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

Given f(x)=x−2x−3let f−1(x)=y⇒ x=f(y)⇒ x=y−2y−3⇒ xy−3x=y−2⇒ xy−y=3x−2⇒ y(x−1)=3x−2⇒ y(x−1)=3x−2⇒ y=3x−2x−1∴ f−1(x)=3x−2x−1and g(x)= 2x−3let g−1(x)=y⇒ x=g(y)⇒ x= 2y−3⇒ y− x+32∴ g−1(x)=x+32Let f−1(x)+g−1(x)=132⇒ 3x−2x−1+x+32=132⇒ 6x−4+(x+3)(x−1)(x−1)2=132⇒ 6x−4+x2−x+3x−=13x−13⇒ x2−5x+6=0⇒ (x−2)(x−3)=0⇒ x= 2, 3∴ Sum of values of x= 2+3⇒ 5

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

Given f(x)=x−2x−3let f−1(x)=y⇒ x=f(y)⇒ x=y−2y−3⇒ xy−3x=y−2⇒ xy−y=3x−2⇒ y(x−1)=3x−2⇒ y(x−1)=3x−2⇒ y=3x−2x−1∴ f−1(x)=3x−2x−1and g(x)= 2x−3let g−1(x)=y⇒ x=g(y)⇒ x= 2y−3⇒ y− x+32∴ g−1(x)=x+32Let f−1(x)+g−1(x)=132⇒ 3x−2x−1+x+32=132⇒ 6x−4+(x+3)(x−1)(x−1)2=132⇒ 6x−4+x2−x+3x−=13x−13⇒ x2−5x+6=0⇒ (x−2)(x−3)=0⇒ x= 2, 3∴ Sum of values of x= 2+3⇒ 5

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$\begin{array}{l} L e t f : R - {3} \to R - {1} b e d e f i n e d b y f (x) = \frac{x - 2}{x - 3} \\ L e t g : R \to R b e g i v e n a s g (x) = 2 x - 3. T h e n, t h e s u m o f a l l t h e v a l u e s o f x f o r w h i c h \\ f^{- 1} (x) + g^{- 1} (x) = \frac{13}{2} i s e q u a l t o . \end{array}$

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Detailed Solution

$\begin{array}{l} G i v e n f (x) = \frac{x - 2}{x - 3} \\ l e t f^{- 1} (x) = y \Rightarrow x = f (y) \\ \Rightarrow x = \frac{y - 2}{y - 3} \\ \Rightarrow x y - 3 x = y - 2 \\ \Rightarrow x y - y = 3 x - 2 \\ \Rightarrow y (x - 1) = 3 x - 2 \\ \Rightarrow y (x - 1) = 3 x - 2 \\ \Rightarrow y = \frac{3 x - 2}{x - 1} \\ ∴ f^{- 1} (x) = \frac{3 x - 2}{x - 1} \\ a n d g (x) = 2 x - 3 \\ l e t g^{- 1} (x) = y \Rightarrow x = g (y) \\ \Rightarrow x = 2 y - 3 \\ \Rightarrow y - \frac{x + 3}{2} \\ ∴ g^{- 1} (x) = \frac{x + 3}{2} \end{array}$

$\begin{array}{l} L e t f^{- 1} (x) + g^{- 1} (x) = \frac{13}{2} \\ \Rightarrow \frac{3 x - 2}{x - 1} + \frac{x + 3}{2} = \frac{13}{2} \\ \Rightarrow \frac{6 x - 4 + (x + 3) (x - 1)}{(x - 1) 2} = \frac{13}{2} \\ \Rightarrow 6 x - 4 + x^{2} - x + 3 x - = 13 x - 13 \\ \Rightarrow x^{2} - 5 x + 6 = 0 \\ \Rightarrow (x - 2) (x - 3) = 0 \\ \Rightarrow x = 2, 3 \\ ∴ S u m o f v a l u e s o f x = 2 + 3 \Rightarrow 5 \end{array}$