** 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) = 8a ^{3} 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) = 8a^{3 }and f(a) = a^{\(\frac{1}{3}\)}

Therefore, fog(a) = f(g(a)) = f(8a^{3}) = (8a^{3})^{\(\frac{1}{3}\) }= (2^{3} a^{3})^{ \(\frac{1}{3}\)} = 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 = I_{a}

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 = I_{Range }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 = I_{R}

**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) = a**

^{2}+ 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) = a^{2} + 4

For one- one function,

Let, g(a) = g(b)

\(\boldsymbol{\Rightarrow }\) a^{2} + 4 = b^{2} + 4

\(\boldsymbol{\Rightarrow }\) a^{2} = b^{2 }

\(\boldsymbol{\Rightarrow }\) a = b

Hence, the function f is one- one function.

For onto function,

For b ɛ [4, ∞), let b = a^{2} + 4

\(\boldsymbol{\Rightarrow }\) a^{2} = 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 = I_{R}

**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) = 9a^{2} + 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(9a^{2} + 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 = I_{R }and gof = b = I_{Range }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, g_{1} and g_{2})

Thus, for every b ɛ Q, we have,

gof_{1 }(b) = I_{Y}(b) = gof_{2}(b)

\(\boldsymbol{\Rightarrow }\) g(f_{1}(b)) = g(f_{2}(b))

\(\boldsymbol{\Rightarrow }\) f_{1}(b) = f_{2}(b) [as g is invertible \(\boldsymbol{\Rightarrow }\) g is on- one]

\(\boldsymbol{\Rightarrow }\) f_{1}= f_{2} [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 = I_{X } and gof = I_{Y}, 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 = I_{X }and hof = I_{Y}, 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 = I_{P} and gof = I_{Q}

So, g^{-1} = f

Then, fog = I_{P} and gof = I_{Q }

\(\boldsymbol{\Rightarrow }\) g^{-1}og= I_{P} and fof^{-1} = I_{Q}

**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) a ^{3}**

**(c) a **

**(d) (3 – a ^{3)}**

** **

**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 = I_{R – \(\left \{- \frac{4}{3} \right \}\)} and gof = I_{Range 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.**