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प्रश्न
Let `f:R - {-4/3} -> R` be a function defined as `f(x) = (4x)/(3x + 4)`. The inverse of f is map g Range `f -> R -{- 4/3}`
(A) `g(y) = (3y)/(3-4y)`
(B) `g(y) = (4y)/(4 - 3y)`
(C) `g(y) = (4y)/(3 - 4y)`
(D) `g(y) = (3y)/(4 - 3y)`
उत्तर
t is given that `f:R -{-4/3} -> R` is defined as `f(x) = (4x)/(3x - 4)`
Let y be an arbitrary element of Range f.
Then, there exists x ∈ `R - {-4/3}` such that `y = f(x)`
`=> y = (4x)/(3x + 4)`
=>3xy + 4y = 4x Let us define g: Range `f -> R -{-4/3} " as " g(y) = (4y)/(4 -3y)`
=> x(4 - 3y) = 4y
Now `(gof)(x) = g(f(x)) = g((4x)/(3x + 4))`
`=> x = (4y)/(4 -3y)`
`= (4(4x/(3x + 4)))/(4 - 3("4x"/(3x + 4))) = "16x"/(12x + 16 - 12x) = "16x"/16 = x`
And `(fog)(y) = f(g(y)) = f((4y)/(4 - 3y))`
`= (4("4y"/(4-3y)))/(3("4y"/(4 - 3y)) + 4) = (16y)/(12y + 16 - 12y) = (16y)/16 = y`
:. `gof = I_(R - (-4/3))` and fog = I_"Range f"
Thus, g is the inverse of f i.e., f−1 = g.
Hence, the inverse of f is the map `g: Range f -> R - {-4/3}` which is given by
`g(y) = "4y"/(4 - 3y)`
The correct answer is B.
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