SEE Mathematics: Function

1. If h(x)= 2x+3 and g(x) = 3x-2, find the hₒg(5). 
Solution:
Here, 
    h (x)=2x+3 and  g(x) =3x-2.
Now,
                        hog (5) = h(g(5))
                                     = h(3×5-2)
                                     =h(13)=2×13+3 
                      ∴ hog(5) =29.
Thus, the value of hog (5) is 29.

2. If f(x) = 2x+32 then, find the value of f^-1(x). 
Solution:
Here,
      f (x) = 2x+32
Let y = f(x)
⇒  y= 2x+32
Interchanging the position of x and y,
      x = 2y+32
⇒ 2x= 2y+3
⇒ 2x- 3=2y
⇒ y = 2x-32
∴ f^-1(x) =2x-32
Thus, the value of f^-1(x) is 2x-32.

3. If f (x)=x+1 and g(x) =2x+1, find gₒf(x).
Solution:
Here,
 f(x)=x+1 and  g(x)=2x+1.
So,
      gₒf(x) = g[f(x)]
                 = g(x+1)
                 =2(x+1)+1
                 =2x+2+1
                 = 2x + 3.
 ∴ gof(x) = 2x+3.

4. If f (x)=3x-2 and fₒg(x) =6x-2, find g(x).
Solution:
Here,
      f(x)=3x-2 and  fₒg(x) = 6x-2.    
So,
      f[g(x)]= 6x-2
⇒ 3g(x)-2 = 6x-2         
⇒ 3g(x) = 6x  
⇒ g(x) = 3x
∴   g(x) = 2x.

5. If f (x) = 4x+3 then, find the value of f^-1(4).
Solution:
Here, 
      f(x) = 4x+3        
 ⇒ y = 4x+3        
Interchanging the position of x and y then,
      x =4y+3
⇒ x-3= 4y
 ⇒ y = (x-3)/4       
 ∴ f^-1(x) =(x-3)/4.
Now, 
     f^-1(4) =  ( 4-3)/4  
               =  1/4 .
Thus, the value of f^-1(4)  is    1/4.

6. If f^-1(x) =x+32, find f(x).
Solution:
Here,
      f^-1(x) =x+32
Let y =x+32
Interchanging x and y then,
x = y+32
⇒ 2x = y+3
⇒ 2x -3= y
∴    y =2x-3.
Thus, f(x) = 2x-3.

7. If f (x)=2x+5 and  g(x) =3x-1, find gₒf(x). 
Solution:
Here,
f(x)=2x+5 and  g(x) =3x-1
So,
   gₒf(x) = g(f(x))
               = g(2x+5)
               =3(2x+5) -1
               = 6x+15-1 
               = 6x + 14.

  ∴ gₒf(x) =6x+14.

8. If f^-1(x) = 2x-3, find f(x).        
Solution
Here,
      f^-1 (x) = 2x-3.
Let
      y = f^-1 (x)
⇒ y= 2x-3.

Interchanging the Value of x and y
     x= 2y-3
⇒ x+3 = 2y 
⇒ y = x+32
Thus,f(x)= x+32 .

9. If g (x) = x-23 and h (x)=3x+2, prove that gh(x) is an identity function.
Solution:
Here,
g(x) = x-23 and h (x)=3x+2.
We have, gh(x) = g (h(x))= g(3x+2)
                          = 3x+2-23
                          = 3x3
              ∴ gh(x)= x.
Thus, gh(x) = x shows that it is an identity function.

10. If f (x)=2x-3  and  g(x) =x² +1, find the value of fₒg(3). 

Here,
      f (x)=2x-3  and  g(x) =x² +1.
We have, fₒg(3)  = f[g(3)]
                                     = f(3²+1) 
                                     =f(10)  
                                     = 2×10-3
                                      =17.
Thus, the value of fₒg(3)  is 17.


11. If f = {x, 5x – 13}, g = {x, 2x+73 } and and         g^-1(x) = ff(x), find the value of x.
Here,
f = {x, 5x – 13},
g = {x, 2x+73 }
and g^-1(x) = ff(x)
Let y = g(x)
or, y = 2x+73
Interchanging the position of x and y, then
x = 2y+73
⇒ 3x = 2y + 7
⇒ 3x – 7 = 2y
⇒ y = 3x-72
∴ g^-1(x) = 3x-72
Again,
  ff(x)  = f(f(x))
              = f(5x – 13)
              = 5(5x – 13) – 13
              = 25x – 65 - 13
∴ ff(x) = 25x – 78
Now, by the question,
      3x-72= 25x – 78
⇒ 3x – 7 = 50x – 156
⇒ 194 = 47x
∴ x = 149/47.
Thus, the value of x is 3.17.

12. If 2f(x) = kx -3, 13 g(x) = 1x+2  and f og^-1(3) = -14, find the value of k. 
Solution:
Here, 
    2f(x) = kx – 3,
    13g(x) = 1x+2 and
     f og^-1(3) = -14
∴ f (x) = kx – 32,
    g(x) = 3x+2

For g^-1(x); 
Let y = g(x) 
 ⇒ y = 3x+2
Interchanging the position of x and y, 
     x = 3y+2
⇒ xy + 2x = 3 
⇒ xy = 3 – 2x 
     y = 3-2xx
∴ g^-1(3) = 3-2 ×33     
              = 3-63    
              = -33
              = -1 
Now,
      f g^-1(3) = -14
⇒ f(g^-1(3)) = -14
⇒  f(-1)    = -14
k-1- 32 = -14
⇒ -k – 3 = -12
⇒ 2k + 6 = 1
⇒ 2k = -5 
∴    k = -52
Thus, the value of k is -52.
 
13. Given that the functions f(x) = 3x – 7 and g(x) = 5x+23. If g^-1f(x) = 8, find the value of x. 
Solution
Here,
f(x) = 3x – 7 and
g(x) = 5x+23
Taking g(x) = 5x3
Let y = g(x) = 5x+23
Interchanging the position of x and y. 
      x = 5y+23
⇒ 3x = 5y + 2 
⇒ 3x – 2 = 5y 
⇒y = 3x-25
∴ g^-1(x) = 3x-25
Now,
         g^-1(f(x)) = g^-1(3x – 7) 
⇒ 8 = 33x-7-25
⇒ 9x – 21 – 2 = 40 
⇒ 9x = 63 
∴    x = 7.
Thus, the value of x is 7. 
 
14. If f(x) = x+12, g(x) = x-52 and fₒg(x) = 6, find the value of x. 
Solution:
Here, 
f(x) = x+12, 
g(x) = x-52
and fₒg(x) = 6
Taking g(x) = x-52
Let y = g(x)
⇒ y = x-52
Interchanging the position of x and y 
     x = y-52
⇒ 2x = y – 5
∴ y = 2x + 5 
∴ g^-1(x) = 2x + 5.

Now, f og-1(x) = 6 
⇒ f(g^-1(x))   = 6 
⇒ f(2x + 5)  = 6 
⇒  2x+5+12 = 6
⇒ 2x + 6 = 12
⇒ 2x = 6 
∴ x = 3.
Thus, the value of x is 3. 
  

15. If f(x) = 3x + 4 and g(x) = 2(x + 1) then

prove that (fₒg)(x) = (gₒf)(x). 

Solution:
Here, 
       f(x) = 3x + 4
and g(x) = 2(x + 1)
               = 2x + 2.
Now,
fₒg(x) = f[g(x)]
            = f(2x + 2) 
            = 3 (2x + 2) + 4
            = 6x + 6 + 4
∴ fₒg(x) = 6x + 10 .
Again, 
gₒf(x) = g(3x + 4) 
           = 2(3x + 4) + 2 
           = 6x + 8 + 2 
           = 6x + 10.
Thus, fₒg(x) = gₒf(x).


 16. If f(x) = 3x – 1 and fₒg(x) = 6x + 5, then find (gₒf)^-1. 
Solution:
Here, 
     f(x) = 3x – 1 and
     fₒg(x) = 6x + 5
⇒ f(g(x)) = 6x + 5
⇒ 3g(x) – 1 = 6x + 5
⇒ 3g(x) = 6x + 6 
∴   g(x) = 2x + 2.
Now, 
gₒf(x) = g[f(x)] = g(3x – 1)
           = 2(3x -1) + 2
           = 6x – 2 + 2 
∴ gₒf = 6x .
Let y = 6x 
Interchanging x and y then, 
x = 6y 
∴ y = x6
Thus, (gₒf)^-1 = x6.

17. If f(x) = 3x – 7, g(x) = 4x-23 and f^-1(x) = g(x), find the value of x. 
Solution:
Here, 
f(x) = 3x – 7 and
g(x) = 4x-23
Let
      y = f(x) 
⇒ y = 3x – 7 
Interchanging x and y, then
x = 3y – 7 
⇒ x + 7 = 3y 
⇒ y = x+73
∴ f^-1(x) = x+73
We have given,
    f^-1(x) = g(x)
⇒  x+73 = 4x-23
⇒ 12x – 6 = 3x + 21
⇒ 9x = 27 
∴   x = 3.
Thus, the value of x is 3.

18. If f(x) = 2x – 7 and fₒg(x) = 4x + 3, find (gₒf)^-1(x). 
Solution:
Here, 
    f(x) = 2x + 7 and
    fₒg(x) = 4x + 3.
So,
      fₒg(x) = 4x + 3
⇒ f(g(x)) = 4x + 3
⇒ 2.g(x) - 7 = 4x + 3 
⇒ 2g(x) = 4x + 10 
∴   g(x) = 2x + 5
We have, 
gₒf(x) = g(f(x))
          = g(2x – 7) 
           = 2(2x – 7) + 5
           = 4x – 14 + 5 
∴ gₒf(x) = 4x – 9
Let y = 4x – 9 and interchanging x and y
then,
     x = 4y – 9 
⇒ x + 9 = 4y 
Thus, (gₒf)^-1 (x) = x+94. 
 
19. If f(x) = 3x – 4 and f^-1g(x) = 3x+23,
find g(x) and f^-1(12).
Solution:
Here, 
     f(x) = 3x – 4 and
     f^-1g(x) = 3x+23
Let,
      y = f(x) 
 ⇒ y = 3x – 4 
Interchanging the position of x and y then, 
     x = 3y – 4 
⇒ x + 4 = 3y 
∴ y = x+43
i.e. f^-1(x) = x+43
By the question, 
f^-1g(x) = 3x+23
So,
    f^-1(g (x)) = 3x+23
⇒ gx+ 43 = 3x+23
⇒ g(x) + 4 = 3x + 2 
∴ g(x) = 3x – 2 
Now, 
f^-1(12) =  12 +43    
           = 9231    
           = 92 × 13
           = 32
Thus,
         g(x) = 3x – 2 and
     f^-1 (12) = 32.

20. If f(x) = 2x + 3, g(x) = 5 and ff(x) = g^-1(x), find the value of x. 
Solution:
Here, 
f(x) = 2x + 3, 
g(x) = x+52
ff(x) = g^-1(x)
So, 
      ff(x) = f(f(x)) 
              = f(2x + 3) 
              = 2(2x + 3) + 3
              = 4x + 6 + 3
              = 4x + 9 .
Again,
Let, y = g(x)
  ⇒ y = x+52
Interchanging the position of x and y, 
       x = y+52
⇒ 2x = y + 5 
⇒ 2x – 5 = y 
∴ y = 2x – 5 
So, g^-1(x) = 2x - 5 .

Now, 
       ff(x) = g^-1(x)
⇒ 4x + 9 = 2x – 5 
⇒ 2x = -14 
∴   x = -7 .
Thus, the value of x is -7. 
 
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