# Ncert Solutions For Class 12 Maths Ex 1.3

## Ncert Solutions For Class 12 Maths Chapter 1 Ex 1.3

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.

Sol:

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)

Sol:

We need to prove that,

(g + f)oh = foh + goh

$$\boldsymbol{\Rightarrow }$$ {(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) = $$x^{\frac{1}{3}}$$

Sol:

(i) g(a) = |a| and f(a) = |5a – 2|

Therefore, the binary operation * is not associative.

fog(a) = f(g(a)) = f(|a|) = $$\left | 5\left | x \right | – 2 \right |$$

Therefore, gof(a) = g(f(a)) = g(|5x – 2|) = $$\left |\left | 5x – 2 \right | \right |$$ = $$\left | 5x – 2 \right |$$

(ii) g(a) = 8a3 and f(a) = a$$\frac{1}{3}$$

Therefore, fog(a) = f(g(a)) = f(8a3) = (8a3)$$\frac{1}{3}$$ = (23 a3) $$\frac{1}{3}$$ = 2a

Therefore, gof(a) = g(f(a)) = g($$a^{\frac{1}{3}}$$) = 8($$a^{\frac{1}{3}}$$)3 = 8a

Q-4: If g(a) = $$\frac{\left(4a + 3 \right)}{\left(6a – 4 \right)}$$, a ≠ $$\frac{2}{3}$$. Prove that gog(a) = a, for every a ≠ $$\frac{2}{3}$$. What will be the inverse of the function g?

Sol:

As per the data given in the question, we have

g(a) = $$\frac{4a + 3}{6a – 4}$$, a ≠ $$\frac{2}{3}$$

(gog)(x) = g(g(x)) = g($$\frac{4a + 3}{6a – 4}$$) = $$\frac{4\left (\frac{4a + 3}{6a – 4} \right) + 3}{6 \left (\frac{4a + 3}{6a – 4} \right) – 4}$$

= $$\frac{16a + 12 + 18a – 12}{24a + 18 – 24a + 16}$$

= $$\frac{34a}{34}$$

Therefore, gog(a) = a, for all of the a ≠ $$\frac{2}{3}$$

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)}

Sol:

(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) = $$\frac{a}{\left(a + 2 \right)}$$ will be one- one. What will be the inverse of the function g : [-2, 2] → Range g.

Sol:

g : [-2, 2] → R which is given by

g(a) = $$\frac{a}{\left(a + 2 \right)}$$

For one- one:

Let, g(a) = g(b)

$$\boldsymbol{\Rightarrow }$$ $$\frac{a}{\left(a + 2 \right)}$$ = $$\frac{b}{\left(b + 2 \right)}$$

$$\boldsymbol{\Rightarrow }$$ a(b + 2) = b (a + 2)

$$\boldsymbol{\Rightarrow }$$ ab + 2a = ab + 2b

$$\boldsymbol{\Rightarrow }$$ 2a = 2b

$$\boldsymbol{\Rightarrow }$$ 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]

$$\boldsymbol{\Rightarrow }$$ b = $$\frac{a}{\left(a + 2 \right)}$$

$$\boldsymbol{\Rightarrow }$$ b(a + 2) = a

$$\boldsymbol{\Rightarrow }$$ ab + 2b = a

$$\boldsymbol{\Rightarrow }$$ a(1 – b) = 2b

$$\boldsymbol{\Rightarrow }$$ a = $$\frac{2b}{\left(1 – b \right)}$$, b ≠ 1

Then,

Let we define f: Range g → [-2, 2] as

f(b) = $$\frac{2b}{\left(1 – b \right)}$$, b ≠ 1

Thus,

(fog)(a) = f(g(a)) = f$$\left(\frac{a}{a + 2} \right)$$= $$\frac{2\left (\frac{a}{a + 2} \right)}{1 – \left (\frac{a}{a + 2} \right)} = \frac{2a}{a + 2 – a} = \frac{2a}{2} = a$$

and,

(gof)(a) = g(f(a)) = g $$\left (\frac{2b}{1 – b} \right) = \frac{\left (\frac{2b}{1 – b} \right)}{\left (\frac{2b}{1 – b} \right) + 2} = \frac{2b}{2b + 2 – 2b} = \frac{2b}{2} = b$$

Hence,

fog = a = $$I_{\left \lfloor -1, 1 \; \right \rfloor}$$ and gof = b = IRange f

Thus,

g-1 = f

g-1(b) = $$\frac{2b}{2}$$, 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.

Sol:

g : R → R which is given by,

g(a) = 4a + 3

For one- one function,

Let, g(a) = g(b)

$$\boldsymbol{\Rightarrow }$$ 4a + 3 = 4b + 3

$$\boldsymbol{\Rightarrow }$$ 4a = 4b

$$\boldsymbol{\Rightarrow }$$ a = b

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

For onto function,

For b ɛ R, let b = 4a + 3.

$$\boldsymbol{\Rightarrow }$$ a = $$\frac{b – 3}{4}$$ ɛ R.

Hence, for every b ɛ R, there does exist a = $$\frac{b – 3}{4}$$ ɛ R, so that

g(a) = g $$\left (\frac{b – 3}{4} \right)$$ = 4 $$\left (\frac{b – 3}{4} \right)$$ + 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) = $$\frac{b – 3}{4}$$

Then,

(fog)(a) = f(g(a)) = f(4a + 3) = $$\frac{\left (4a + 3 \right) – 3}{4} = \frac{4x}{4} = x$$

And,

(gof)(b) = g(f(x)) = g$$g\left (\frac{b – 3}{4} \right) = 4\left (\frac{b – 3}{4} \right) + 3 = b – 3 + 3 = b$$

So, fog = gof = IR

Therefore, the function f will be invertible and have inverse which is given by g-1(b) = f(b) = $$\frac{b – 3}{4}$$

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) = $$\sqrt{b – 4}$$, where R+ will be the set of all the non- negative real number.

Sol:

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

For one- one function,

Let, g(a) = g(b)

$$\boldsymbol{\Rightarrow }$$ a2 + 4 = b2 + 4

$$\boldsymbol{\Rightarrow }$$ a2 = b2

$$\boldsymbol{\Rightarrow }$$ a = b

Hence, the function f is one- one function.

For onto function,

For b ɛ [4, ∞), let b = a2 + 4

$$\boldsymbol{\Rightarrow }$$ a2 = b – 4 ≥ 0

$$\boldsymbol{\Rightarrow }$$ a = $$\sqrt{b – 4}$$ ≥ 0

Hence, for every b ɛ [4, ∞), there exists a = $$\sqrt{b – 4}$$ ɛ R+, so that

G(a) = g($$\sqrt{b – 4}$$) = ($$\sqrt{b – 4}$$)2 + 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, ∞) → R+ by f(b) = $$\sqrt{b – 4}$$

Then,

(fog)(a) = f(g(a)) = f($$x^{2} + 4$$) = $$\sqrt{\left (a^{2} + 4 \right) – 4} = \sqrt{a^{2}}$$ = a

And,

(gof)(b) = g(f(b)) = g$$\left (\sqrt{b – 4} \right)$$ = $$\left (\sqrt{b – 4} \right)^{2}$$ + 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) = $$\sqrt{b – 4}$$

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) = $$\left (\frac{\left (\sqrt{b + 6} \right) – 1}{3} \right)$$.

Sol:

g: R+ → [-5, ∞) which is given by g(a) = $$9a^{2} + 6a – 5$$

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

Let,

b = $$9a^{2} + 6a – 5$$

$$\boldsymbol{\Rightarrow }$$ b = (3a + 1)2 – 1 – 5 = (3a + 1)2 – 6

$$\boldsymbol{\Rightarrow }$$ b + 6 = (3a + 1)2

$$\boldsymbol{\Rightarrow }$$ 3a + 1 = $$\sqrt{b + 6}$$ [as b ≥ -5 $$\boldsymbol{\Rightarrow }$$ b + 6 > 0]

$$\boldsymbol{\Rightarrow }$$ a = $$\frac{\left (\sqrt{b + 6} \right) – 1}{3}$$

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

Let us consider, f : [-5, ∞) → R+ as f(b) = $$\frac{\left (\sqrt{b + 6} \right) – 1}{3}$$

Then,

(fog)(a) = f(g(a)) = f(9a2 + 6a – 5) = g ((3a + 1)2 – 6) = $$\frac{\sqrt{\left (3a + 1 \right)^{2} – 6 + 6} – 1}{3}$$

=$$\frac{3a + 1 – 1}{3}$$ = $$\frac{3a}{3}$$ = a

And,

(gof)(b) = g(f(b)) = g$$\left (\frac{\sqrt{b + 6} – 1}{3} \right) = \left [3 \left (\frac{\sqrt{b + 6} – 1}{3} \right) + 1 \right]^{2} – 6$$

= $$\left (\sqrt{b + 6} \right)^{2} – 6$$ = b + 6 – 6 = b

Hence, fog = a = IR and gof = b = IRange g

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

g-1(b) = f(b) = $$\left (\frac{\left (\sqrt{y + 6} \right)^{2} – 1}{3} \right)$$

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

Sol:

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,

gof1 (b) = IY(b) = gof2(b)

$$\boldsymbol{\Rightarrow }$$ g(f1(b)) = g(f2(b))

$$\boldsymbol{\Rightarrow }$$ f1(b) = f2(b) [as g is invertible $$\boldsymbol{\Rightarrow }$$ g is on- one]

$$\boldsymbol{\Rightarrow }$$ 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.

Sol:

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

So, the inverse of the function of f exists and f-1 = h $$\boldsymbol{\Rightarrow }$$ (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.

Sol:

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

$$\boldsymbol{\Rightarrow }$$ 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) = $$\left (3 – a^{3} \right)^{\frac{1}{3}}$$, then gog(a) is

(a) $$\frac{1}{a^{3}}$$

(b) a3

(c) a

(d) (3 – a3)

Sol:

g : R → R which is given by g(a) = $$\left (3 – a^{3} \right)^{\frac{1}{3}}$$

Then,

gog(a) = g(g(a)) = g$$\left (\left (3 – a^{3} \right)^{\frac{1}{3}} \right)$$ = $$\left [3 – \left (\left (3 – a^{3} \right)^{\frac{1}{3}} \right)^{3} \right]^{\frac{1}{3}}$$

= $$\left [3 – \left (3 – a^{3} \right) \right]^{\frac{1}{3}}$$ = $$\left (a^{3} \right)^{\frac{1}{3}} = a$$

Hence, gog(a) = a

Therefore, the correct answer is C.

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

(a) f(b) = $$\frac{3b}{3 – 4b}$$ (b) f(b) = $$\frac{4b}{4 – 3b}$$

(c) f(b) = $$\frac{4b}{3 – 4b}$$ (d) f(b) = $$\frac{3b}{4 – 3b}$$

Sol:

As per the data given in the question, we have

g : R – $$\left \{-\frac{4}{3} \right \}$$ → R which is a function defined as g(a) = $$\frac{4a}{3a + 4}$$.

Let, b be the arbitrary element having range g.

Thus, there exists a ɛ R – $$\left \{-\frac{4}{3} \right \}$$ so that, b = f(a).

$$\boldsymbol{\Rightarrow }$$ b = $$\frac{4a}{3a + 4}$$

$$\boldsymbol{\Rightarrow }$$ b(3a + 4) = 4a

$$\boldsymbol{\Rightarrow }$$ 3ab + 4b = 4a

$$\boldsymbol{\Rightarrow }$$ a = $$\frac{4b}{4 – 3b}$$

Let, f : Range g → R – $$\left \{-\frac{4}{3} \right \}$$ so that, f(b) = $$\frac{4b}{4 – 3b}$$

Thus,

fog(a) = f(g(a) = f$$\left (\frac{4a}{3a + 4} \right) = \frac{4\left (\frac{4a}{3a + 4} \right)}{4 – 3\left (\frac{4a}{3a + 4} \right)}$$

= $$\frac{16a}{12a + 16 – 12a} = \frac{16a}{16}$$ = a

And,

gof(b) = g(f(b)) = g$$\left (\frac{4b}{4 – 3b} \right) = \frac{4\left (\frac{4b}{4 – 3b} \right)}{3\left (\frac{4b}{4 – 3b} \right) + 4}$$

= $$\frac{16b}{12b + 16 – 12b} = \frac{16b}{16} = b$$

Thus, fog = IR – $$\left \{- \frac{4}{3} \right \}$$ and gof = IRange f

Therefore, the inverse of the function g is the map f: Range g → R – $$\left \{- \frac{4}{3} \right \}$$, which will be given by

g(b) = $$\frac{4b}{4 – 3b}$$

Hence, the correct answer is b.