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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.

Here f : A → B and g : B → C are onto functions ∴ g o f is defined from A to C. ∴ g is an onto mapping from B → C ∴ to each z ∈ C, there exists y ∈ B such that g (y) = z Again f is an onto mapping from A to B
∴ to each y ∈ B, there exists y ∈ A such that f(x) = y ∴ to each z ∈ C, there exists x ∈ A such that z = g (y) = g (f(x)) = (g o f) (x) ∴ g o f is onto.

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