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Let : f:[0, ] A ; f(x)= ⁻¹ x + - ⁻¹ x is an onto function, then:

Options

  1. Af(x) is injective
  2. Bf(x) is many-one
  3. Cset A is [ , 2 )
  4. Dset A is [ , 2 )

Correct answer

C. set A is [ , 2 )

Step-by-step solution

⁻¹ x=t ( - 2 , 2 ) and f(x) is defined when 0 t 0 so injective. Also, at t=0 f(x)= 0 + = at t= 2 f(x)= 2 + 2 =2 2 = 2

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