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RD Sharma Solutions for Class 12 Chapter 2 – Function Exercise 2.2

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Exercise 2.2 Page No: 2.46

1. Find gof and fog when f: R → R and g : R → R is defined 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) = x^1/3

Solution:

(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) = x^1/3

(gof) (x) = g (f (x))

= g (8x³)

= (8x³)^1/3

= [(2x)³]^1/3

= 2x

Now, (fog) (x) = f (g (x))

= f (x^1/3)

= 8 (x^1/3)³

= 8x

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.

Solution:

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-domain of 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-domain of 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)}

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.

Solution:

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-domain of 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-domain of g is not the same as the domain of f.

So, fog does not exist.

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.

Solution:

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 f and 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 the 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 image in 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 find fog,

we know that the co-domain of g is the 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 find gof,

Co-domain of f is the 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)}

5. Find fog (2) and gof (1) when f: R → R; f(x) = x² + 8 and g: R → R; g(x) = 3x³ + 1.

Solution:

Given f: 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

6. Let R⁺ be the set of all non-negative real numbers. If f: R⁺ → R⁺ and g : R⁺ → R⁺ are defined as f(x)=x² and g(x)=+ √x, find fog and gof. Are they equal functions.

Solution:

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 find fog 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

7. Let f: R → R and g: R → R be defined by f(x) = x² and g(x) = x + 1. Show that fog ≠ gof.

Solution:

Given f: 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