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# Let f : N→N be a function defined as f(x)=9x2+6x−5. Show that f : N→S, where S is the range of f, is invertible. Find the inverse of f and hence find f−1(43) and f−1(163). - CBSE (Science) Class 12 - Mathematics

#### Question

Let f : N→N be a function defined as f(x)=9x^2+6x−5. Show that f : N→S, where S is the range of f, is invertible. Find the inverse of f and hence find f^-1(43) and f^−1(163).

#### Solution 1

Given: f : N→ N is a function defined as           f(x)=9x2+6x−5

Let y=f(x) = 9x2+6x−5

⇒ y = 9x^2+6x-5

⇒ y =9x^2+6x+1-1-5

⇒ y =(9x^2+6x+1) - 6

⇒ y =(3x+1)^2 - 6

⇒ y+6=(3x+1)^2

⇒sqrt( y+6)=3x+1                  (∵ y ∈ N)

⇒sqrt(y+6)-1=3x

⇒x=(sqrt(y+6)-1)/3

⇒g (y) = (sqrt(y+6)-1)/3             [Let x = g (y)]

Now,

fog (y) = f [g (y) ]

=f ((sqrt( y+6)-1)/3)

=9((sqrt (y+6)-1)/3)^2 +6((sqrt (y+6)-1)/3)-5

=9((y+6-2sqrt(y+6)+1)/9)+2(sqrt(y+6)-1)-5

=y+6-2sqrt (y+6)+1+2sqrt(y+6)-2-5

=y

=I_Y, Identify funtion

gof (x) = g [f(x)]

=g (9x^2+6x-5)

=(sqrt((9x^2+6x-5)+6)-1)/3

=(sqrt((9x^2+6x+1))-1)/3

=(sqrt((3x+1)^2)-1)/3

=((3x+1)-1)/3

=(3x)/3

=x

=I_x, Identify function

Since, fog(y) and gof(x) are identify function .

Thus, f is invertible.

So,f^-1(x)=g (x)=(sqrt (x+6)-1)/3.

Now,

f^-1(43)=(sqrt(43+6)-1)/3=(sqrt(49)-1)/3 =(7-1)/3=6/3=2 And f^1(163)=(sqrt(163+6)-1)/3=(sqrt(169)-1)/3= (13-1)/3=12/3=4.

#### Solution 2

We have,

f : N→ is a function defined as f (x) = 9x2 + 6x-5.

Let y = f (x) = 9x2 + 6x - 5

⇒ y = 9x2+ 6x − 5

⇒ y = 9x2+ 6x + 1−1−5

⇒ y = (9x2+ 6x +1) − 6

⇒ y = ( 3x+1)2 − 6

⇒ y + 6 = (3x+1)2

⇒ sqrt(y+6) = 3x +1 (∵ y ∈ N)

⇒ sqrt (y+6) -1 = 3x

⇒ x =(sqrt(y + 6) - 1)/3

⇒ g (y) = (sqrt(y+6)-1)/3 [Let x = g (y) ]

Now,

fog (y) = f [ g (y) ]

= f ((sqrt(y+6) -1)/3)

= 9((sqrt(y+6) -1)/3)^2 + 6 ((sqrt(y+6) -1)/3) -5

= 9 ((y +6 - 2 sqrt(y +6) +1 )/9) +2 (sqrt(y+6)-1)-5

=  y+6 -2 sqrt(y+6)+1+2 sqrt(y+6)-2 -5

= y

= IY, Identity function

gof (x) = g [f(x)]

= g (9x^2 + 6x -5)

= (sqrt((9x^2 + 6x -5)+6)-1)/3

= (sqrt((9x^2 + 6x +1))-1)/3

= (sqrt((3x +1)^2) -1)/3

= ((3x +1) -1)/3

= (3x)/3

= x

= IX, Identity function

Since, fog(y) and gof(x) are identity function.

Thus, f is invertible.

So , f^-1(x)= g (x) = (sqrt(x+6)-1)/3

Now,

f^-1 (43)=(sqrt(43+6)-1)/3 = (sqrt(49)-1)/3 = (7-1)/3 = 6/3 =2

And f^-1 (163) = (sqrt (163+6) -1)/3 = (sqrt(169)-1)/3 = (13- 1)/3 = 12/3 =4

Is there an error in this question or solution?
Solution Let f : N→N be a function defined as f(x)=9x2+6x−5. Show that f : N→S, where S is the range of f, is invertible. Find the inverse of f and hence find f−1(43) and f−1(163). Concept: Inverse of a Function.
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