EX. NO. 1.5, RELATION AND FUNCTIONS, 10TH NEW SYLLABUS, Tamil Nadu, TN,

EX. NO. 1.5, RELATION AND FUNCTIONS, 10TH NEW SYLLABUS, Tamil Nadu, TN,

Question 1 : 
Using the functions f and g given below, find f o g and g o f . Check whether  f o g = g o f .
(i) f (x) = x −6, g(x) = x2
Solution : 
f o g (x)  = f[g(x)]
We apply the function given for g(x).
f o g (x)  = f[x2]
Instead of x, we have x2, so we apply  x2 instead of x in f(x).
f o g (x)  = f(x2) = x2 − 6   ---(1)
g o f (x)  = g[f(x)]
=  g [x- 6]
Instead of x, we have x - 6, so we apply  x - 6 instead of x in g(x).
g o f (x)  = g(x - 6)  = (x - 6)2  ---(2)
f o g (x) g o f (x)
(ii) f (x) = 2/x, g(x) = 2x2 - 1 
Solution : 
f o g (x)  = f[g(x)]
We apply the function given for g(x).
f o g (x)  = f[2x2 - 1]
Instead of x, we have 2x2 - 1, so we apply  2x2 - 1 instead of x in f(x).
f o g (x)  = f(2x2 - 1) = 2/(2x2 - 1)   ---(1)
g o f (x)  = g[f(x)]
=  g [2/x]
Instead of x, we have 2/x, so we apply  2/x instead of x in g(x).
g o f (x)  = g(2/x) = 2(2/x)2 - 1
=  2(4/x2) - 1
=  (8/x2) - 1   ---(2)
f o g (x) g o f (x)
(iii) f (x) = (x + 6)/3,  g(x) = 3 - x 
Solution : 
f o g (x)  = f[g(x)]
We apply the function given for g(x).
f o g (x)  = f[3 - x]
Instead of x, we have 3 - x, so we apply  3 - x instead of x in f(x).
f(3 - x)  = (3-x+6)/3
=  (9-x)/3  ---(1)
g o f (x)  = g[f(x)]
=  g [(x + 6)/3]
Instead of x, we have [(x + 6)/3], so we apply  [(x + 6)/3] instead of x in g(x).
 g [(x + 6)/3]  = 3 - [(x + 6)/3]
  =  (9 - x - 6)/3
=  (3 - x)/3  ---(2)
f o g (x) g o f (x)
(iv) f (x) = 3 + x,  g(x) = x - 4
Solution : 
f o g (x)  = f[g(x)]
We apply the function given for g(x).
f o g (x)  = f[x - 4]
Instead of x, we have x - 4, so we apply x - 4 instead of x in f(x).
f(x - 4)  = 3 + x - 4
=  x - 1  ---(1)
g o f (x)  = g[f(x)]
=  g [3 + x]
Instead of x, we have 3 + x, so we apply 3 + x instead of x in g(x).
 g [3 + x]  = 3 + x - 4 
  = x - 1  ---(2)
(1)  = (2)
f o g (x)  = g o f (x)
(v)  f (x) = 4x2 − 1, g(x) = 1 + x
Solution :
f o g (x)  = f[g(x)]
We apply the function given for g(x).
f o g (x)  = f[1 + x]
Instead of x, we have 1 + x, so we apply 1 + x instead of x in f(x).
f(1 + x)  = 4(1+x)2 - 1
  =  4(1 + x2 + 2x) - 1
  =  4 + 4x2 + 8x - 1
f o g (x)  = 4x2 + 8x + 3   ---(1)
g o f (x)  = g[f(x)]
=  g [4x2 − 1]
Instead of x, we have 4x2 − 1, so we apply 4x2 − 1 instead of x in g(x).
 g [4x2 − 1]  = 1 + 4x2 − 1
  = 4x2  ---(2)
f o g (x) g o f (x)

Question 2 :
Find the value of k, such that f o g = g o f
(i) f (x) = 3x +2, g(x) = 6x −k
Solution :
Given that 
f o g = g o f
f o g  = f[g(x)]
= f[6x - k] 
Now we are going to apply 6x - k instead of x in f(x).
f o g  = 3(6x - k) + 2 ----(1)
g o f  = g[f(x)]
=  g[3x + 2]
Apply 3x + 2 instead of x in the function g(x).
g o f  = 6(3x + 2) - k  -----(2)
(1)  = (2)
3(6x - k) + 2  = 6(3x + 2) - k
18x - 3k + 2  = 18x + 12 - k
-3k + k  = -2 + 12
-2k  = 10
k  = -5
Hence the value of k is -5.

(ii) f (x) = 2x −k, g(x) = 4x + 5
Given that 
f o g = g o f
f o g  = f[g(x)]
= f[4x + 5] 
Now we are going to apply 4x + 5 instead of x in f(x).
f o g  = 2(4x + 5) - k ----(1)
g o f  = g[f(x)]
=  g[2x - k]
Apply 2x - k instead of x in the function g(x).
g o f  = 4(2x - k) + 5  -----(2)
(1)  = (2)
2(4x + 5) - k  = 4(2x - k) + 5
8x + 10 - k  = 8x - 4k + 5
-k + 4k  = 5 - 10
3k  = -5
k  = -5/3
Hence the value of k is -5/3.

Question 3 :
If f (x) = 2x −1, g (x) = (x + 1)/2, show that f o g = g o f = x
Solution :
f o g  = f[g(x)]
=  f[(x + 1)/2]
Now we apply (x + 1)/2 instead of x in f(x).
  =  2((x + 1)/2) - 1
  =  x + 1 - 1
  =  x --(1)
g o f  = g[f(x)]
=  g[2x - 1]
=  (2x - 1 + 1)/2
=  2x/2
  =  x --(2)
(1)  = (2)
Question 4 :
(i) If f (x) = x2 −1, g(x) = x −2 find a, if g o f (a) = 1 .
Solution :
f(a)  = a2 −1 and g(a) = a −2
g o f (a)  = g[f(a)]
  =  g[a2 −1]
  =  (a2 −1) - 2
g o f (a)  = a2 −3  
Given that 
 g o f (a) = 1 
a2 −3 = 1
a2 = 4
a  = ±2

(ii) Find k, if f (k) = 2k −1 and f o f (k) = 5.
Solution :
f o f (k)  = f[f(k)]
  =  f[2k - 1]
Now we apply 2k - 1 instead of k in f(k)
  =  2(2k - 1) - 1
  =  4k - 2 - 1
f o f (k)  = 4k - 3
Given that :
f o f (k)  = 5
4k - 3  = 5
4k  = 5 + 3
4k  = 8
k  = 2 
Hence the value of k is 2.
Question 5 :
Let A, B, C ⊆ N and a function f : A -> B be defined by f(x) = 2x + 1 and g : B -> C be defined by g(x) = x2 . Find the range of f o g and g o f .
Solution :
f o g  = f[g(x)]
=  f[x2]
Now we apply x2 instead of x in f(x).
f o g  = 2 x2 + 1
y = 2x2 + 1
Range :
{y | y = 2x2 + 1 and x ∊ N}
g o f  = g[f(x)]
=  g[2x + 1]
Now we apply 2x + 1 instead of x in g(x).
g o f  = (2x + 1)2
Range :
{y | y = (2x + 1)2 and x ∊ N}
Question 6 :
Let f (x) = x2 −1 . Find (i) f o f (ii) f o f o f
Solution :
(i) f o f  
  =  f[f(x)]
  =  f[x2 −1]
Now we apply x2 −1 instead of x in f(x).
  =  (x2 −1)2 - 1
  =  x4 - 2x2 + 1 - 1
f o f  = x4 - 2x2 
(ii) f o f o f
f o f  = x4 - 2x2 
f o f o f  = f [f o f]
  =  f[x4 - 2x2]
Now we apply x4 - 2x2 instead of x in f(x).
  =  (x4 - 2x2)2 - 1
Question 7 :
If f : R -> R and g : R -> R are defined by f(x) = x5 and g(x) = x4 then check if f, g are one-one and f o g is one-one?
Solution :
 f(x) = x5
For every positive and negative values of x, we get positive and negative values of y.
Every element in x is associated with different elements of y. Hence it is one to one function.
g(x) = x4 
For every positive and negative values of x, we get only positive values of y.
Negative values of y is not associated with any elements of x. Hence it is not one to one function.
fog(x)  = f[g(x)]
  =  f[x4]
now, we apply x4 instead of x in f(x)
f[x4]  = (x5)4
fog(x)  = x20
fog is not one to one function.
Question 8 :
Consider the functions f (x), g(x), h(x) as given below. Show that (f o g) o h = f o (g o h) in each case.
(i) f(x) = x −1, g(x) = 3x +1 and h(x) = x2
Solution :
(f o g) o h
fog(x)  = f[g(x)]
  =  f[3x + 1]
  =  3x + 1 - 1
fog(x)  = 3x
(f o g) o h  = (f o g) [h(x)]
  =  (f o g) [x2]
(f o g) o h  = 3x2     --------(1)
f o (g o h)
(g o h)  = g[h(x)]
  =  g[x2]
  =  3x2 +1 
f o (g o h)  = f [goh]
  =  f[3x2 +1] 
=  3x2 + 1 - 1
f o (g o h)  = 3x2 -------(2)
(1)  = (2)
(f o g) o h  = f o (g o h)
(ii) f (x) = x2, g(x) = 2x and h(x) = x + 4
Solution :
(f o g) o h
fog(x)  = f[g(x)]
  =  f[2x]
  =  (2x)2
fog(x)  = 4x2
(f o g) o h  = (f o g) [h(x)]
  =  (f o g) [x + 4]
(f o g) o h  = 4(x+4)2
=  4(x2 + 8x + 16)
=  4x2 + 32x + 64     --------(1)
f o (g o h)
(g o h)  = g[h(x)]
  =  g[x + 4]
  =  2(x + 4)
(g o h)  = 2x + 8
f o (g o h)  = f [goh]
  =  f[2x + 8] 
f o (g o h)  = (2x + 8)2
 =  (2x)2 + 2(2x)(8) + 82
=  4x2 + 32 x + 64-------(2)
(1)  = (2)
(f o g) o h  = f o (g o h)
(iii) f (x) = x −4, g(x) = x2 and h(x) = 3x −5
Solution :
(f o g) o h
fog(x)  = f[g(x)]
  =  f[x2]
  =  x2 −4
fog(x)  = x2 −4
(f o g) o h  = (f o g) [h(x)]
  =  (f o g) [3x −5]
(f o g) o h  = (3x - 5)2 - 4
=  (3x)2 - 2 (3x)(5) + 52 - 4
=  9x2 - 30x + 25 - 4
=  9x2 - 30x + 21   --------(1)
f o (g o h)
(g o h)  = g[h(x)]
  =  g[3x −5]
  =  (3x −5)2
=  (3x)2 - 2 (3x)(5) + 52
(g o h)  = 9x2 - 30x + 25
f o (g o h)  = f [goh]
  =  f[9x2 - 30x + 25] 
f o (g o h)  = 9x2 - 30x + 25 - 4
=  9x2 - 30x + 21   --------(2)
(1)  = (2)
Hence proved.

Question 9 :
Let f = {(−1, 3),(0,−1),(2,−9)} be a linear function from Z into Z . Find f (x).
Solution :
Let the linear function be "y = ax + b"


By applying the value of b in (1)
-a + (-1)  = 3
-a  = 3 + 1
-a  = 4
a  = -4
By applying the value of "a" and "b", we get 
y = -4x - 1
Hence the required linear equation is -4x - 1.
Question 10 :
In electrical circuit theory, a circuit C(t) is called a linear circuit if it satisfies the superposition principle given by C(at1 + bt2) = aC(t1) + bC(t2), where a,b are constants. Show that the circuit C(t) = 3t is linear.
Solution :
Let's take two points  t1 and t2 from domain of C(t).

now,  c(at1 + bt2)  = 3(at1 + bt2)
c(at1) =  3at1
ac(t1) =  3at1
bc(t2) =  3bt2
3(at1 + bt2)  = 3at1 + 3at2 
(or) C(at1 + bt2) = aC(t1) + bC(t2),
Hence c(t) is linear.

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