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Let f(x) = x, g(x) | x | where f : N → Z and g : Z → Z Now g(– 1) = | – | | = 1, g(1) = | 1 | = 1 g is not one-one i.e. injective. Now f : N → Z and g : Z → Z ∴ g o f : N → Z Let x 1 , x 2 ∈ N such that (g o f) (x 1 ) = (g o f)( x 2 ) ∴ g (f(x 1 )) = g(f(x 2 )) ∴ g(x 1 ) = g(x 2 ) ∴ |x 1 | = | x 2 | ⇒ x 1 = x 2 [∵ x 1 > 0, x 2 > 0 ] ∴ (g o f)(x 1 ) = (g o f) (x 2 ) ⇒ x 1 = x 2 ∴ g o f is onto.

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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.

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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.

Let    f(x) = x, g(x) | x |
where f : N → Z and g : Z → Z
Now    g(– 1) = | – | | = 1, g(1) = | 1 | = 1
g is not one-one i.e. injective.
Now f : N → Z and g : Z → Z
∴ g o f : N → Z
Let x₁ , x₂ ∈ N such that
(g o f) (x₁) = (g o f)( x₂) ∴ g (f(x₁)) = g(f(x₂))
∴ g(x₁) = g(x₂)
∴ |x₁| = | x₂|
⇒ x₁ = x₂    [∵ x₁ > 0, x₂ > 0 ]
∴ (g o f)(x₁) = (g o f) (x₂) ⇒ x₁ = x₂ ∴ g o f is onto.

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