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Definite Integration question

2019 · 8 Apr · Shift 1 · Q32
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Definite Integration question

2019 · 8 Apr · Shift 1 · Q32

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1
If f(x)=2−xcos⁡x2+xcos⁡xf(x) = {{2 - x\cos x} \over {2 + x\cos x}}f(x)=2+xcosx2−xcosx​ and g(x) = logex, (x > 0) then the value of integral ∫−π4π4g(f(x))dx\int\limits_{ - {\pi \over 4}}^{{\pi \over 4}} {g\left( {f\left( x \right)} \right)dx{\rm{ }}}−4π​∫4π​​g(f(x))dx is
  1. A
    loge3
  2. B
    loge2
  3. C
    loge1
  4. D
    logee
View written solutionFree

Correct answer: C

  1. We need to evaluate I=∫−π/4π/4g(f(x)) dx=∫−π/4π/4ln⁡ ⁣(2−xcos⁡x2+xcos⁡x)dx.I=\int_{-\pi/4}^{\pi/4} g(f(x))\,dx=\int_{-\pi/4}^{\pi/4} \ln\!\left(\frac{2-x\cos x}{2+x\cos x}\right)dx.I=∫−π/4π/4​g(f(x))dx=∫−π/4π/4​ln(2+xcosx2−xcosx​)dx.

  2. Let h(x)=ln⁡ ⁣(2−xcos⁡x2+xcos⁡x).h(x)=\ln\!\left(\frac{2-x\cos x}{2+x\cos x}\right).h(x)=ln(2+xcosx2−xcosx​). We check whether h(x)h(x)h(x) is odd.

Since cos⁡(−x)=cos⁡x\cos(-x)=\cos xcos(−x)=cosx, we have f(-x)=\frac{2-(-x)\cos(-x)}{2+(-x)\cos(-x)}= rac{2+x\cos x}{2-x\cos x}. Therefore, f(−x)=1f(x).f(-x)=\frac{1}{f(x)}.f(−x)=f(x)1​.

Now apply logarithm: h(−x)=ln⁡(f(−x))=ln⁡(1f(x))=−ln⁡(f(x))=−h(x).h(-x)=\ln(f(-x))=\ln\left(\frac{1}{f(x)}\right)=-\ln(f(x))=-h(x).h(−x)=ln(f(−x))=ln(f(x)1​)=−ln(f(x))=−h(x). So h(x)h(x)h(x) is an odd function.

  1. The limits of integration are symmetric about 000: [−π4,π4].\left[-\frac{\pi}{4},\frac{\pi}{4}\right].[−4π​,4π​]. The integral of an odd function over symmetric limits is 000. Hence, I=0.I=0.I=0.

  2. Now match with options: 0=ln⁡1.0=\ln 1.0=ln1. So the correct option is ln⁡1.\boxed{\ln 1}.ln1​.

  3. Comparison with stored answer: Stored correct answer is C, which corresponds to ln⁡1\ln 1ln1. This matches our result.

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