RELATIONS AND FUNCTIONS FOR GRADE 10

Question 1 :

The functions f and g are defined by f (x) = 6x + 8; g (x)  = (x - 2)/3

(i) Calculate the value of gg (1/2)

(ii) Write an expression for gf (x) in its simplest form.

Solution :

g (x)  = (x - 2)/3

g(1/2)  =  ((1/2) - 2)/3

  =  (-3/2)/3

g(1/2)   =  -1/2

gg(1/2)  =  ((-1/2) - 2)/3

=  (-5/2)/3

gg(1/2)  =  -5/6

(ii) Write an expression for gf (x) in its simplest form.

gf (x)  =  g[6x +8]

Now we have to apply 6x + 8 instead of x in g(x).

  =  [(6x + 8) - 2]/3

  =  (6x + 6)/3

  =  2x + 2  

=  2 (x + 1)

Question 2 :

Write the domain of the following real functions.

(i) f (x)  =  (2x + 1)/(x - 9)

Solution :

To find the domain, let us equate the denominator to 0

x - 9 = 0 

x  =  9

The function is defined for all real values of x except 9.

Hence the required domain is R - {9}.

(ii)  p(x)  =  -5/(4x² + 1)

Solution :

In the denominator, we have x2. For all real values of x, we get positive values. Hence the required domain is R. 

(iii)  g(x)  =  √(x - 2)

Solution :

Since the given function is in radical sign, we should not get negative answer. For that we have to apply the values grater than 2.

Hence the required domain for he given function is [2, ∞)

(iv) h(x)  =  x + 6

Solution :

For all real values of x, we get defined values of h(x). Hence the domain is R.

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