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.

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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 g ∘ f : A → C 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 g ∘ f : A → C 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 g ∘ f : A → C is one-one and onto, hence bijective.

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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 $$(g \circ f)^{-1} = f^{-1} \circ g^{-1}$$

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

i) To prove that f ∘ I_A​ = 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 I_B​ ∘ 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 $$I_B \circ f = f$$

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

i) To prove that f ∘ f⁻¹ = I_B​

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 I_B​, 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⁻¹ ∘ f = I_A

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 I_A​, 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 $$f^{-1} \circ f = I_A$$