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Question
Consider the function f : R+ → [-9 , ∞ ]given by f(x) = 5x2 + 6x - 9. Prove that f is invertible with f -1 (y) = `(sqrt(54 + 5y) -3)/5` [CBSE 2015]
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Solution
We have ,
f (x) = 5x2+ 6x − 9
Let y = 5x2+ 6x − 9
` = 5 (x^2 + 6/5x - 9/5)`
` = 5(x^2 + 2 xx x xx 3/5 + 9 /25 - 9/25 - 9/5)`
`= (( x + 3/5)^2 - 9/25 - 9/25)`
`=(x+ 3/5)^2 - 9/5 - 9 `
`= 5 (x + 3/5)^2 - 54/5`
⇒ `y + 54/5 = 5 (x+3/5)^2`
⇒ `(5y + 54)/25 (x + 3/5)^2`
⇒ `sqrt (5y +54)/25 = x +3/5`
⇒ `x = sqrt (5y +54)/5 - 3/5`
⇒ `x = (sqrt (5y +54)-3)/5 `
Let g (y) =` (sqrt(5y +54) -3)/5`
Now,
fog (y) = f (g (y))
= f `((sqrt (5y+54)-3)/5)`
= 5 `((sqrt (5y+54)-3)/5)^2 + 6 ((sqrt (5y+54)-3)/5) = - 9 `
`= 5 ((5y + 54 +9 - 6 sqrt (5y +54))/25) + ((6 sqrt(5y + 54) -18)/5) -9`
`= (5y + 63 - 6 sqrt (5y + 54))/5 +(6 sqrt (5y + 54)- 18)/5 -9`
=` (5y + 63 - 18 - 45) /5`
= y
= IY, Identity function
Also, gof (x) = g (f(x))
= g (5x2 + 6x - 9 )
`= (sqrt(5(5x^2 + 6x - 9)+ 54)-3)/5`
`= (sqrt(25x^2 + 30x - 45 +54) -3)/5`
`=(sqrt(25 x^2 + 30x + 9) -3)/5`
`= (sqrt((5x + 3)^2) - 3)/5`
`= (5x +3 -3)/5`
= x
= IX , Identity function
So, f is invertible .
Also, `f^-1 (y) = g (y) = (sqrt(5y +54) -3)/5`
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