# NCERT Solutions for Class 12 Maths Chapter 1 – Exercise 1.3

With all our research we gather information on NCERT Solutions for Class 12 Maths Chapter 1 – Exercise 1.3. This solution will help CBSE Board candidate to solve Class 12th Maths subject of Exercise 1.3. Here we have given question from Relations and Functions Chapter of Class 12th Maths subject along with their respective answers.

NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.3 Overview
Book National Council of Educational Research and Training (NCERT)
Class 12th Class
Subject NCERT Solution Class 12th Maths
Chapter Chapter 1 – Relations and Functions

## Chapter 1: Relations and Functions – Exercise 1.3

 Important Notification Apply for BTech Program at Manipal Institute of Technology - Apply Now SRMJEEE BTech Admission 2019 Opens - Learn More Looking for Engineering Admission 2019 Join Amrita University - Apply Here UPES B.Tech Admissions Open for 2019 BATCH - Apply Here Chandigarh Fast Track Admission 2019 - Apply Here

Q-1: Let us consider, g : {3, 5, 6} → {2, 4, 7} and f : {2, 4, 7} → {3, 5} will begiven by g = {(3, 4), (5, 7), (6, 3)} and f = {(3, 5), (4, 5), (7, 3)}. Find fog.

Solution

Solution

The given functions g : {3, 5, 6} → {2, 4, 7} and f : {2, 4, 7} → {3, 5} will be defined by g = {(3, 4), (5, 7), (6, 3)} and f = {(3, 5), (4, 5), (7, 3)}

fog(3) = f[g(3)] = f(4) = 5 [as, g(3) = 4 and f(4) = 5]

fog(5) = f[g(5)] = f(7) = 3 [as, g(5) = 7 and f(7) = 3]

fog(6) = f[g(6)] = f(3) = 5 [as, g(6) = 3 and f(3) = 5]

Hence,

fog = {(3, 5), (5, 3), (6, 5)}

Q-2: Let us consider

f, g and h be the functions from R to R. Prove that:

(g + f)oh = goh + foh

(g.f)oh = (goh).(foh)

Solution

Solution

We need to prove that,

(g + f)oh = foh + goh

{(g + f)oh}(x) = {goh + foh}(x)

Now,

LHS = {(g + f)oh}(x)

= (g + f)[h(x)]

= g[h(x)] + f[h(x)]

= (goh)(x) + (foh)(x)

= {(goh) + (foh)}{x)

= RHS

Therefore, {(g + f)oh} = goh + foh

Hence, proved.

We need to prove:

(g.f)oh = (goh).(foh)

LHS = [(g.f)oh](x)

= (g.f)[h(x)]

= g[h(x)]. f[h(x)]

= (goh)(x) . (foh)(x)

= {(goh).(foh)}(x)

= (goh).(foh)

Hence, LHS = RHS

Therefore, (g.f)oh = (goh).(foh)

Q-3: Find fog and gof, if

(i) g(a) = |a| and f(a) = |5a – 2|

(ii) g(a) = 8a3 and f(a) = x1/3

Solution

Solution

(i) g(a) = |a| and f(a) = |5a – 2|

Therefore, the binary operation * is not associative.

fog(a) = f(g(a)) = f(|a|) = |5|x|2|

Therefore, gof(a) = g(f(a)) = g(|5x – 2|) = ||5x2|| = |5x2|

(ii) g(a) = 8aand f(a) = a1/3

Therefore, fog(a) = f(g(a)) = f(8a3) = (8a3)1/3 = (23 a3) 1/3 = 2a

Therefore, gof(a) = g(f(a)) = g(a1/3) = 8(a1/3)3 = 8a

Q-4: If g(a) = (4a+3) / (6a4), a ≠ 2/3. Prove that gog(a) = a, for every a ≠ 2/3. What will be the inverse of the function g?

Solution

Solution

As per the data given in the question, we have

g(a) = 4a+3/6a4, a ≠ 2/3

(gog)(x) = g(g(x)) = g(4a+3/6a4) = 4(4a+3/6a4)+3      /       6(4a+3/6a4)4

16a+12+18a12      /       24a+1824a+16

34a / 34

Therefore, gog(a) = a, for all of the a ≠ 2/3

So, gog = Ia

Therefore, the function f given is invertible and the inverse of the function f is f itself.

Q-5: Explain with suitable reason that, whether the following functions has any inverse

(i) g : {2, 3, 4, 5} → {11} with

g = {(2, 11), (3, 11), (4, 11), (5, 11)}

(ii) f : {6, 7, 8, 9} → {2, 3, 4, 5} with

f = {(6, 5), (7, 4), (8, 5), (9, 3)}

(iii) h : {3, 4, 5, 6} → {8, 10, 12, 14} with

h = {(3, 8), (4, 10), (5, 12), (6, 14)}

Solution

Solution

(i) g : {2, 3, 4, 5} → {11} which is defined as g = {(2, 11), (3, 11), (4, 11), (5, 11)}

From the definition of f, given in the question, we note that, the function f is a many one function as,

g(2) = g(3) = g(4) = g(5) = 11

Hence, the function g is not one- one.

Therefore, in this case, the function g doesn’t have any inverse.

(ii) f : {6, 7, 8, 9} → {2, 3, 4, 5} which is defined as f = {(6, 5), (7, 4), (8, 5), (9, 3)}

From the definition of f, given in the question, we note that, the function f is a many one function as,

f(6) = f(7) = 4

Hence, the function f is not one- one.

Therefore, in this case, the function f doesn’t have any inverse.

(iii) h : {3, 4, 5, 6} → {8, 10, 12, 14} which is defined as

h = {(3, 8), (4, 10), (5, 12), (6, 14)}

Here, we can see that, all the distinct elements from the set {3, 4, 5, 6} have distinct images under h.

Hence, the function h is one- one.

Also, the function h is onto as each element y from the set {8, 10, 12, 14}, there exists an element, say a, in the set {3, 4, 5, 6} so that h(a) = b.

Hence, h is one- one as well as onto function.

Therefore, the function h will have an inverse.

Q-6: Prove that:

g : [-2, 2] → R,

which is given by g(a) = a/(a+2) will be one- one. What will be the inverse of the function g : [-2, 2] → Range g.

Solution

Solution

g : [-2, 2] → R which is given by

g(a) = a/(a+2)

For one- one:

Let, g(a) = g(b)

a/(a+2) = b/(b+2)

a(b + 2) = b (a + 2)

ab + 2a = ab + 2b

2a = 2b

a = b

Hence, the function g is one- one function.

We can see that, g : [-2, 2] → Range g is onto.

Hence, g : [-2, 2] → Range g here is one- one and onto, and thus, the inverse of the function f : [-2, 2] → Range g exists.

Let us consider f : Range g → [-2, 2] be the inverse for g.

Let, b be any arbitrary element of the range f.

As, f : [-2, 2] → Range g is onto, here we have

b = g(a) for some values of a ɛ [-2, 2]

b = a/(a+2)

b(a + 2) = a

ab + 2b = a

a(1 – b) = 2b

a = 2b/(1b), b ≠ 1

Then,

Let we define f: Range g → [-2, 2] as

f(b) = 2b(1b), b ≠ 1

Thus,

(fog)(a) = f(g(a)) = f(a/a+2)2(a/a+2)    /    1(a/a+22a / a+22a / a

and,

(gof)(a) = g(f(a)) = g (2b/1b)=(2b/1b)    /     (2b/1b)+2  =  2b / 2b+222b/b

Hence,

fog = a = I1,1 and gof = b = IRange f

Thus,

g-1 = f

g-1(b) = 2b/2, b ≠ 1.

Q-7: Let us consider g : R → R which is given by g(a) = 4a + 3. Prove that the function g is invertible. Also find the inverse of g.

Solution

Solution

g : R → R which is given by,

g(a) = 4a + 3

For one- one function,

Let, g(a) = g(b)

4a + 3 = 4b + 3

4a = 4b

a = b

Hence, the function g is a one- one function.

For onto function,

For b ɛ R, let b = 4a + 3.

a = b3 / 4   ɛ R.

Hence, for every b ɛ R, there does exist a = b34 ɛ R, so that

g(a) = g (b3/4) = 4 (b3/4) + 3 = b

Hence, the function g is onto.

Therefore, the function g is one- one and onto and hence, g-1 will exist.

Let us consider, f: R → R by f(x) = b3/4

Then,

(fog)(a) = f(g(a)) = f(4a + 3) = (4a+3)3/4=4x/4=x

And,

(gof)(b) = g(f(x)) = gg(b3/4)=4(b3/4)+3=b3+3=b

So, fog = gof = IR

Therefore, the function f will be invertible and have inverse which is given by g-1(b) = f(b) = b34

Q-8: Let us consider a function, g : R+ → [4, ) which is given by g(a) = a2 + 4. Prove that the function g is invertible with the inverse g-1 of the given function g by g-1 (b) = √b-4, where R+will be the set of all the non- negative real number.

Solution

Solution

g : R→ [4, ∞) which is given by g(a) = a2 + 4

For one- one function,

Let, g(a) = g(b)

a2 + 4 = b2 + 4

a2 = b2

a = b

Hence, the function f is one- one function.

For onto function,

For b ɛ [4, ∞), let b = a2 + 4

a2 = b – 4 ≥ 0

a = b4‾‾‾‾√ ≥ 0

Hence, for every b ɛ [4, ∞), there exists a = b4‾‾‾‾√ ɛ R+, so that

G(a) = g(b4‾‾‾‾√) = (b4‾‾‾‾√)+ 4 = b – 4 + 4 = b

Hence, the function g is onto.

So, the function g is one- one and onto and hence, g-1 exists.

Let we define, f : [4, ∞) → Rby f(b) = b4

Then,

(fog)(a) = f(g(a)) = f(x2+4) = (a2+4)a2 = a

And,

(gof)(b) = g(f(b)) = g(b4) = (b4)2 + 4 = b – 4 + 4 = b

Hence, gof = fog = IR

Therefore, the function g is invertible and it will be the inverse of g which is given by g-1(b) = f(b) = b4

Q-9: Let us consider a function, g: R→ [-5, ∞) which is given by g(a) = 9a2 + 6a – 5. Prove that, the function g is invertible with g-1(b) = (b+6)1/3).

Solution

Solution

g: R→ [-5, ∞) which is given by g(a) = 9a2+6a5

Let, b be an arbitrary element of [-5, ∞)

Let,

b = 9a2+6a5

b = (3a + 1)2 – 1 – 5 = (3a + 1)2 – 6

b + 6 = (3a + 1)2

3a + 1 = b+6 [as b ≥ -5  b + 6 > 0]

a = (b+6)13

Hence, the function g is onto, which means having range g = [-5, ∞)

Let us consider, f : [-5, ∞) → R+ as f(b) = (b+6)1 / 3

Then,

(fog)(a) = f(g(a)) = f(9a2 + 6a – 5) = g ((3a + 1)2 – 6) = (3a+1)26+6  –  1   / 3

=3a+11 / 3 = 3a / 3 = a

And,

(gof)(b) = g(f(b)) = g(b+6  1 / 3)=[3(b+6  –  1 / 3)+1]26

(b+6)26 = b + 6 – 6 = b

Hence, fog = a = Iand gof = b = IRange g

Therefore, g is invertible and the inverse of g will be given by:

g-1(b) = f(b) = ((y+6)21   /   3)

Q-10: Let us consider a function g : P → Q be an invertible function. Prove that the function g have a unique inverse

Solution

Solution

Let, the function, g : P → Q be an invertible function.

Let, function g have two inverses (say, g1 and g2)

Thus, for every b ɛ Q, we have,

gof(b) = IY(b) = gof2(b)

g(f1(b)) = g(f2(b))

f1(b) = f2(b) [as g is invertible  g is on- one]

f1= f2 [as f is one- one]

Therefore, the function g has a unique inverse.

Q-11. Let us consider the function g: {2, 3, 4} → {x, y, z}, generally given by g(2) = x, g(3) = y, g(4) = z. Find g-1 and prove that (g-1)-1 = g.

Solution

Solution

The given function g : {2, 3, 4} → {x, y, z} which is given by g(2) = , g(3) = y, and g(4) = z

If we will define f: {x, y, z} → {2, 3, 4}, since f(x) = 2, f(y) = 3, f(z) = 4.

(gof)(x) = g(f(x)) = g(2) = x

(gof)(y) = g(f(y)) = g(3) = y

(gof)(z) = g(f(z)) = g(4) = z

And,

(fog)(2) = f(g(2)) = f(x) = 2

(fog)(3) = f(g(3)) = f(y) = 3

(fog)(4) = f(g(4)) = f(x) = 4

Thus,

fog = Iand gof = IY, such that P = {2, 3, 4} and Q = {x, y, z}

Hence, the inverse of the function g exists and g-1 = f

Thus,

g-1 :{x, y, z} → {2, 3, 4} which is given by g-1(x) = 2, g-1(y) = 3 and g-1(z) = 4

Let us now obtain the inverse for g-1, i.e., finding the inverse of the function f.

Let us define h : {2, 3, 4} → {x, y, z} such that h(2) = x, h(3) = y and h(4) = z

Now,

(foh)(2) = f(h(2)) = f(x) = 2

(foh)(3) = f(h(3)) = f(x) = 3

(foh)(4) = f(h(4)) = f(x) = 4

And,

(hof)(x) = h(f(x)) = h(2) = x

(hof)(y) = h(f(y)) = h(3) = y

(hof)(z) = h(f(z)) = h(4) = z

Hence, foh = Iand hof = IY, P = {2, 3, 4} and Q = {x, y, z}.

So, the inverse of the function of f exists and f-1 = h  (g-1)-1 = h

We can see that, h = g

Hence, (f-1)-1 = f.

Q-12: Let us consider g : P → Q be an invertible function. Prove that the inverse of g-1 is g, i.e., (g-1)-1 = g.

Solution

Solution

As per the given condition,

g : P → Q is an invertible function.

Thus, there exists a function say, f : Q → P so that, fog = IP and gof = IQ

So, g-1 = f

Then, fog = IP and gof = IQ

g-1og= IP and fof-1 = IQ

Therefore, g-1 : Q → P will be invertible and the function g is the inverse of g-1, i.e., (g-1)-1 = g

Q-13: If g : R → R is given by g(a) = (3–a3)1/3, then gog(a) is

(a) 1 / a3

(b) a3

(c) a

(d) (3 – a3)

Solution

Solution

g : R → R which is given by g(a) = (3a3)1/3

Then,

gog(a) = g(g(a)) = g((3a3)1/3) = [3–   ((3a3)1/3)3]1/3

[3(3a3)]1/3 = (a3)1/3=a

Hence, gog(a) = a

Therefore, the correct answer is C.

Q-14: Let us consider a function g: R – {4/3} → R as g(a) = 4a/3a+4. The inverse of the function g is map f: Range g → R – {4/3} is given by

(a) f(b) = 3b/34b (b) f(b) = 4b/43b

(c) f(b) = 4b/34b (d) f(b) = 3b/43b

Solution

Spoiler title

As per the data given in the question, we have

g : R – {4/3} → R which is a function defined as g(a) = 4a/3a+4.

Let, b be the arbitrary element having range g.

Thus, there exists a ɛ R – {4/3} so that, b = f(a).

b = 4a/3a+4

b(3a + 4) = 4a

3ab + 4b = 4a

a = 4b/43b

Let, f : Range g → R – {4/3} so that, f(b) = 4b/43b

Thus,

fog(a) = f(g(a) = f(4a / 3a+4)=(4a/3a+4)   /   43(4a/3a+4)

16a/12a+1612a  =  16a / 16 = a

And,

gof(b) = g(f(b)) = g(4b/43b)=4(4b/43b) / 3(4b/43b)+4

16b / 12b+161216b / 16=b

Thus, fog = IR – {4/3} and gof = IRange f

Therefore, the inverse of the function g is the map f: Range g → R – {4/3}, which will be given by

g(b) = 4b/43b

Hence, the correct answer is b.