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Let f : A → B and g : B → C be two functions such that g o f : A ∴ C is defined. We are given that g of : A → C is one-one. We are to prove that f is one-one If possible, suppose that f is not one-one. there exists x 1 , x 2 ∈ A such that x 1 ≠ x 2 but f (x 1 ) = f (x 2 ) But f(x 1 ) = f (x 2 ) ⇒ g (f(x 1 )) = g (f(x 2 )) ⇒ (gof) (x 1 ) = (gof) (x 2 ) ∴ x 1 , x 2 ∈ A such that x 1 ≠ x 2 but (gof) (x 1 ) = (gof) (x 2 ) ∴ gof is not one-one, which is against the given hypothesis that g of is one-one our supposition is wrong. ∴ f is one-one. Consider f : {1, 2, 3, 4} ≠ {1, 2, 3, 4, 5, 6} defined as f (x) = x, ∀ x and g : {1, 2, 3, 4, 5, 6} → {1, 2, 3, 4, 5, 6} as g. (x) = x, for x = 1, 2, 3, 4 and g (5) = g (6) = 5. Then, g o f (x) = x ∀ x, which shows that g o f is one-one. But g is clearly not one-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.

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277. Show that f : A → B and g : B → C are onto, then g of : A → C is also onto.

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Long Answer Type

278. Consider functions f and g such that composite g of is defined and is one-one. Are f and g both necessarily one-one.

Let f : A → B and g : B → C be two functions such that g o f : A ∴ C is defined.
We are given that g of : A → C is one-one.
We are to prove that f is one-one If possible, suppose that f is not one-one.
there exists x₁, x₂ ∈ A such that x₁ ≠ x₂ but f (x₁) = f (x₂)
But f(x₁) = f (x₂) ⇒ g (f(x₁)) = g (f(x₂))
⇒ (gof) (x₁) = (gof) (x₂)
∴ x₁, x₂ ∈ A such that x₁ ≠ x₂ but (gof) (x₁) = (gof) (x₂)
∴ gof is not one-one, which is against the given hypothesis that g of is one-one our supposition is wrong.
∴ f is one-one.
Consider f : {1, 2, 3, 4} ≠ {1, 2, 3, 4, 5, 6} defined as f (x) = x, ∀ x and g : {1, 2, 3, 4, 5, 6} → {1, 2, 3, 4, 5, 6} as g. (x) = x, for x = 1, 2, 3, 4 and g (5) = g (6) = 5. Then, g o f (x) = x ∀ x, which shows that g o f is one-one. But g is clearly not one-one.

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Short Answer Type

279. Are/and g both necessarily onto, if g o f is onto ?

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280. Give examples of two functions f : N → Z and g : Z → Z such that g o f is injective but g is not injective.

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