Solved Questions For MTH202 Final Exam

1.

Solution:
f: R R
g: R R
f(x) = x²  g(x) = 3x + 1
         (fog)(x) = f [g(x)]
                    = f [3x + 1]
                    = (3x + 1)²
                    = 9x² + 6x + 1
         (gof)(x) = g [f(x)]
                    = g [x²]
                    = 3(x²) + 1
                    = 3x² + 1
We observe that fog gof, that is the commutative law does not hold for the composition of functions.

2.

Solution:
f: R R
    f(x) = x² - 3x + 2
(fof)(x) = f [f(x)]
          = f (x² - 3x + 2)
          = (x² - 3x + 2)² - 3(x² - 3x +2) + 2
          = x⁴ + 9x² + 4 - 6x³ - 12x + 4x²
              - 3x² + 9x - 6 + 2
          = x⁴ - 6x³ + 10x² - 3x.

3.

f: R R defined by f(x) = x² + 3x + 1
g: R R defined by g(x) = 2x - 3

         i.            fog(x) = f [g(x)]

         = f [2x - 3]

         = (2x - 3)² + 3(2x - 3) + 1

         = 4x² - 12x + 9 + 6x - 9 + 1

         = 4x² - 6x + 1

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

         = g [x² + 3x + 1]

         = 2 (x² + 3x + 1) - 3

         = 2x² + 6x + 2 - 3

         = 2x² + 6x - 1

      iii.             fof(x) = f [f(x)]

         = f [x² + 3x + 1]

         = (x² + 3x + 1)² + 3(x² + 3x + 1) + 1

         = x⁴ + 11x² + 1 + 6x³ + 6x + 3x² + 9x + 3 + 1

         = x⁴ + 6x³ + 14x² + 15x + 5

4. gog(x) = g [g(x)]

          = g [2x - 3]

          = 2 (2x - 3) - 3

          = 4x - 6 - 3

          = 4x - 9

4.

f: R R defined by f(x) = x + 1
g: R R defined by g(x) = x - 1
 fog(x) = f [g(x)]
          = f (x - 1)
          = (x - 1) + 1
          = x
gof (x) = g [f(x)]
         = g [x + 1]
         = (x + 1) -1
         = x
          fog = gof = I_R
         I_R is the identity function.
          I_R(x) = x.

5.

f: N Z₀ defined by f(x) = 2x
g: Z₀ Q defined by g(x) = 1/x
h: Q R defined by h(x) = 5x + 2
To verify associativity we have to prove that ho(gof) = (hog)of.
Consider
  [ho(gof)](x) = h [(gof) (x)]
                  = h [g (f (x))]
                  = h [g (2x)]
                  = h [1/2x]
                  = 5 * 1/2x + 2
                  = 5/2x + 2
[(hog)of] (x) = (hog) [f(x)]
                  = (hog) (2x)
                  = h [g (2x)]
                  = h (1/2x)
                  = 5 * 1/2x + 2
                  = 5/2x + 2

6.

f: R R is the identity function
f(x) = x
fof (x) = f [f(x)]
         = f (x)
         = x
(ff) (x) = f (x) * f (x)
         = x * x
         = x²

What is the smallest integer N such that

a.         éN/7ù = 5         b. éN/9ù = 6

SOLUTION:

a.  N = 7 × (5 – 1) + 1 = 7 × 4 + 1 = 29

b.  N = 9 × (6 – 1) + 1 = 9 × 5 + 1 = 46

EXAMPLE:

                    Use the Euclidean algorithm to find gcd(330, 156)

Solution:

1.Divide 330 by 156:

                              This gives 330 = 156 · 2 + 18

2.Divide 156 by 18:

                              This gives 156 = 18 · 8 + 12

3.Divide 18 by 12:

                              This gives 18 = 12 · 1 + 6

4.Divide 12 by 6:

                              This gives 12 = 6 · 2 + 0

Hence gcd(330, 156) = 6.

By ADEEL ABBAS, Bhakkar. AdeelAbbasbk@gmail.com