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Definite Integration question

2022 · Shift 1 · Q27
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Definite Integration question

2022 · Shift 1 · Q27

JEE AdvancedMathematicsDefinite IntegrationMultiple correct+4 / −2
Consider the equation ∫1e(log⁡ex)1/2x(a−(log⁡ex)3/2)2dx=1,a∈(−∞,0)∪(1,∞)\int_{1}^{e} \frac{\left(\log _{\mathrm{e}} x\right)^{1 / 2}}{x\left(a-\left(\log _{\mathrm{e}} x\right)^{3 / 2}\right)^{2}} d x=1, \quad a \in(-\infty, 0) \cup(1, \infty)∫1e​x(a−(loge​x)3/2)2(loge​x)1/2​dx=1,a∈(−∞,0)∪(1,∞) Which of the following statements is/are TRUE?
  1. A
    No aaa satisfies the above equation
  2. B
    An integer aaa satisfies the above equation
  3. C
    An irrational number aaa satisfies the above equation
  4. D
    More than one aaa satisfy the above equation
View written solutionFree

Correct answer: C, D

  1. Evaluate the integral in terms of aaa

Given

∫1e(ln⁡x)1/2x(a−(ln⁡x)3/2)2 dx=1,a∈(−∞,0)∪(1,∞).\int_{1}^{e} \frac{(\ln x)^{1/2}}{x\left(a-(\ln x)^{3/2}\right)^2}\,dx = 1, \qquad a\in(-\infty,0)\cup(1,\infty).∫1e​x(a−(lnx)3/2)2(lnx)1/2​dx=1,a∈(−∞,0)∪(1,∞).

Let

t=(ln⁡x)3/2.t=(\ln x)^{3/2}.t=(lnx)3/2.

Then

dt=32(ln⁡x)1/2⋅1x dx⇒(ln⁡x)1/2x dx=23 dt.dt = \frac{3}{2}(\ln x)^{1/2}\cdot \frac{1}{x}\,dx \quad\Rightarrow\quad \frac{(\ln x)^{1/2}}{x}\,dx = \frac{2}{3}\,dt.dt=23​(lnx)1/2⋅x1​dx⇒x(lnx)1/2​dx=32​dt.

Now change limits:

  • When x=1x=1x=1, ln⁡1=0⇒t=0\ln 1=0 \Rightarrow t=0ln1=0⇒t=0.
  • When x=ex=ex=e, ln⁡e=1⇒t=1\ln e=1 \Rightarrow t=1lne=1⇒t=1.

So the integral becomes

23∫01dt(a−t)2=1.\frac{2}{3}\int_0^1 \frac{dt}{(a-t)^2}=1.32​∫01​(a−t)2dt​=1.
  1. Integrate

Since

∫dt(a−t)2=1a−t,\int \frac{dt}{(a-t)^2} = \frac{1}{a-t},∫(a−t)2dt​=a−t1​,

we get

23[1a−t]01=1.\frac{2}{3}\left[\frac{1}{a-t}\right]_0^1 = 1.32​[a−t1​]01​=1.

Thus

23(1a−1−1a)=1.\frac{2}{3}\left(\frac{1}{a-1}-\frac{1}{a}\right)=1.32​(a−11​−a1​)=1.

Simplify:

1a−1−1a=a−(a−1)a(a−1)=1a(a−1).\frac{1}{a-1}-\frac{1}{a} = \frac{a-(a-1)}{a(a-1)}=\frac{1}{a(a-1)}.a−11​−a1​=a(a−1)a−(a−1)​=a(a−1)1​.

Hence

23⋅1a(a−1)=1.\frac{2}{3}\cdot \frac{1}{a(a-1)}=1.32​⋅a(a−1)1​=1.

So

a(a−1)=23.a(a-1)=\frac{2}{3}.a(a−1)=32​.

That is,

a2−a−23=0.a^2-a-\frac{2}{3}=0.a2−a−32​=0.

Multiply by 333:

3a2−3a−2=0.3a^2-3a-2=0.3a2−3a−2=0.
  1. Solve for aaa

Using the quadratic formula,

a=3±9+246=3±336.a=\frac{3\pm\sqrt{9+24}}{6}=\frac{3\pm\sqrt{33}}{6}.a=63±9+24​​=63±33​​.

So the two solutions are

a1=3+336,a2=3−336.a_1=\frac{3+\sqrt{33}}{6}, \qquad a_2=\frac{3-\sqrt{33}}{6}.a1​=63+33​​,a2​=63−33​​.
  1. Check domain

We need a∈(−∞,0)∪(1,∞)a\in(-\infty,0)\cup(1,\infty)a∈(−∞,0)∪(1,∞).

  • For a1=3+336a_1=\dfrac{3+\sqrt{33}}{6}a1​=63+33​​, since 33>3\sqrt{33}>333​>3, a1>1,a_1>1,a1​>1, so it is allowed.

  • For a2=3−336a_2=\dfrac{3-\sqrt{33}}{6}a2​=63−33​​, since 33>3\sqrt{33}>333​>3, a2<0,a_2<0,a2​<0, so it is also allowed.

Thus both values satisfy the equation.

Also, both are irrational because they involve 33\sqrt{33}33​.

  1. Evaluate the options

  • A: No aaa satisfies the above equation
    False, since there are two solutions.

  • B: An integer aaa satisfies the above equation
    False, both solutions are irrational.

  • C: An irrational number aaa satisfies the above equation
    True.

  • D: More than one aaa satisfy the above equation
    True, there are two valid solutions.

Therefore, the correct options are:

C,D\boxed{C, D}C,D​
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