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Definite Integration question

2021 · 20 Jul · Shift 2 · Q35
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  5. /2021 · 20 Jul · Shift 2 · Q35

Definite Integration question

2021 · 20 Jul · Shift 2 · Q35

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1
Let g(t)=∫−π/2π/2cos⁡(π4t+f(x))dxg(t) = \int_{ - \pi /2}^{\pi /2} {\cos \left( {{\pi \over 4}t + f(x)} \right)} dxg(t)=∫−π/2π/2​cos(4π​t+f(x))dx, where f(x)=log⁡e(x+x2+1),x∈Rf(x) = {\log _e}\left( {x + \sqrt {{x^2} + 1} } \right),x \in Rf(x)=loge​(x+x2+1​),x∈R. Then which one of the following is correct?
  1. A
    g(1) = g(0)
  2. B
    2g(1)=g(0)\sqrt 2 g(1) = g(0)2​g(1)=g(0)
  3. C
    g(1)=2g(0)g(1) = \sqrt 2 g(0)g(1)=2​g(0)
  4. D
    g(1) + g(0) = 0
View written solutionFree

Correct answer: B

  1. Given integral

We need to evaluate

g(t)=∫−π/2π/2cos⁡(π4t+f(x))dx,g(t)=\int_{-\pi/2}^{\pi/2} \cos\left(\frac{\pi}{4}t+f(x)\right)dx,g(t)=∫−π/2π/2​cos(4π​t+f(x))dx,

where

f(x)=log⁡(x+x2+1).f(x)=\log\left(x+\sqrt{x^2+1}\right).f(x)=log(x+x2+1​).

Recall that

log⁡(x+x2+1)=arsinh⁡(x).\log\left(x+\sqrt{x^2+1}\right)=\operatorname{arsinh}(x).log(x+x2+1​)=arsinh(x).

So f(x)f(x)f(x) is an odd function:

f(−x)=−f(x).f(-x)=-f(x).f(−x)=−f(x).
  1. Use angle addition formula

Let

a=π4t.a=\frac{\pi}{4}t.a=4π​t.

Then

cos⁡(a+f(x))=cos⁡acos⁡f(x)−sin⁡asin⁡f(x).\cos(a+f(x))=\cos a\cos f(x)-\sin a\sin f(x).cos(a+f(x))=cosacosf(x)−sinasinf(x).

Hence

g(t)=cos⁡a∫−π/2π/2cos⁡f(x) dx−sin⁡a∫−π/2π/2sin⁡f(x) dx.g(t)=\cos a\int_{-\pi/2}^{\pi/2}\cos f(x)\,dx-\sin a\int_{-\pi/2}^{\pi/2}\sin f(x)\,dx.g(t)=cosa∫−π/2π/2​cosf(x)dx−sina∫−π/2π/2​sinf(x)dx.

Now, since f(x)f(x)f(x) is odd:

  • sin⁡(f(x))\sin(f(x))sin(f(x)) is odd,
  • cos⁡(f(x))\cos(f(x))cos(f(x)) is even.

Therefore,

∫−π/2π/2sin⁡f(x) dx=0.\int_{-\pi/2}^{\pi/2}\sin f(x)\,dx=0.∫−π/2π/2​sinf(x)dx=0.

So,

g(t)=cos⁡(πt4)∫−π/2π/2cos⁡f(x) dx.g(t)=\cos\left(\frac{\pi t}{4}\right)\int_{-\pi/2}^{\pi/2}\cos f(x)\,dx.g(t)=cos(4πt​)∫−π/2π/2​cosf(x)dx.

Let

I=∫−π/2π/2cos⁡f(x) dx.I=\int_{-\pi/2}^{\pi/2}\cos f(x)\,dx.I=∫−π/2π/2​cosf(x)dx.

Then

g(t)=Icos⁡(πt4).g(t)=I\cos\left(\frac{\pi t}{4}\right).g(t)=Icos(4πt​).
  1. Compute g(0)g(0)g(0) and g(1)g(1)g(1)

For t=0t=0t=0,

g(0)=Icos⁡0=I.g(0)=I\cos 0=I.g(0)=Icos0=I.

For t=1t=1t=1,

g(1)=Icos⁡(π4)=I⋅12.g(1)=I\cos\left(\frac{\pi}{4}\right)=I\cdot \frac{1}{\sqrt2}.g(1)=Icos(4π​)=I⋅2​1​.

Thus,

g(1)=g(0)2.g(1)=\frac{g(0)}{\sqrt2}.g(1)=2​g(0)​.

So,

2 g(1)=g(0).\sqrt2\,g(1)=g(0).2​g(1)=g(0).
  1. Check options

  • A: g(1)=g(0)g(1)=g(0)g(1)=g(0)
    False, since g(1)=g(0)2g(1)=\dfrac{g(0)}{\sqrt2}g(1)=2​g(0)​.

  • B: 2 g(1)=g(0)\sqrt2\,g(1)=g(0)2​g(1)=g(0)
    True.

  • C: g(1)=2 g(0)g(1)=\sqrt2\,g(0)g(1)=2​g(0)
    False.

  • D: g(1)+g(0)=0g(1)+g(0)=0g(1)+g(0)=0
    False.

  1. Final answer

The correct option is

B.\boxed{\text{B}}.B​.
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