11 + Interview Questions in COMPOSITE FUNCTIONS IN GENERAL AWARENESS Page 1 InterviewSolution

1.	If A = {a,b,c}, B = {u, v, w}. if f: A → B and g : B → A, defined as f = {(a, v), (b, u), (c, w)}, g = {(u, b), (v, a), (w, c)} then find (fog) and (gof).
Answer» Given, f= {(a, v), (b, u), (c, w)} g= {(u, b), (v, a), (w, c)} f(a)= v and g(u) = b f(b)= u and g(v) = a f(c)= w and g(w) = C So, from fog(x) = f(g(x)] 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)} gof(a) = g[f(a)] = g(v) = a gof(b) = g[fb)] = g(u) = b gof(c) = g[(c)] = g(w) = C gof = {(a, a), (b, b), (c, c)}

1.

If A = {a,b,c}, B = {u, v, w}. if f: A → B and g : B → A, defined as f = {(a, v), (b, u), (c, w)}, g = {(u, b), (v, a), (w, c)} then find (fog) and (gof).

Answer»

Given, f= {(a, v), (b, u), (c, w)}

g= {(u, b), (v, a), (w, c)}

f(a)= v and g(u) = b

f(b)= u and g(v) = a

f(c)= w and g(w) = C

So, from fog(x) = f(g(x)]

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)}

gof(a) = g[f(a)] = g(v) = a

gof(b) = g[fb)] = g(u) = b

gof(c) = g[(c)] = g(w) = C

gof = {(a, a), (b, b), (c, c)}

Discussion

2.	If f : R → R, f(x) = 2x + 1 and g : R → R, g(x) = x3, then (gof)-1(27) is equal to :(a) 2(b) 1(c) -1(d) 0
Answer» Answer is (b) Let (gof)^-1(27) = x ∴ (gof)(x) = 27 g{f(x)} = 27 g{2x + 1} = 27 (2x + 1)³ = 27 2x + 1 = 27^1/3 2x + 1 = 3^{3 x 1/3} 2x + 1 = 3 ∴ 2x = 2 x = 1

2.

If f : R → R, f(x) = 2x + 1 and g : R → R, g(x) = x3, then (gof)-1(27) is equal to :(a) 2(b) 1(c) -1(d) 0

Answer»

Answer is (b)

Let (gof)^-1(27) = x

∴ (gof)(x) = 27

g{f(x)} = 27

g{2x + 1} = 27

(2x + 1)³ = 27

2x + 1 = 27^1/3

2x + 1 = 3^{3 x 1/3}

2x + 1 = 3

∴ 2x = 2

x = 1

Discussion

3.	If f: R → R and g: R+ R, where f(x) = 2x + 3 and g(x) = x2 + 1, then (gof)(2):(a) 38(b) 42(c) 46(d) 50
Answer» Answer is (d) Let (gof)(2) = y ∴ y = g{{{2}} = g{2 x 2 + 3} = g(7) = (7)² + 1 = 49 + 1 = 50

3.

If f: R → R and g: R+ R, where f(x) = 2x + 3 and g(x) = x2 + 1, then (gof)(2):(a) 38(b) 42(c) 46(d) 50

Answer»

Answer is (d)

Let (gof)(2) = y

∴ y = g{{{2}}

= g{2 x 2 + 3}

= g(7)

= (7)² + 1

= 49 + 1

= 50

Discussion

4.	If functions f and g be defined as below, then find (gof)(x) : f : R → R, f(x) = 2x + x-2 g : R → R, g(x) = x4 + 2x + 4
Answer» Given, f: R → R f(x) = 2x + x^-2 g: R → M R, g(x)= x⁴ + 2x + 4. ∴ (gof)(x) = g(f(x)} = g{2x + x^-2} = (2x + x^-2)⁴ + 2(2x + x^-2) + 4

4.

If functions f and g be defined as below, then find (gof)(x) : f : R → R, f(x) = 2x + x-2 g : R → R, g(x) = x4 + 2x + 4

Answer»

Given, f: R → R

f(x) = 2x + x^-2

g: R → M R, g(x)= x⁴ + 2x + 4.

∴ (gof)(x) = g(f(x)} = g{2x + x^-2}

= (2x + x^-2)⁴ + 2(2x + x^-2) + 4

Discussion

5.	If f, g, h be three functions from R to R, defined as f(x) = x2, g(x) = cos x and h(x) = 2x + 3, then find {ho(gof)} (√2π).
Answer» Given function, f(x) = x², g(x) = cos x, h(x) = 2x + 3 ∴ {ho(gof)}(x) = hog{f(x)} = h[g{f(x)}] = h[g(x²)] = h(cos x²) = 2 cos x² + 3 {ho(gof)}√2π = 2 cos (√2π)² + 3 = 2 cos 2π + 3 = 2 x 1 + 3 = 5

5.

If f, g, h be three functions from R to R, defined as f(x) = x2, g(x) = cos x and h(x) = 2x + 3, then find {ho(gof)} (√2π).

Answer»

Given function,

f(x) = x², g(x) = cos x, h(x) = 2x + 3

∴ {ho(gof)}(x) = hog{f(x)}

= h[g{f(x)}]

= h[g(x²)] = h(cos x²)

= 2 cos x² + 3

{ho(gof)}√2π = 2 cos (√2π)² + 3

= 2 cos 2π + 3

= 2 x 1 + 3 = 5

Discussion

6.	Which of the following operations is commutative in R:(a) ab = a2b(b) ab = ab(c) ab = a – b + ab(d) a b = a + b + a2b
Answer» (a) ∵ ab=a^b and ba = b²a ∴ ab ≠ ba So, operation is not commutative. (b) ∵ ab = ab and ba = b^a ∴ ab ≠ ba So, operation is not commutative. (c) ∵ ab = a – a + ab and ba = b – a + ba ∴ ab ≠ ba So, operation is not commutative. (d) ∵ ab= a + b + a²b and : ba = b + a + b²a ∴ ab* ≠ b*a So, operation is not commutative. Hence, any of these operation is not commutative.

6.

Which of the following operations is commutative in R:(a) ab = a2b(b) ab = ab(c) ab = a – b + ab(d) a b = a + b + a2b

Answer»

(a) ∵ a*b=a^b and b*a = b²a

∴ a*b ≠ b*a

So, operation is not commutative.

(b) ∵ a*b = ab and b*a = b^a

∴ a*b ≠ b*a

So, operation is not commutative.

(c) ∵ a*b = a – a + ab

and b*a = b – a + ba

∴ a*b* ≠ b*a

So, operation is not commutative.

(d) ∵ a*b= a + b + a²b

and : b*a = b + a + b²a

∴ a*b* ≠ b*a

So, operation is not commutative.

Hence, any of these operation is not commutative.

Discussion

7.	Subtraction is an operation on Z, which is(a) commutative and associative(b) associative but not commutative(c) neither commutative and nor associative(d) commutative but not associative
Answer» Answer is (c) Let a, b ∈ z Commutativity : As we know, a-b ≠ b – a Associativity : (a – b) – c ≠ a – (b – c)

7.

Subtraction is an operation on Z, which is(a) commutative and associative(b) associative but not commutative(c) neither commutative and nor associative(d) commutative but not associative

Answer»

Answer is (c)

Let a, b ∈ z

Commutativity : As we know,

a-b ≠ b – a

Associativity : (a – b) – c ≠ a – (b – c)

Discussion

8.	If f: R+ → R+ and g : R+ → R+, defined as f(x) = x2, g(x) = √x, then find gof and fog wheather are they equivalent ?
Answer» Given, f: R⁺ → R⁺, f(x) = x² g : R⁺ → R⁺, g(x)= √x Then, (fog): R⁺ → R⁺ and (gof): R⁺ → R⁺ are defined ∴ (gof)(x) = g[f(x)] = g(x²) = √x² = x (fog)(x) = f(g(x)] = f(√x)= (√5)² = x Based on above, (gof) and (fog) are of same domain and co-domain. (fog)(x) = (gof)(x) ∀ x ∈ R⁺ So, (fog) and (gof) are equivalent functions.

8.

If f: R+ → R+ and g : R+ → R+, defined as f(x) = x2, g(x) = √x, then find gof and fog wheather are they equivalent ?

Answer»

Given, f: R⁺ → R⁺, f(x) = x²

g : R⁺ → R⁺, g(x)= √x

Then, (fog): R⁺ → R⁺ and (gof): R⁺ → R⁺ are defined

∴ (gof)(x) = g[f(x)]

= g(x²) = √x² = x

(fog)(x) = f(g(x)]

= f(√x)= (√5)² = x

Based on above, (gof) and (fog) are of same domain and co-domain.

(fog)(x) = (gof)(x) ∀ x ∈ R⁺

So, (fog) and (gof) are equivalent functions.

Discussion

9.	If A = {1, 2, 3,4}, B = {a,b,c,d}, then define four bijection from A to B and also find their inverse functions.
Answer» Given A = {1, 2, 3, 4), B = {a, b, c, d} (a) f₁ = {(1, a),(2, b), (3, c), (4, d)} f₁^-1 = {(a, 1), (1, 2), (c, 3), (d, 4)} (b) f₂ = {(1, a), (2, c), (3, b), (4, d)} f₂^-1 = {(a, 1), (C, 2), (6, 3), (d, 4)} (c) f₃ = {(1, b), (2, a), (3, d), (4, b)} f₃^-1 = {(b, 1), (a, 2), (d, 3), (6,4)} (d) f₄ = {(1, c), (2, d), (3, a),(4, b)} f₄^-1 = {(c, 1), (d, 2), (a, 3), (b, 4)}

9.

If A = {1, 2, 3,4}, B = {a,b,c,d}, then define four bijection from A to B and also find their inverse functions.

Answer»

Given

A = {1, 2, 3, 4), B = {a, b, c, d}

(a) f₁ = {(1, a),(2, b), (3, c), (4, d)}

f₁^-1 = {(a, 1), (1, 2), (c, 3), (d, 4)}

(b) f₂ = {(1, a), (2, c), (3, b), (4, d)}

f₂^-1 = {(a, 1), (C, 2), (6, 3), (d, 4)}

(c) f₃ = {(1, b), (2, a), (3, d), (4, b)}

f₃^-1 = {(b, 1), (a, 2), (d, 3), (6,4)}

(d) f₄ = {(1, c), (2, d), (3, a),(4, b)}

f₄^-1 = {(c, 1), (d, 2), (a, 3), (b, 4)}

Discussion

10.	If f : R → R,f(x) = cos (x + 2), is f-1 exists.
Answer» Given function f : R → R, f(x) = cos (x + 2). Putting x = 2π f(2π) = cos (2π+ 2) = cos (2) Putting x = 0 f(0) = cos (0 + 2) = cos 2 Here, only one image is obtained for 0 and 2π. So, ‘f’ is not one-one. Thus, ‘f’ is not one-one onto. Hence, f^-1 : R → R does not exist.

10.

If f : R → R,f(x) = cos (x + 2), is f-1 exists.

Answer»

Given function

f : R → R, f(x) = cos (x + 2).

Putting x = 2π

f(2π) = cos (2π+ 2)

= cos (2)

Putting x = 0

f(0) = cos (0 + 2) = cos 2

Here, only one image is obtained for 0 and 2π.

So, ‘f’ is not one-one.

Thus, ‘f’ is not one-one onto.

Hence, f^-1 : R → R does not exist.

Discussion

11.	If A = {1, 2, 3, 4}, B = {3, 5, 7, 9}, C = {7, 23, 47, 79} and f: A → B, g : B → C such that f(x) = 2x + 1 and g(x) = x2 – 2, then find (gof)-1 and f-1og-1 in ordered form.
Answer» A = {1, 2, 3, 4}, B = {3, 5, 7,9} and C = {7, 23, 47, 79} f : A → B, f (x) = 2x + 1 g : B → C, g(x) = x² – 2 Now, gof (x)=g{f(x)} = g(2x + 1) = (2x + 1)² – 2 = 4x² + 4x – 1 ∴ gof (x) = 4x² + 4x – 1 On putting x = 1, 2, 3, 4 gof = {(1,7), (2, 23), (3,47),(4, 79)} ∵ gof is bijection function. ∴ Its inverse is possible ⇒ (gof)^-1 = {(7, 1), (23, 2), (47,3), (79,4)} ⇒ f^-1og^-1 = {(7,1), (23, 2) (47, 3), (79,4)} ∵ (gof)^-1 = f^-1og^-1 by theorem.

11.

If A = {1, 2, 3, 4}, B = {3, 5, 7, 9}, C = {7, 23, 47, 79} and f: A → B, g : B → C such that f(x) = 2x + 1 and g(x) = x2 – 2, then find (gof)-1 and f-1og-1 in ordered form.

Answer»

A = {1, 2, 3, 4}, B = {3, 5, 7,9} and

C = {7, 23, 47, 79}

f : A → B, f (x) = 2x + 1

g : B → C, g(x) = x² – 2

Now, gof (x)=g{f(x)} = g(2x + 1)

= (2x + 1)² – 2 = 4x² + 4x – 1

∴ gof (x) = 4x² + 4x – 1

On putting x = 1, 2, 3, 4

gof = {(1,7), (2, 23), (3,47),(4, 79)}

∵ gof is bijection function.

∴ Its inverse is possible

⇒ (gof)^-1 = {(7, 1), (23, 2), (47,3), (79,4)}

⇒ f^-1og^-1 = {(7,1), (23, 2) (47, 3), (79,4)}

∵ (gof)^-1 = f^-1og^-1 by theorem.

Discussion

Explore topic-wise InterviewSolutions in .

If A = {a,b,c}, B = {u, v, w}. if f: A → B and g : B → A, defined as f = {(a, v), (b, u), (c, w)}, g = {(u, b), (v, a), (w, c)} then find (fog) and (gof).

If f : R → R, f(x) = 2x + 1 and g : R → R, g(x) = x3, then (gof)-1(27) is equal to :(a) 2(b) 1(c) -1(d) 0

If f: R → R and g: R+ R, where f(x) = 2x + 3 and g(x) = x2 + 1, then (gof)(2):(a) 38(b) 42(c) 46(d) 50

If functions f and g be defined as below, then find (gof)(x) : f : R → R, f(x) = 2x + x-2 g : R → R, g(x) = x4 + 2x + 4

If f, g, h be three functions from R to R, defined as f(x) = x2, g(x) = cos x and h(x) = 2x + 3, then find {ho(gof)} (√2π).

Which of the following operations is commutative in R:(a) a*b = a2b(b) a*b = ab(c) a*b = a – b + ab(d) a * b = a + b + a2b

Subtraction is an operation on Z, which is(a) commutative and associative(b) associative but not commutative(c) neither commutative and nor associative(d) commutative but not associative

If f: R+ → R+ and g : R+ → R+, defined as f(x) = x2, g(x) = √x, then find gof and fog wheather are they equivalent ?

If A = {1, 2, 3,4}, B = {a,b,c,d}, then define four bijection from A to B and also find their inverse functions.

If f : R → R,f(x) = cos (x + 2), is f-1 exists.

If A = {1, 2, 3, 4}, B = {3, 5, 7, 9}, C = {7, 23, 47, 79} and f: A → B, g : B → C such that f(x) = 2x + 1 and g(x) = x2 – 2, then find (gof)-1 and f-1og-1 in ordered form.

Which of the following operations is commutative in R:(a) ab = a2b(b) ab = ab(c) ab = a – b + ab(d) a b = a + b + a2b