[csc(πa)]=π[−csc(πa)cot(πa)](π).

Thus,

g′(a)=−π2csc⁡(πa)cot⁡(πa).g'(a)=-\pi^2\csc(\pi a)\cot(\pi a).g′(a)=−π2csc(πa)cot(πa).
  1. Evaluate at a=12a=\tfrac12a=21​

At a=12a=\tfrac12a=21​,

csc⁡(π2)=1,cot⁡(π2)=0.\csc\left(\frac{\pi}{2}\right)=1, \qquad \cot\left(\frac{\pi}{2}\right)=0.csc(2π​)=1,cot(2π​)=0.

Hence,

g′(12)=−π2⋅1⋅0=0.g'\left(\frac12\right)=-\pi^2\cdot 1\cdot 0=0.g′(21​)=−π2⋅1⋅0=0.
  1. Check the options

The value is

0.0.0.

So the correct option is D.

  1. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

They agree.

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