RD Chapter 2 Functions Ex 2.2 Solutions
Question - 1 : - Find gof and fog when f: R → R and g : R → R isdefined by
(i) f(x) = 2x + 3 and g(x) = x2 + 5.
(ii) f(x) = 2x + x2 and g(x)= x3
(iii) f (x) = x2 + 8 and g(x) =3x3 + 1
(iv) f (x) = x and g(x) = |x|
(v) f(x) = x2 + 2x − 3 and g(x) =3x − 4
(vi) f(x) = 8x3 and g(x) = x1/3
Answer - 1 : -
(i)Given, f: R → R and g: R → R
So, gof: R → R and fog: R → R
Also given that f(x) =2x + 3 and g(x) = x2 + 5
Now, (gof) (x) = g (f (x))
= g (2x +3)
= (2x + 3)2 + 5
= 4x2+ 9 + 12x +5
=4x2+ 12x + 14
Now, (fog) (x) = f (g(x))
= f (x2 + 5)
= 2 (x2 + 5) +3
= 2 x2+ 10 + 3
= 2x2 + 13
(ii)Given, f: R → R and g: R → R
so, gof: R → R and fog: R → R
f(x) =2x + x2 and g(x) = x3
(gof) (x)= g (f (x))
= g (2x+x2)
= (2x+x2)3
Now, (fog) (x)= f (g (x))
= f (x3)
= 2 (x3)+ (x3)2
= 2x3 +x6
(iii)Given, f: R → R and g: R → R
So, gof: R → R and fog: R → R
f(x) = x2 +8 and g(x) = 3x3 + 1
(gof) (x)= g (f (x))
= g (x2 + 8)
= 3 (x2+8)3 + 1
Now, (fog) (x)= f (g (x))
= f (3x3 + 1)
= (3x3+1)2 + 8
= 9x6 + 6x3 + 1+ 8
= 9x6 +6x3 + 9
(iv)Given, f: R → R and g: R → R
So, gof: R → R and fog: R → R
f(x)= x and g(x) = |x|
(gof) (x)= g (f (x))
= g (x)
= |x|
Now (fog) (x)= f (g (x))
= f (|x|)
= |x|
(v)Given, f: R → R and g: R → R
So, gof: R → R and fog: R → R
f(x) = x2 +2x − 3 and g(x) = 3x − 4
(gof) (x)= g (f(x))
= g (x2 +2x − 3)
= 3 (x2 +2x − 3) − 4
= 3x2 + 6x − 9 − 4
= 3x2 +6x − 13
Now, (fog) (x)= f (g (x))
= f (3x − 4)
= (3x − 4)2 +2 (3x − 4) −3
= 9x2 +16 − 24x + 6x – 8 − 3
= 9x2 −18x + 5
(vi)Given, f: R → R and g: R → R
So, gof: R → R and fog: R → R
f(x) = 8x3 and g(x)= x1/3
(gof) (x)= g (f (x))
= g (8x3)
= (8x3)1/3
= [(2x)3]1/3
= 2x
Now, (fog) (x)= f (g (x))
= f (x1/3)
= 8 (x1/3)3
= 8x
Question - 2 : - Let f = {(3, 1), (9, 3), (12, 4)} and g = {(1, 3), (3, 3) (4, 9) (5, 9)}. Show that gof and fog are both defined. Also, find fog and gof.
Answer - 2 : -
Given f = {(3, 1), (9,3), (12, 4)} and g = {(1, 3), (3, 3) (4, 9) (5, 9)}
f : {3, 9, 12} → {1, 3, 4} and g : {1, 3, 4, 5} → {3, 9}
Co-domainof f is a subset of the domain of g.
So, gof exists and gof: {3, 9, 12} → {3, 9}
(gof) (3) =g (f (3)) = g (1) = 3
(gof) (9) =g (f (9)) = g (3) = 3
(gof) (12) =g (f (12)) = g (4) = 9
⇒ gof ={(3, 3), (9, 3), (12, 9)}
Co-domainof g is a subset of the domain of f.
So, fog exists and fog: {1, 3, 4, 5} → {3, 9, 12}
(fog) (1) =f (g (1)) = f (3) = 1
(fog) (3) =f (g (3)) = f (3) = 1
(fog) (4) =f (g (4)) = f (9) = 3
(fog) (5) =f (g (5)) = f (9) = 3
⇒ fog ={(1, 1), (3, 1), (4, 3), (5, 3)}
Question - 3 : - Let f = {(1, −1), (4, −2), (9, −3), (16, 4)} and g = {(−1, −2), (−2, −4), (−3, −6), (4, 8)}. Show that gof is defined while fog is not defined. Also, find gof.
Answer - 3 : -
Given f = {(1,−1), (4, −2), (9, −3), (16, 4)} and g = {(−1, −2), (−2, −4), (−3, −6),(4, 8)}
f: {1, 4, 9, 16} → {-1, -2, -3, 4} and g: {-1, -2, -3, 4} → {-2, -4, -6,8}
Co-domainof f = domain of g
So, gof exists and gof: {1, 4, 9, 16} → {-2, -4, -6, 8}
(gof) (1) = g (f (1)) = g (−1) = −2
(gof) (4) = g (f (4))= g (−2) = −4
(gof) (9) = g (f (9)) = g (−3) = −6
(gof) (16) =g (f (16)) = g (4) = 8
So, gof = {(1, −2), (4, −4), (9, −6), (16, 8)}
But the co-domainof g is not same as the domain of f.
So, fog does not exist.
Question - 4 : - Let A = {a, b, c}, B = {u, v, w} and let f and g be two functions from A to B and from B to A, respectively, defined as: f = {(a, v), (b, u), (c, w)}, g = {(u, b), (v, a), (w, c)}.
Show that f and g both are bijections and find fog and gof.
Answer - 4 : -
Given f ={(a, v), (b, u), (c, w)}, g = {(u, b),(v, a), (w, c)}.
Also given that A = {a, b, c}, B = {u, v, w}
Now we have to show fand g both are bijective.
Consider f ={(a, v), (b, u), (c, w)} and f: A → B
Injectivity of f: No two elements of A have the same image in B.
So, f is one-one.
Surjectivity of f: Co-domain of f = {u, v, w}
Range of f = {u, v, w}
Both are same.
So, f is onto.
Hence, f is a bijection.
Now consider g ={(u, b), (v, a), (w, c)} and g: B → A
Injectivity of g: No two elements of B have the same imagein A.
So, g is one-one.
Surjectivity of g: Co-domain of g = {a, b, c}
Range of g = {a, b, c}
Both are the same.
So, g is onto.
Hence, g is a bijection.
Now we have to findfog,
we know that Co-domain of g is same as the domain of f.
So, fog exists and fog:{u v, w} → {u, v, w}
(fog) (u) = f (g (u)) = f (b) = u
(fog) (v) = f (g (v)) = f (a) = v
(fog) (w) = f (g (w)) = f (c) = w
So, fog = {(u, u), (v, v), (w, w)}
Now we have to findgof,
Co-domain of f is same as the domain of g.
So, fog exists and gof:{a, b, c} → {a, b, c}
(gof) (a) = g (f (a)) = g (v) = a
(gof) (b) = g (f (b)) = g (u) = b
(gof) (c) = g (f (c)) = g (w) = c
So, gof = {(a, a), (b, b), (c, c)}
Question - 5 : - Find fog (2) and gof (1) whenf: R → R; f(x) = x2 + 8and g: R → R; g(x) = 3x3 + 1.
Answer - 5 : -
Givenf: R → R; f(x) = x2 + 8 and g: R → R; g(x)= 3x3 + 1.
Consider(fog) (2) = f (g (2))
= f (3 × 23 +1)
= f(3 × 8 + 1)
= f (25)
= 252 +8
= 633
(gof) (1) = g (f (1))
= g (12 +8)
= g (9)
= 3 × 93 +1
= 2188
Question - 6 : - Let R+ be the set of all non-negative real numbers.If f: R+ → R+ and g : R+ → R+ aredefined as f(x)=x2 and g(x)=+ √x, find fog and gof.Are they equal functions.
Answer - 6 : -
Given f: R+ → R+ and g: R+ → R+
So, fog: R+ → R+ and gof: R+ → R+
Domains of fog and gof are the same.
Now we have to findfog and gof also we have to check whether they are equal or not,
Consider(fog) (x) = f (g (x))
= f (√x)
= √x2
= x
Now consider(gof) (x) = g (f (x))
= g (x2)
= √x2
= x
So, (fog) (x) = (gof) (x), ∀x ∈ R+
Hence, fog = gof
Question - 7 : - Let f: R → R and g: R → R bedefined by f(x) = x2 and g(x) = x + 1.Show that fog ≠ gof.
Answer - 7 : -
Givenf: R → R and g: R → R.
So, the domains of f and g are the same.
Consider(fog) (x) = f (g (x))
= f (x + 1)= (x + 1)2
= x2 +1 + 2x
Again consider(gof) (x) = g (f (x))
= g (x2)= x2 + 1
So, fog ≠ gof
Question - 8 : - .
Answer - 8 : -
Question - 9 : -
.
Answer - 9 : -
Question - 10 : - Consider f : N → N, g : N → N and h : N → R defined as f(x) = 2x, g(y) = 3y + 4 and
h(z) = sin z for all x, y, z ϵ N.
show tht ho (gof) = (hog) of.
Answer - 10 : -