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Indefinite Integrals question

2017 · 9 Apr · Shift 1 · Q41
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Indefinite Integrals question

2017 · 9 Apr · Shift 1 · Q41

JEE MainMathematicsIndefinite IntegralsMCQ+4 / −1
If    \,\,\, f (3x−43x+4)\left( {{{3x - 4} \over {3x + 4}}} \right)(3x+43x−4​)= x + 2, x e−43e -{4 \over 3}e−34​, and ∫\int {}∫ f(x) dx = A log ∣\left| {} \right.∣ 1 −-− x ∣\left| {} \right.∣ + Bx + C, then the ordered pair (A, B) is equal to : (where C is a constant of integration)
  1. A
    (83,23)\left( {{8 \over 3},{2 \over 3}} \right)(38​,32​)
  2. B
    (−83,23)\left( { - {8 \over 3},{2 \over 3}} \right)(−38​,32​)
  3. C
    (−83,−23)\left( { - {8 \over 3}, - {2 \over 3}} \right)(−38​,−32​)
  4. D
    (83,−23)\left( { {8 \over 3}, - {2 \over 3}} \right)(38​,−32​)
View written solutionFree

Correct answer: B

  1. Given relation

We are given

f(3x−43x+4)=x+2f\left(\frac{3x-4}{3x+4}\right)=x+2f(3x+43x−4​)=x+2

for x≠−43x\ne -\frac43x=−34​.

We need to find f(t)f(t)f(t) first, then integrate it.

  1. Let
t=3x−43x+4t=\frac{3x-4}{3x+4}t=3x+43x−4​

We solve this for xxx in terms of ttt.

t(3x+4)=3x−4t(3x+4)=3x-4t(3x+4)=3x−4 3tx+4t=3x−43tx+4t=3x-43tx+4t=3x−4 3tx−3x=−4−4t3tx-3x=-4-4t3tx−3x=−4−4t 3x(t−1)=−4(1+t)3x(t-1)=-4(1+t)3x(t−1)=−4(1+t) x=−4(1+t)3(t−1)=4(1+t)3(1−t)x=\frac{-4(1+t)}{3(t-1)}=\frac{4(1+t)}{3(1-t)}x=3(t−1)−4(1+t)​=3(1−t)4(1+t)​

So,

f(t)=x+2=4(1+t)3(1−t)+2f(t)=x+2=\frac{4(1+t)}{3(1-t)}+2f(t)=x+2=3(1−t)4(1+t)​+2

Now simplify:

f(t)=4(1+t)3(1−t)+6(1−t)3(1−t)f(t)=\frac{4(1+t)}{3(1-t)}+\frac{6(1-t)}{3(1-t)}f(t)=3(1−t)4(1+t)​+3(1−t)6(1−t)​ f(t)=4+4t+6−6t3(1−t)f(t)=\frac{4+4t+6-6t}{3(1-t)}f(t)=3(1−t)4+4t+6−6t​ f(t)=10−2t3(1−t)f(t)=\frac{10-2t}{3(1-t)}f(t)=3(1−t)10−2t​

Thus,

f(x)=10−2x3(1−x)f(x)=\frac{10-2x}{3(1-x)}f(x)=3(1−x)10−2x​
  1. Rewrite for easy integration

Factor numerator:

f(x)=2(5−x)3(1−x)f(x)=\frac{2(5-x)}{3(1-x)}f(x)=3(1−x)2(5−x)​

Now express in partial form:

10−2x3(1−x)=23⋅5−x1−x\frac{10-2x}{3(1-x)}=\frac{2}{3}\cdot\frac{5-x}{1-x}3(1−x)10−2x​=32​⋅1−x5−x​

Let

5−x1−x=k+m1−x\frac{5-x}{1-x}=k+\frac{m}{1-x}1−x5−x​=k+1−xm​

Then

5−x=k(1−x)+m=k−kx+m5-x=k(1-x)+m=k-kx+m5−x=k(1−x)+m=k−kx+m

Comparing coefficients:

  • coefficient of xxx: −k=−1⇒k=1-k=-1\Rightarrow k=1−k=−1⇒k=1
  • constant term: k+m=5⇒1+m=5⇒m=4k+m=5\Rightarrow 1+m=5\Rightarrow m=4k+m=5⇒1+m=5⇒m=4

Hence,

5−x1−x=1+41−x\frac{5-x}{1-x}=1+\frac{4}{1-x}1−x5−x​=1+1−x4​

So,

f(x)=23(1+41−x)=23+83(1−x)f(x)=\frac{2}{3}\left(1+\frac{4}{1-x}\right)=\frac{2}{3}+\frac{8}{3(1-x)}f(x)=32​(1+1−x4​)=32​+3(1−x)8​
  1. Integrate
∫f(x) dx=∫(23+83(1−x))dx\int f(x)\,dx=\int \left(\frac{2}{3}+\frac{8}{3(1-x)}\right)dx∫f(x)dx=∫(32​+3(1−x)8​)dx =23x+83∫11−x dx=\frac{2}{3}x+\frac{8}{3}\int \frac{1}{1-x}\,dx=32​x+38​∫1−x1​dx

Now,

∫11−x dx=−log⁡∣1−x∣\int \frac{1}{1-x}\,dx=-\log|1-x|∫1−x1​dx=−log∣1−x∣

Therefore,

∫f(x) dx=23x−83log⁡∣1−x∣+C\int f(x)\,dx=\frac{2}{3}x-\frac{8}{3}\log|1-x|+C∫f(x)dx=32​x−38​log∣1−x∣+C

Comparing with

∫f(x)dx=Alog⁡∣1−x∣+Bx+C\int f(x)dx=A\log|1-x|+Bx+C∫f(x)dx=Alog∣1−x∣+Bx+C

we get

A=−83,B=23A=-\frac{8}{3},\qquad B=\frac{2}{3}A=−38​,B=32​

So,

(A,B)=(−83,23)(A,B)=\left(-\frac{8}{3},\frac{2}{3}\right)(A,B)=(−38​,32​)
  1. Option check

This matches Option B.

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