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Functions (LAQs)

Maths-1A | 1. Functions – LAQs:
Welcome to LAQs in Chapter 1: Functions. This page contains the most Important FAQs for Long Answer Questions in this Chapter. Each answer is provided in simple and easy-to-understand steps. This will support your preparation and help you secure top marks in your exams.


LAQ-1 : If f:A→B, g:B→C are two bijective functions then prove that gof:A→C is also a bijective function.

Given that f, g are bijective functions, So f, g are both one-one and onto functions.

i) To prove that gf : AC is one-one

Let (g \circ f)(a_1) = (g \circ f)(a_2) [ \text{for } a_1, a_2 \in A]

\Rightarrow g[f(a_1)] = g[f(a_2)]

\Rightarrow f(a_1) = f(a_2)

\Rightarrow a_1 = a_2

g \circ f : A \rightarrow C

ii) To prove that gf : AC is onto.

Given f : A \rightarrow B is onto, then f(a) = b\ldots(1)

Given g : B \rightarrow C is onto, then g(b) = c\ldots(2)

Now (g \circ f)(a) = g[f(a)] = g(b) = c { \text{From } (2) \text{ and } (1)}

g \circ f : A \rightarrow C

Hence, we proved that gf : AC is one-one and onto, hence bijective.


LAQ-2 : If f : A → B, g : B → C are two bijective functions then prove that (gof)(-1) = f(-1) og(-1)

Part-1:

Given that f: A \rightarrow B g: B \rightarrow C are two bijective functions, then

g \circ f: A \rightarrow C is bijection \Rightarrow (g \circ f)^{-1}: C \rightarrow A is also a bijection

f^{-1}: B \rightarrow A g^{-1}: C \rightarrow B are both bijections \Rightarrow (f^{-1} \circ g^{-1}): C \rightarrow A is also a bijection.

So, (g \circ f)^{-1} and f^{-1} \circ g^{-1} both have the same domain ‘C’.

Part-2:

Given f: A \rightarrow B is bijection, then f(a) = b \Rightarrow a = f^{-1}(b)

g: B \rightarrow C is bijection, then g(b) = c \Rightarrow b = g^{-1}(c)

g \circ f: A \rightarrow C is bijection, then g \circ f(a) = c \Rightarrow a = (g \circ f)^{-1}(c)

Now, (f^{-1} \circ g^{-1})(c) = f^{-1}[g^{-1}(c)] = f^{-1}(b) = a

Thus, (g \circ f)^{-1}(c) = (f^{-1} \circ g^{-1})(c)

Hence, we proved that (g \circ f)^{-1} = f^{-1} \circ g^{-1}


LAQ-3 : If f : A → B is a function and IA, IB are identity functions on A,B respectively then prove that foIA = f = IB of

i) To prove that fIA​ = f

Part-1:

Given f: A \rightarrow B is a function.

We know I_A: A \rightarrow A

f \circ I_A: A \rightarrow B

So, f \circ I_A​ and f, both have the same domain A.

Part-2:

For a \in A (f \circ I_A)(a) = f[I_A(a)]

= f(a)

Hence, we proved that f \circ I_A = f

ii) To prove that IB​ ∘ f = f

Part-1:

Given f: A \rightarrow B is a function.

We know I_B: B \rightarrow B

I_B \circ f: A \rightarrow B

So, I_B \circ f and f, both have the same domain A.

Part-2:

For a \in A (I_B \circ f)(a) = I_B[f(a)]

= f(a)

Hence, we proved that I_B \circ f = f


LAQ-4 : If f : A → B is a bijective function then prove that (i) fof(-1) = IB (ii) f(-1) of = IA

i) To prove that ff−1 = IB

Part-1:

Given f: A \rightarrow B is a bijective function, then f^{-1}: B \rightarrow A is also a bijection

f \circ f^{-1}: B \rightarrow B

We know, I_B: B \rightarrow B

So, f \circ f^{-1} and IB​, both have the same domain B.

Part-2:

For b \in B (f \circ f^{-1})(b) = f[f^{-1}(b)]

Since f and f−1 are inverses, applying f−1 to b retrieves the original element a in A that f maps to b, hence f[f^{-1}(b)] = b

Therefore, (f \circ f^{-1})(b) = b = I_B(b)

Hence, we proved that f \circ f^{-1} = I_B​

ii) To prove that f−1f = IA

Part-1:

Given f: A \rightarrow B is a bijective function, then f^{-1}: B \rightarrow A is also a bijection

f^{-1} \circ f: A \rightarrow A

We know I_A: A \rightarrow A

So, f^{-1} \circ f and IA​, both have the same domain A.

Part-2:

For a \in A (f^{-1} \circ f)(a) = f^{-1}[f(a)]

Since f and f−1 are inverses, applying f to a and then f−1 to the result retrieves the original element a, hence f^{-1}[f(a)] = a

Therefore, (f^{-1} \circ f)(a) = a = I_A(a)

Hence, we proved that f^{-1} \circ f = I_A