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Here f : A → B and g : B → C are one-to-one functions ∴ g o f is a function from A to C. Let x 1 , x 2 ∈ A Now (g o f) (x 1 ) = (g o f) (x 2 ) ⇒ g (f (x 1 )) = g (f(x 2 )) ⇒ f(x 1 ) = f(x 2 ) [∵ g is one-to-one] ⇒ x 1 = x 2 [∵ f is one-to-one] ⇒ (g o f) (x 1 ) = (g o f) (x 2 ) ⇒ x 1 = x 2 , ∀ x 1 ,x 2 x 1 ,x 2 ∈ A ∴ gof is one-to-one.

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Relations and Functions

Short Answer Type

271.

Consider f : N → N, g : N → N and h : N → R defined as f (x) = 2 x, g(y) = 3 y + 4 and h(z) = sin z ∀ x, y and z in N. Show that h o (g o f) = (h o g) o f.

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272. Let f: (I, 2. 3} → a b. c be one-one and onto function given by f(I) = a,f(2) = b and f(3) = c. Show that there exists a function g : {a, b. c} → {1, 2, 3} such that g o f = 1_x and f o g = 1_Y where, X = {1, 2, 3} and Y= {a, b, c}.

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273.

Let f, g and h be function from R to R. Show that (f + g) o h = f o h + g o h

(f . g) o h = (f o h) . (g o h)

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274. Let f : Z → Z be defined by f (x) = x + 2. Find g : Z → Z such that gof = I_z.

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275. If f : R → R is defined by f (x) = x²– 3 x + 2, find f (f (x)).

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276. Show that if f : A → B and g : B → C are one-one, then g o f : A → C is also one-one.

Here f : A → B and g : B → C are one-to-one functions ∴ g o f is a function from A to C.
Let x₁, x₂ ∈ A
Now (g o f) (x₁) = (g o f) (x₂)
⇒ g (f (x₁)) = g (f(x₂))
⇒ f(x₁) = f(x₂) [∵ g is one-to-one]
⇒ x₁ = x₂ [∵ f is one-to-one]
⇒ (g o f) (x₁) = (g o f) (x₂) ⇒ x₁ = x₂ , ∀ x₁,x₂ x₁,x₂ ∈ A
∴ gof is one-to-one.