Verified Solution • 2025–26 Session moderate

QJEEM21Mafdfc: JEE Main 2021 Mathematics - Class 11 JEEM21Mafdfc Solution: Let f : R {3} R {1} be defined by f(x) = . Let g : R R be given …

Let f : R −- {3} →\to R −- {1} be defined by f(x) = x−2x−3{{x - 2} \over {x - 3}} .

Let g : R →\to R be given as g(x) = 2x −- 3. Then, the sum of all the values of x for which f −- 1(x) + g −- 1(x) = 132{{13} \over 2} is equal to :

Multiple Choice Options

  • A 3
  • B 5
  • C 2
  • D 7

Step-by-Step Verified Solution

Finding inverse of f(x)

y=x−2x−3⇒xy−3y=x−2⇒x(y−1)=3y−2y = {{x - 2} \over {x - 3}} \Rightarrow xy - 3y = x - 2 \Rightarrow x(y - 1) = 3y - 2

∴\therefore f−1(x)=3x−2x−1{f^{ - 1}}(x) = {{3x - 2} \over {x - 1}}

Similarly for g−1(x){g^{ - 1}}(x)

y=2x−3⇒x=y+32⇒g−1(x)=x+32y = 2x - 3 \Rightarrow x = {{y + 3} \over 2} \Rightarrow {g^{ - 1}}(x) = {{x + 3} \over 2}

∴\therefore 3x−2x−1+x+32=132{{3x - 2} \over {x - 1}} + {{x + 3} \over 2} = {{13} \over 2}

⇒6x−4+x2+2x−3=13x−13\Rightarrow 6x - 4 + {x^2} + 2x - 3 = 13x - 13

⇒x2−5x+6=0\Rightarrow {x^2} - 5x + 6 = 0

⇒(x−2)(x−3)=0\Rightarrow (x - 2)(x - 3) = 0

⇒\Rightarrow x = 2 or 3

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