SEE Mathematics: Polynomial

1. If 2x³ – 4x² + kx + 10 is divided by (x + 2), the remainder is 4. Find the value of k using remainder theorem.
Solution:
Here,
     Let  p(x) = 2x³ – 4x² + kx + 10, 
       d(x) = x + 2 and R = 4 .
Comparing x + 2 with x - a , we get
        a = -2 
Now, Using remainder theorem,
     p(-2)  = 2 × (-2)³ – 4 × (-2)² + k(-2) + 10
⇒ 4 = - 16 – 16 – 2k + 10 
⇒ 26 = -2k 
∴    k = - 13 .
Thus, the value of k is -13. 

2. If x – 3 is a factor of 2x³ – x² + 10x – k, then find the value of k.
Solution:
Here,
p(x) = 2x³ – x² + 10x – k and a factor = (x – 3).
Comparing x - 3 with  x - a, we get
 a = 3 .
Now,
Using factor  theorem.
    p(3) = 0
⇒ 2  × 3³ – 3² + 10  × 3 – k
⇒ 0 = 54 – 9 + 30 – k
∴    k = 75.
Thus, the value of k is 75. 

3. If x – 5 is a factor of x³ + px² + 4x + 5, find the value of p.
Solution:
Here,

p(x) = x³ + px² + 4x + 5 and a factor = (x – 5)

Comparing x - 5 with x - a, we get

x = 5 . 

Now,

Using factor theorem, 

              p(5) =  0

⇒  5³ + p × 5² + 4 × 5 + 5=0

⇒  125 + 25p + 25= 0

⇒   25p = -150

∴     p = - 6

Thus, the value of p is -6.

4. Using remainder theorem, find the remainder when 8x³ – 4x² + 2x – 5 is divided by 2x – 1.

Solution:

Here,

p(x) = 8x³ – 4x² + 2x – 5 and d(x) = 2x – 1= 2( x - 1/2)

Comparing x - 1/2 with x- a, we get,

a = 1/2. 

Now,Using remainder theorem   

  R = p(½)

⇒R = 8×(½)³ - 4(½)² + 2(½) - 5

⇒ R = 8 × 1 - 4 × 1 + 1 – 5

⇒ R = 1 – 1 + 1 – 5

∴   R = - 4.

Thus, the remainder is -4.

5. If x – 5 is a factor of 2x³ – 7px + (p – 12), find the value of p.

Solution:

Here,

p(x) = 2x³ – 7px + (p – 12) and factor = x – 5.

 Comparing x - 5 with x - a, we get, a = 5.

 Now,By using factor theorem

 p(5) = 0

 ⇒ 2 × 5³ – 7p × 5 + p – 12 = 0

 ⇒ 250 – 35p + p – 12  = 0 

 ⇒ 34p = 238 

  ∴    p = 7 .

 Thus, the value of p is 7.

6. If x³ – 19x – 30 = (x + 2). Q(x), find Q(x) by using synthetic division method.

Solution:
Here,
x³ – 19x – 30 = (x + 2). Q(x) 
Comparing (x + 2) with (x – a), then a = -2. 
Using synthetic division, 


∴ Q(x) = x² – 2x – 15.

7. State factor theorem. Use factor theorem to determine whether x + 3 is a factor of the polynomial x³ – 8x + 3. 
Solution:
i)
Statement of factor theorem:
"If a polynomial p(x) is divided by (x – a) and f(a) = R = 0 then (x – a) is a factor of p(x).
ii)
Here, p(x) = x³ – 8x + 3 and d(x) = x + 3
Comparing (x + 3) with (x – a) then, a = -3
Now,
        P(a) = P(3)
                 = (-3)³ – 8 ×  (-3) + 3 
                 = -27 + 24 + 3
                 = 0 .
Since P(3) = 0 shows that (x + 3)  is a factor  of P(x). 

 8. If x³ – 21x – 20 = (x + 1). Q(x), find Q(x) by using synthetic division method.
Solution: 
 Here,
x³ – 19x – 30 = (x + 2). Q(x) 
Comparing (x - a) with (x + 1), then a = -1. 
Using synthetic division, 

∴ Q(x) = x² – x – 20.

9. If the polynomial x² + 6x² + kx + 10 is divided by (x + 2), the remainder is 4, find the value of k using the remainder theorem.
Solution:
Here, p(x) = x³ + 6x² + kx + 10, 
d(x) = x + 2 and R = 4 
Using remainder theorem,
     p(-2) =4
⇒ (-2)³ + 6 × (-2)²+ k (-2) + 10 = 4.
⇒ 2k = 22 
⇒ 2k = 22 
∴   k = 11.
 
Thus, the value of k is 11. 
 
10. If (x + 2) is a factor of x³ – 19x – p, find the value of p.
Solution:
Here,
p(x) = x³ – 19x – p and x + 2 is factor of p(x),
Comparing x + 2 with x - a, we get  a  = - 2.
By factor theorem,
So, p(-2) = 0 
⇒(-2)³ – 19 (-2) – p = 0
⇒ -8 + 38 – p = 0
∴   p = 30 .
Thus, the value of p is 30. 

 11. Solve: 3x³ = 7x² – 4
Solution:
Here,
     3x³ = 7x² – 4
i.e.  3x³ – 7x² + 0.x + 4 = 0.
Let f(x)=3x³ – 7x² + 0.x + 4 = 0
Possible roots of f(x) are ±1,±2,±3,±4,±6,±12
For, x = 1
f(1)= 3.1³ - 7.1² + 4
       = 0
∴ 1 is a root of f(x).
Dividing f(x) by (x-1) by synthetic division method:
 = R.

∴ (x – 1) (3x² – 4x – 4) = 3x³ – 7x² + 0.x + 4
i.e. (x – 1) (3x² – 4x – 4) = 0
⇒ (x – 1) (3x² – 6x + 2x – 4) = 0
⇒ (x – 1) {3x(x – 2) + 2(x – 2)} = 0
⇒ (x – 1) (x – 2) (3x + 2) = 0
⇒ Either, x – 1 = 0 i.e. x = 1
Or, x – 2 = 0 i.e. x = 2
Or, 3x + 1 = 0 i.e. x = - 2/3.
Thus, x = 1 or 2 or - 23 is the solution.

13. Solve: y3 – 6y² + 11y – 6 = 0
Solution:
Here, 
Let f(y) y³ – 6y² + 11y – 6 = 0     
Possible factors of f(x) are ±1,±2,±3,±6
For y = 1, 
f(1)= 1³ - 6.1² + 11.1 - 6
       = 0
∴ 1 is a root of f(x).
Dividing f(y) by (y-1) by synthetic division method:


∴ (y – 1) (y² – 5y + 6) = y³ – 6y² + 11y – 6
i.e. y³ – 6y² + 11y – 6 = 0 
⇒ (y – 1) (y² – 5y + 6) = 0 
⇒ (y – 1) (y² – 3y – 2y + 6) = 0 
⇒ (y – 1) {y(y – 3) - 2(y – 3)} = 0 
⇒ (y – 1) (y – 3) (y - 2) = 0 
⇒  Either, y – 1 = 0 i.e. y=1
Or, y – 3 = 0 i.e. y=3
Or, y - 2 = 0 i.e. y=2
Thus, y = 1 or 3 or 2 is the solutions of the given equation. 

Exercise
14. Solve: 6x³ + x² – 19x + 6 = 0
15. Solve: 2x³ + 6 - 3x² – 11x = 0
16. Solve: x³ - 3x² – 10x + 24 = 0  
17. Solve: 6x³ = 4 - 13x²
18. Solve: 2x³ + 3x² – 11x - 6 = 0
19. Solve: x³ = 7x² - 36
20. Solve: x³ -21x – 20 = 0 
21. Solve: x³ - 4x² + x + 6 = 0 

 

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. 
 
⇒