ML Aggarwal Class 12 Maths Solutions Section A Chapter 1 Relations and Functions Ex 1.4

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ML Aggarwal Class 12 Maths Solutions Section A Chapter 1 Relations and Functions Ex 1.4

Question 1.
If f = {2, 3, 4, 5} → {3, 4, 5, 9} and g : {3, 4, 5, 9} → {7, 11, 15} are functions defined as f (2) = 3, f (3) = 4, f (4) = f (5) = 5 and g (3) = g (4) = 7, g (5) = g (9) = 11, find gof.
Solution:
D_f = {2, 3, 4, 5} ;
R_f = {3, 4, 5, 9} ;
D_g = {3,4, 5, 9} ;
R_g = {7, 11, 15}
Since R_f ⊂ D_g
∴ g of exists or well defined and be a function from {2, 3, 4, 5} to {7, 11, 15}
(gof) (2) = g (f(2)) = g (3) = 7
(gof) (3) = g (f(3)) = g (4) = 7
(gof) (4) = g (f(4)) = g (5)= 11
(gof) (5 ) = g (f(5)) = g (5) = 11
Thus, gof = {(2, 7), (3, 7), (4, 11), (5, 11)}

Question 2.
Let A = {3, 4, 5, 6}, B = {13, 14, 15, 16} and C = {23, 24, 25}. If f : A → B and g : B → C are defined by f = {(3, 13), (4, 14), (5, 15), (6, 16)} and g = {(13, 23), (14, 23), (15, 24), (16, 25)}, then find gof : A → C.
Solution:
Clearly D_f = A = {3, 4, 5, 6} ;
R_f = B = {13, 14, 15, 16}
D_g = {13, 14, 15, 16} ;
R_g ={23, 24, 25}
Clearly gof is well defined and be a function from {3, 4, 5, 6} to {23, 24, 25}.
Since R_f = D_g
Given f = {(3, 13), (4, 14,), (5, 15), (6, 16)}
∴ f (3) = 13 ; f (4) = 14 ;
f (5) = 15 and f (6) = 16
and g = {(13, 23), (14, 23), (15, 24), (16, 25)}
∴ g (13) = 23 ; g (14) = 23 ;
g (15) = 24; and g (16) = 25
∴ gof (3) = g (f (3)) = g (13) = 23 ;
gof (5) = g (f(5)) = g(15) = 24
gof (4) = g (f(4)) = g(14) = 23 ;
gof (6) = g (f(6)) = g (16) = 25
Thus, gof = {(3, 23), (4, 23), (5, 24), (6, 25)}.

Question 3.
Let A = {1, 2, 3, 4}. If the functions f : A → A and g : A → A are defined by f = {(1, 4), (2, 1), (3, 3), (4, 2)} and g = {1, 3), (2, 1), (3, 2), (4, 4)}, then find
(i) gof
(ii) fog
(iii) fof
Solution:
Clearly D_f = A = R_f;
D_g = A = R_f = (1, 2, 3, 4}
Given f = {(l, 4), (2, 1), (3, 3), (4, 2)}
∴ f (1) = 4 ;
f (2) = 1 ;
f (3) = 3,
f (4) = 2 and
g = {(1,3), (2, 1), (3,2), (4, 4)}
∴ g (1) = 3;
g (2) = 1 ;
g (3) = 2;
g (4) = 4

(i) Since R_f ⊂ D_g
∴ gof exists and be a function from A to A.
Thus, (gof) (1) = g (f( 1)) = g (4) = 4
(gof) (2) = g (f (2)) = g (1) = 3
(gof) (3) = g (f (3)) = g (3) = 2
(gof) (4) = g (f (4)) = g (2) = 1
Thus, gof = {(1, 4), (2, 3), (3, 2), (4, 1)}

(ii) fog be well defined since R_g ⊂ D_f
∴ (fog) (1) = f(g (1)) = f (3) = 3
(fog) (2) = f(g (2)) = f (1) = 4
(fog) (3) = f(g (3)) = f (2) = 1
(fog) (4) = f(g (4)) = f ( 4) = 2
∴ fog = {(1, 3), (2, 4), (3, 1), (4,2)}

(iii) fof be well defined and R_f ⊂ D_f
∴ (fof) (1) = f (f(1)) = f(4) = 2
(fof) (2) = f (f(2)) = f(1) = 4
(fof) (3) = f (f(3)) = f(3) = 3
(fof) (4) = f (f(4)) = f(2) = 1
Thus, fof = {(1, 2), (2, 4), (3, 3), (4, 1)}

Question 4.
If the functions f and g are given by f = ((1, 2), (3, 5), (4, 1)) and g = {(2, 3), (5, 1), (1, 3)},
(i) find the domain of f and g
(ii) find the range of f and g
(iii) write fog and gof as sets of ordered pairs.
Solution:
Given f = {(1, 2), (3, 5), (4, 1)} (i) and (ii)
∴ f (1) = 2 ;
f (3) = 5 ;
f (4) = 1
and g = {(2, 3), (5, 1), (1, 3))
∴ g (2) = 3;
g (5) = 1;
g (1) = 3
∴ D_g = {2, 5, 1} ; R_g = {3, 1}

(iii) Since R_g ⊂ D_f
∴ fog is well defined and be a function from {2, 5, 1) to {1, 2, 5).
(fog) (1) = f (g (1)) = f (3) = 5;
(fog) (2) = f (g (2)) = f (3) = 5 ;
(fog) (5) = f(g (5)) = f(1) = 2
∴ fog = {(1, 5), (2, 5), (5, 2))
Since, R_f ⊂ D_g
∴ gof is well defined and be a function from {1, 3, 4) to (1, 3)
∴ (gof) (1) = g (f (1)) = g (2) = 3
(gof) (4) = g (f (4)) = g (1) = 3
(gof) (3) = g (f(3)) = g (5) = 1
∴ gof = {(1, 3), (3, 1), (4, 3)}.

Question 5.
Let R be the set of all real numbers. Find gof and fog when f : R → R and g : R → R are defined by
(i) f (x) = 2x – 1 and g(x) = x² + 3 for all
(ii) f (x) = x + 1 and g (x) = |x| for all x ∈ R
(iii) f (x) = x³ – 1 and g (x) = x² + 2 for all x ∈ R
(iv) f (x) = sin x and g (x) = 5x² for at x ∈ R.
Solution:
(i) Given f : R → R defined by
f(x) = 2x – 1 ∀ x ∈ R
g : R → R defined by g (x) = x² + 3 ∀ x ∈ R
Here fog and gof both exists, and fog : R → R given by
(fog) (x) = f (g (x)) = f (x² + 3)
= 2 (x² + 3) – 1
= 2x² + 5
Also, gof : R → R given by
gof (x) = g (f(x)) = g (2x – 1)
= (2x – 1)² + 3
= 4x² – 4x + 4

(ii) Given f : R → R defined by f (x) = x + 1 ∀ x ∈ R and
g : R → R defined by g (x) = | x | ∀ x ∈ R
Clearly gof and fog exists.
gof : R → R is given by
(gof) (x) = g (f(x)) = g (x + 1) = | x + 1 | and
fog : R → R is given by
(fog) (x) = f(g (x)) = f ( | x | ) = | x | + 1

(iii) f(x) = x³ and g(x) = x² + 1 for all x ∈ R. (ISC 2019)
Solution:
Given f : R → R defined by
f(x) = x³ ∀ x ∈ R
and g : R → R defined by
g(x) = 2x² + 1 ∀ x ∈ R
∴ gof : R → R and fog : R → R
Thus (gof) (x) = g (f(x)) = g(x³)
= 2(x³)² + 1
= 2x⁶ + 1
(fog) (x) = f (g(x))
= f(2x² + 1)
= (2x² + 1)³
= 8x⁶ + 1 + 12x⁴ + 6x²

(iii) (old) Given f : R → R defined by f(x) = x³ – 1 ∀ x ∈ R and
g : R → R defined by g (x) = x² + 2 ∀ x ∈ R
gof : R → R is given by
(gof) (x) = g(f (x)) = g (x³ – 1)
= (x³ – 1)² + 2
= x⁶ – 2x³ + 3
and fog : R → R is given by (fog) (x) = f(g (x)) = f (x² + 2)
= (x² + 2)³ – 1
= x⁶ + 8 + 6x⁴ + 12x² – 1
= x⁶ + 6x⁴ + 12x² + 7

(iv) Given, f : R → R defined by f (x) = sin x ∀ x ∈ R and
g : R → R defined by g (x) = 5x² ∀ x ∈ R
∴ fog and gof exists.
gof : R → R is given by
(gof) (x) = g (f (x)) = g (sin x)
= 5 (sin x)² = 5 sin² x
and (fog) (x) = f (g (x))
= f (5x²)
= sin (5x²)

Question 6.
If f : R → R is defined by f (x) = x² – 3x + 2 for all x ∈ R, then find fof. (NCERT)
Solution:
Given f : R → R defined by f(x) = x² – 3x + 2 ∀ x ∈ R
Clearly fof is well defined as R_f ⊂ D_f
Thus fof given by
(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

Question 7.
Let R be the set of all real numbers. If the functions f : R → R and g : R → R are given by f (x) = x² + 3x + 1 and g (x) = 2x – 3 for all x ∈ R, then find
(i) fog
(ii) gof
(iii) fof
(iv) gof (NCERT Exemplar)
Solution:
Given f : R → R defined by
f(x) = x² + 3x + 1 ∀ x ∈ R and
g : R → R defined by g(x) = 2x – 3 ∀ x ∈ R
Since D_f = R_f = D_g = R_g = R
Thus, fog, gof, fof and gog exists and well defined.

(i) Then fog : R → R is given by (fog) (x) = f (g (x)) = f (2x – 3)
= (2x – 3)² + 3 (2x – 3) + 1
= 4x² – 6x + 1

(ii) gof : R → R is given by
(gof) (x) = g (f (x)) = g (x² + 3x + 1)
= 2 (x² + 3x + 1) – 3
= 2x² + 6x – 1

(iii) fof : R → R is given by
(fof) (x) = f(f (x))
= f (x² + 3x + 1)
= (x² + 3x + 1)² + 3 (x² + 3x + 1) + 1
= x⁴ + 9x² + 1 + 6x³ + 6x + 2x² + 3x² + 9x + 3 + 1
= x⁴ + 6x³ + 14x² + 15x + 5

(v) gog : R → R is given by
(gog) (x) = g (g (x)) = g (2x- 3)
= 2 (2x – 3) – 3
= 4x – 9

Question 8.
If f : R → R and g : R → R are defined by f (x) = x² and g (x) = x + 1 for all x ∈ R, show that gof ≠ fog.
Solution:
Given f : R → R defined by f (x) = x²
and g : R → R defined by g (x) = x + 1
Clearly R_f ⊂ D_g and R_g ⊂ D_f
∴ fog and gof exists ∀ x ∈ R, we have
(gof) (x) = g(f(x))
= g (x²) = x² + 1 and
(fog) (x) = f(g(x))
= g (x + 1) = (x + 1)²
= x² + 2x + 1
∴ gof ≠ fog

Question 9.
If f : R → R and g : R → R are defined by f (x) = x + 1 and g (x) = x – 1 for all x ∈ R, show that gof = fog = IR.
Solution:
Given f : R → R defined as f (x) = x + 1 and
g : R → R defined as g (x) = x – 1
Clearly R_f ⊂ D_g = R and R_g ⊂ D_f = R
∴ gof and fog exists.
∀ x ∈ R we have
(gof) (x) = g (f(x)) = g (x + 1)
= x + 1 – 1 = x
and (fog) (x) = f (g(x))
= f (x- 1)
= x – 1 + 1 = x
∴ gof : R → R be a function such that (gof) (x) = x ∀ x ∈ R
∴ gof = I_R
Also, fog : R → R be a function s.t.
(fog) (x) = x ∀ x ∈ R
∴ fog = I_R
Hence fog = gof= I_R.

Question 10.
If f : R → R and g : R → R are defined by f (x) = 3x² – 2 and g (x) = sin 2x respectively, find
(i) gof
(ii) fog
(iii) fof
Solution:
Given f : R → R defined by f (x) = 3x² – 2∀ x ∈ R and
g : R → R defined by g (x) = sin 2x ∀ x ∈ R
Thus, gof, fog and fof exists

(i) gof : R → R is given by
(gof) (x) = g (f (x)) = g (3x² – 2)
= sin 2 (3x² – 2)
= sin (6x² – 4)

(ii) fog : R → R is given by
(fog) (x) = f (g (x)) = f (sin 2x)
= 3 sin² 2x – 2

(iii) (fof) (x) = f (f(x))
= f (3x² – 2)
= 3 (3x² – 2)² – 2
= 3 (9x⁴ – 12x² + 4) – 2
= 27x⁴ – 36x² + 10

Question 11.
If f : R → R is defined by f (x) = (3 – x³)^1/3, then find (fof) (x).
Solution:
Given f : R → R defined by f (x) = (3 – x³)^1/3 ∀ x ∈ R
Since R_f = D_f = R
∴ fof exists.
∴ fof : R → R is given by
(fof) (x) = f (f(x)) = f { (3 – x³)^1/3 }
= [3 – {(3 – x³)^1/3}³]^1/3
= [3 – (3 – x³)]^1/3
= [x³]^1/3 = x
Thus (fof) (x) = x ∀ x ∈ R

Question 12.
If f : R → R is given by f (x) = (5 – x⁵)^1/5, then find fof.
Solution:
Given f : R → R defined by f(x) = (5 – x⁵)^1/5
Since R_f ⊂ D_f = R
∴ fof exists.
Thus fof : R → R is given by
(fof) (x) = f (f(x)) = f{(5 – x⁵)^1/5}
= [5 – {(5 – x⁵)^1/5}⁵]^1/5
= [5 – (5 – x⁵)]^1/5
= (x⁵)^1/5
= x ∀ x ∈ R
∴ fof = I_R.

Question 13.
If f (x) = | x | and g (x) = [ x ] for all x ∈ R, find
(i) gof (\(\frac{- 5}{3}\))
(ii) (fog) (\(\frac{- 5}{3}\))
Solution:
Given f(x) = | x | and g (x) = [x] ∀ x ∈ R

(i) (gof) (\(\frac{- 5}{3}\)) = g (f ((\(\frac{- 5}{3}\)))
= g(\(|\frac{- 5}{3}|\))
= g(\(\frac{- 5}{3}\))
= [latex]\frac{- 5}{3}[/latex] = 1

(ii) (fog) (\(\frac{- 5}{3}\))
= f (g (\(\frac{- 5}{3}\)))
= f (- 2)
= | – 2 |
= 2

Question 14.
If f (x) = x + 7 and g (x) = x – 7, find (fog) (7).
Solution:
Given f(x) = x + 7;
g(x) = x – 7
(fog) (7) = f (g (7))
= f (7 – 7)
= f (0)
= 0 + 7 = 7