Let f : 0 , 1 → - 1 , 1 and g : - 1 , 1 → 0 , 2 be two functions such that g is injective and g o f : 0 , 1 → 0 , 2 is surjective. Then,

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Let f : 0 , 1 &#8594; - 1 , 1 and g : - 1 , 1 &#8594; 0 , 2 be two functions such that g is injective and g o f : 0 , 1 &#8594; 0 , 2 is surjective. Then,

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Correct answer: B. f must be surjective but need not be injective

Let h x = g f x where, g x is injective and h x is surjective. h x is surjective. &#8756; Codomain of h x = Range of h x Range of h x = 0 , 2 which is also codomain of g . So, must be surjective. Now, domain of g = - 1 , 1 which must be range of f . But codomain of f = - 1 , 1 So, it must be surjective.

Let f : 0 , 1 → - 1 , 1 and g : - 1 , 1 → 0 , 2 be two functions such that g is injective and g o f : 0 , 1 → 0 , 2 is surjective. Then,

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Let f : 0 , 1 &#8594; - 1 , 1 and g : - 1 , 1 &#8594; 0 , 2 be two functions such that g is injective and g o f : 0 , 1 &#8594; 0 , 2 is surjective. Then,

Options

Correct answer

Step-by-step solution

More from Functions

Let f : 0 , 1 → - 1 , 1 and g : - 1 , 1 → 0 , 2 be two functions such that g is injective and g o f : 0 , 1 → 0 , 2 is surjective. Then,