JEE MAIN - Mathematics (2021 - 18th March Evening Shift - No. 13)
Let f : R $$-$$ {3} $$ \to $$ R $$-$$ {1} be defined by f(x) = $${{x - 2} \over {x - 3}}$$.
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) = $${{13} \over 2}$$ is equal to :
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) = $${{13} \over 2}$$ is equal to :
3
5
2
7
Explanation
Finding inverse of f(x)
$$y = {{x - 2} \over {x - 3}} \Rightarrow xy - 3y = x - 2 \Rightarrow x(y - 1) = 3y - 2$$
$$ \therefore $$ $${f^{ - 1}}(x) = {{3x - 2} \over {x - 1}}$$
Similarly for $${g^{ - 1}}(x)$$
$$y = 2x - 3 \Rightarrow x = {{y + 3} \over 2} \Rightarrow {g^{ - 1}}(x) = {{x + 3} \over 2}$$
$$ \therefore $$ $${{3x - 2} \over {x - 1}} + {{x + 3} \over 2} = {{13} \over 2}$$
$$ \Rightarrow 6x - 4 + {x^2} + 2x - 3 = 13x - 13$$
$$ \Rightarrow {x^2} - 5x + 6 = 0$$
$$ \Rightarrow (x - 2)(x - 3) = 0$$
$$ \Rightarrow $$ x = 2 or 3
$$y = {{x - 2} \over {x - 3}} \Rightarrow xy - 3y = x - 2 \Rightarrow x(y - 1) = 3y - 2$$
$$ \therefore $$ $${f^{ - 1}}(x) = {{3x - 2} \over {x - 1}}$$
Similarly for $${g^{ - 1}}(x)$$
$$y = 2x - 3 \Rightarrow x = {{y + 3} \over 2} \Rightarrow {g^{ - 1}}(x) = {{x + 3} \over 2}$$
$$ \therefore $$ $${{3x - 2} \over {x - 1}} + {{x + 3} \over 2} = {{13} \over 2}$$
$$ \Rightarrow 6x - 4 + {x^2} + 2x - 3 = 13x - 13$$
$$ \Rightarrow {x^2} - 5x + 6 = 0$$
$$ \Rightarrow (x - 2)(x - 3) = 0$$
$$ \Rightarrow $$ x = 2 or 3
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