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Definite Integration question

2021 · 17 Mar · Shift 1 · Q25
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  5. /2021 · 17 Mar · Shift 1 · Q25

Definite Integration question

2021 · 17 Mar · Shift 1 · Q25

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1
Which of the following statements is correct for the function g(α\alphaα) for α∈\alpha\inα∈ R such that g(α)=∫π6π3sin⁡αxcos⁡αx+sin⁡αxdxg(\alpha ) = \int\limits_{{\pi \over 6}}^{{\pi \over 3}} {{{{{\sin }^\alpha }x} \over {{{\cos }^\alpha }x + {{\sin }^\alpha }x}}dx}g(α)=6π​∫3π​​cosαx+sinαxsinαx​dx
  1. A
    g(α)g(\alpha )g(α) is a strictly increasing function
  2. B
    g(α)g(\alpha )g(α) is an even function
  3. C
    g(α)g(\alpha )g(α) has an inflection point at α\alphaα=−12-{1 \over 2}−21​
  4. D
    g(α)g(\alpha )g(α) is a strictly decreasing function
View written solutionFree

Correct answer: B

  1. Given function

We need to study

g(α)=∫π/6π/3sin⁡αxcos⁡αx+sin⁡αx dx,α∈R.g(\alpha)=\int_{\pi/6}^{\pi/3} \frac{\sin^\alpha x}{\cos^\alpha x+\sin^\alpha x}\,dx, \qquad \alpha\in\mathbb R.g(α)=∫π/6π/3​cosαx+sinαxsinαx​dx,α∈R.

We must determine which statement is correct.

  1. Rewrite the integrand

Let

I(x,α)=sin⁡αxcos⁡αx+sin⁡αx.I(x,\alpha)=\frac{\sin^\alpha x}{\cos^\alpha x+\sin^\alpha x}.I(x,α)=cosαx+sinαxsinαx​.

Divide numerator and denominator by sin⁡αx\sin^\alpha xsinαx:

I(x,α)=11+(cos⁡xsin⁡x)α=11+cot⁡αx.I(x,\alpha)=\frac{1}{1+\left(\frac{\cos x}{\sin x}\right)^\alpha} =\frac{1}{1+\cot^\alpha x}.I(x,α)=1+(sinxcosx​)α1​=1+cotαx1​.

So,

g(α)=∫π/6π/311+cot⁡αx dx.g(\alpha)=\int_{\pi/6}^{\pi/3} \frac{1}{1+\cot^\alpha x}\,dx.g(α)=∫π/6π/3​1+cotαx1​dx.
  1. Use the substitution x↦π2−xx\mapsto \frac{\pi}{2}-xx↦2π​−x

Consider

g(−α)=∫π/6π/311+cot⁡−αx dx.g(-\alpha)=\int_{\pi/6}^{\pi/3} \frac{1}{1+\cot^{-\alpha}x}\,dx.g(−α)=∫π/6π/3​1+cot−αx1​dx.

Since cot⁡−αx=tan⁡αx\cot^{-\alpha}x=\tan^\alpha xcot−αx=tanαx,

g(−α)=∫π/6π/311+tan⁡αx dx.g(-\alpha)=\int_{\pi/6}^{\pi/3} \frac{1}{1+\tan^\alpha x}\,dx.g(−α)=∫π/6π/3​1+tanαx1​dx.

Now use the identity

11+tan⁡αx=11+cot⁡α(π2−x).\frac{1}{1+\tan^\alpha x}=\frac{1}{1+\cot^\alpha\left(\frac\pi2-x\right)}.1+tanαx1​=1+cotα(2π​−x)1​.

Let

t=π2−x⇒dx=−dt.t=\frac\pi2-x \quad \Rightarrow \quad dx=-dt.t=2π​−x⇒dx=−dt.

When x=π/6x=\pi/6x=π/6, t=π/3t=\pi/3t=π/3; when x=π/3x=\pi/3x=π/3, t=π/6t=\pi/6t=π/6. Thus

g(−α)=∫π/3π/611+cot⁡αt(−dt)=∫π/6π/311+cot⁡αt dt=g(α).g(-\alpha)=\int_{\pi/3}^{\pi/6} \frac{1}{1+\cot^\alpha t}(-dt) =\int_{\pi/6}^{\pi/3} \frac{1}{1+\cot^\alpha t}\,dt =g(\alpha).g(−α)=∫π/3π/6​1+cotαt1​(−dt)=∫π/6π/3​1+cotαt1​dt=g(α).

Hence,

g(−α)=g(α)\boxed{g(-\alpha)=g(\alpha)}g(−α)=g(α)​

for all real α\alphaα.

So g(α)g(\alpha)g(α) is an even function.

Therefore, Option B is correct.

  1. Check monotonicity statements A and D

Since g(α)g(\alpha)g(α) is even, it cannot be strictly increasing on all of R\mathbb RR, and it also cannot be strictly decreasing on all of R\mathbb RR.

Indeed, an even function satisfies

g(α)=g(−α),g(\alpha)=g(-\alpha),g(α)=g(−α),

so distinct inputs generally give equal values. That rules out both strict increase and strict decrease on all real numbers.

Thus:

  • A is false
  • D is false
  1. Check option C

Option C claims that g(α)g(\alpha)g(α) has an inflection point at α=−12\alpha=-\tfrac12α=−21​.

There is no immediate symmetry or structural reason forcing an inflection specifically at −12-\tfrac12−21​. In contrast, we have rigorously proved the even-function property, which directly confirms option B. Since this is a single-correct MCQ, C must be false.

  1. Final answer

The correct statement is

B: g(α) is an even function.\boxed{\text{B: } g(\alpha) \text{ is an even function}.}B: g(α) is an even function.​

This matches the stored correct answer.

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