Verified Solution • 2025–26 Session moderate

QJEEM17Mabbfa: JEE Main 2017 Mathematics - Class 12 JEEM17Mabbfa Solution: If f = x + 2, x , and f(x) dx = A log 1 x + Bx + C, then the…

If    \,\,\, f (3x−43x+4)\left( {{{3x - 4} \over {3x + 4}}} \right) = x + 2, x ≠\ne −- 43{4 \over 3} , 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)

Multiple Choice Options

  • A (83,23)\left( {{8 \over 3},{2 \over 3}} \right)
  • B (−83,23)\left( { - {8 \over 3},{2 \over 3}} \right)
  • C (−83,−23)\left( { - {8 \over 3}, - {2 \over 3}} \right)
  • D (83,−23)\left( { {8 \over 3}, - {2 \over 3}} \right)

Step-by-Step Verified Solution

Given,

f (3x−43x+4)\left( {{{3x - 4} \over {3x + 4}}} \right) = x + 2,    x   ≠\ne   −- 43{4 \over 3}

Let, 3x−43x+4{{3x - 4} \over {3x + 4}} = t

⇒\Rightarrow    \,\,\, 3x −- 4 = 3tx + 4t

⇒\Rightarrow    \,\,\, 3x −- 3tx = 4t + 4

⇒\Rightarrow    \,\,\, x = 4t+43−3t{{4t + 4} \over {3 - 3t}}

So, f(t) = 4t+43−3t{{4t + 4} \over {3 - 3t}} + 2 = 10−2t3−3t{{10 - 2t} \over {3 - 3t}}

∴   \therefore\,\,\, f (x) = 10−2x3−3x{{10 - 2x} \over {3 - 3x}}

∴   \therefore\,\,\, ∫f(x) dx\int {f(x)\,dx}

= ∫2x−103x−3 dx\int {{{2x - 10} \over {3x - 3}}} \,dx

= ∫2x3x−3−10∫dx3x−3\int {{{2x} \over {3x - 3}} - 10\int {{{dx} \over {3x - 3}}} }

= 23∫x−1x+1dx+23∫dxx−1−103∫dxx−1{2 \over 3}\int {{{x - 1} \over {x + 1}}} dx + {2 \over 3}\int {{{dx} \over {x - 1}} - {{10} \over 3}} \int {{{dx} \over {x - 1}}}

= 23{2 \over 3} x + 23{2 \over 3} log ∣x−1∣\left| {x - 1} \right| −- 103{{10} \over 3} log ∣x−1∣\left| {x - 1} \right| + C

= 23{2 \over 3} x −- 83{8 \over 3} log ∣x−1∣\left| {x - 1} \right| + C

∴   \therefore\,\,\, A = −- 83{8 \over 3} and B = 23{2 \over 3}

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