Let f- -3- -1 be defined by fx - x-2x-3-Let g- be given as gx-2x-3- Then- the sum of all values of x for which f-1x-g-1x-132 is equal to

F(x) = x 2x 3 ⇒ y = nbsp;x 2x 3 ⇒ xy 3y = x 2 ⇒ x(y 1)=3y 2 ⇒ x = 3y 2y 1 ∴ f^ 1(x) = nbsp;3x 2x 1 Similarly, nbsp; g^ 1(x) = x+32 Now, f^ 1(x)+g^ 1(x)=3 ...

Let f: ℝ-3→ℝ ...

Question

Let $f : R - {3} \to R - {1}$ be defined by $f (x) = \frac{x - 2}{x - 3} .$
Let $g : R \to R$ be given as $g (x) = 2 x - 3.$ Then, the sum of all values of $x$ for which $f^{- 1} (x) + g^{- 1} (x) = \frac{13}{2}$ is equal to

7

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Solution

The correct option is B $5$
$f (x) = \frac{x - 2}{x - 3} \Rightarrow y = \frac{x - 2}{x - 3} \Rightarrow x y - 3 y = x - 2 \Rightarrow x (y - 1) = 3 y - 2 \Rightarrow x = \frac{3 y - 2}{y - 1} ∴ f^{- 1} (x) = \frac{3 x - 2}{x - 1}$
Similarly,
$g^{- 1} (x) = \frac{x + 3}{2}$

Now,
$f^{- 1} (x) + g^{- 1} (x) = \frac{3}{2} \Rightarrow \frac{3 x - 2}{x - 1} + \frac{x + 3}{2} = \frac{13}{2} \Rightarrow 2 (3 x - 2) + (x - 1) (x + 3) = 13 (x - 1) \Rightarrow x^{2} - 5 x + 6 = 0 \Rightarrow x = 2 or 3$

Hence, the required sum is $5.$

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

The correct option is B 5f(x)=x−2x−3⇒y=x−2x−3⇒xy−3y=x−2⇒x(y−1)=3y−2⇒x=3y−2y−1∴f−1(x)=3x−2x−1 Similarly, g−1(x)=x+32 Now, f−1(x)+g−1(x)=32⇒3x−2x−1+x+32=132⇒2(3x−2)+(x−1)(x+3)=13(x−1)⇒x2−5x+6=0⇒x=2 or 3 Hence, the required sum is 5.

Let $f : R - {3} \to R - {1}$ be defined by $f (x) = \frac{x - 2}{x - 3} .$
Let $g : R \to R$ be given as $g (x) = 2 x - 3.$ Then, the sum of all values of $x$ for which $f^{- 1} (x) + g^{- 1} (x) = \frac{13}{2}$ is equal to