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Q.If A={1,2,3}A = \{1, 2, 3\}, B={α,β,γ}B = \{\alpha, \beta, \gamma\}, C={p,q,r}C = \{p, q, r\} and f:A→Bf : A \to B, g:B→Cg : B \to C are defined by f={(1,α),(2,γ),(3,β)}f = \{(1, \alpha), (2, \gamma), (3, \beta)\}, g={(α,q),(β,r),(γ,p)}g = \{(\alpha, q), (\beta, r), (\gamma, p)\}, then show that f and g are bijective functions and (gof)−1=f−1og−1(gof)^{-1} = f^{-1}og^{-1}

Telangana TsbieTelangana Board of Intermediate Education (Intermediate 1st Year) 2024Subjective· 7mImportance★★★★★
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Check each of ff and gg is one-one and onto (hence bijective), compute gofgof and its inverse directly, then compute f−1og−1f^{-1}og^{-1} and show the two results are identical.

Given: A={1,2,3}A=\{1,2,3\}, B={α,β,γ}B=\{\alpha,\beta,\gamma\}, C={p,q,r}C=\{p,q,r\}, f={(1,α),(2,γ),(3,β)}f=\{(1,\alpha),(2,\gamma),(3,\beta)\}, g={(α,q),(β,r),(γ,p)}g=\{(\alpha,q),(\beta,r),(\gamma,p)\}.

Step 1. ff is bijective. The images α,γ,β\alpha,\gamma,\beta of 1,2,31,2,3 are all distinct ⇒\Rightarrow ff is one-one (injective). Every element of B={α,β,γ}B=\{\alpha,\beta,\gamma\} appears as an image ⇒\Rightarrow ff is onto (surjective). Hence ff is bijective.

Step 2. gg is bijective. The images q,r,pq,r,p of α,β,γ\alpha,\beta,\gamma are all distinct ⇒\Rightarrow gg is one-one. Every element of C={p,q,r}C=\{p,q,r\} appears as an image ⇒\Rightarrow gg is onto. Hence gg is bijective.

Step 3. Compute gof:A→Cgof:A\to C:

gof(1)=g(f(1))=g(α)=q,gof(2)=g(f(2))=g(γ)=p,gof(3)=g(f(3))=g(β)=rgof(1)=g(f(1))=g(\alpha)=q,\quad gof(2)=g(f(2))=g(\gamma)=p,\quad gof(3)=g(f(3))=g(\beta)=r

gof={(1,q),(2,p),(3,r)}gof = \{(1,q),(2,p),(3,r)\}

Step 4. Compute (gof)−1(gof)^{-1} (swap each pair):

(gof)−1={(q,1),(p,2),(r,3)}(gof)^{-1} = \{(q,1),(p,2),(r,3)\}

Step 5. Compute f−1f^{-1} and g−1g^{-1}:

f−1={(α,1),(γ,2),(β,3)},g−1={(q,α),(r,β),(p,γ)}f^{-1} = \{(\alpha,1),(\gamma,2),(\beta,3)\}, \qquad g^{-1} = \{(q,\alpha),(r,\beta),(p,\gamma)\}

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