Important Questions - Answers
 

1. If the function f is defined by f(x) = x + 2,   x > 1

                                                                          = 2,         -1 ≤ x ≤ 1

                                                                          = x -1,      -3 < x < 1

      find   (i) f (3)   (ii) f(0)   (iii) f(-1.5)   (iv) f(2) + f(-2)   (v) f(-5)

Sol: (i) f(x) = x + 2,   x > 1

         f(3) = 3 + 2 = 5

         (ii) f(x) = 2, -   1 ≤ x ≤ 1

           f(0) = 2

        (iii) f(x) = x - 1,  -3 < x < 1

           f(-1.5) = -1.5 - 1 = -2.5

        (iv) f(x) = x + 2

           f(x) = 2 + 2 = 4

         f(x) = x - 1

           f(-2) = -2 - 1 = -3

            f(2) + f(-2) = 4 - 3 = 1

          v) f(-5) cannot be found out by using the given definition of f(x).
 

2. If f: R  R is defined by f(x) =  , then show that f (tan ) = cos 2 

Sol:

3. If f : R - {±1}   R is defined by f(x) = log  ,

     then show that = 2f(x).

Sol:

4. If A = {1, 2, 3, 4} and f : A  R is defined by f(x) =  , find the range of f.

Sol:

5. Verify if f : R  (0,  ) defined by f(x) = 2^x is a bijection or not?

Sol:

6. Is g = {(1, 1), (2, 3), (3, 5), (4, 7)} a function from A to B, where A = { 1, 2, 3, 4} and B =

     {1, 3, 5, 7}. If g(x) is defined as g(x) = ax + b, then find 'a' and 'b'.

Sol: A = {1, 2, 3, 4}; B = {1, 3, 5, 7}

       g = {(1, 1), (2, 3), (3, 5), (4, 7)}

        g (1) = 1, g(2) = 3, g(3) = 5, g(4) = 7

       This suits the definition of a function

        f is a function from A to B

       g(x) = ax + b, x 

 A

       g(1) = a(1) + b = a + b         a + b = 1  (1)

       g(2) = a(2) + b = 2a + b      2a + b = 3  (2)

       g(3) = a(3) + b = 3a + b      3a + b = 5 

(3)

       g(4) = a(4) + b = 4a + b        4a + b = 7  (4)

       Solving any two equations, we get a = 2, b = -1.

7. If f(x) = e^x and g(x)= log_e^x, then show that gof= fog and find f^-1 and g^-1.

Sol:   f(x) = e^x, g(x) = log_e^x

                   (gof)(x) = g[f(x)]

                                = g[e^x]

                                = log_e e^x

                                = x (log_ee)

                                = x  (1)

       (fog)(x) = f[g(x)]

                     = f(log_ex)

                     = elog_ex

                     = x  (2)

      (1), (2)  (gof)(x) = (fog)(x)   

gof = fog

      Let y = f(x) = e^x

         f^-1(y) = x;      y = e^x

       log_ey = x 

                 x = x

       f^-1 (y) = log_ey

       f^-1 (x) = log_ex  (3)

      Let z = g(x) = logex

         g^-1 (z) = x, z = logex

                            e^z = x

          x = x

     

 g^-1(z) = ez

        g^-1 (x) = ex

        (1), (2), (3), (4)  The result.