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$f : R \to R defined by f (x) = \frac{1}{2} x | x | + \cos x + 1 is$

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a

one-one and onto

b

one-one and into

c

many-one and onto

d

many-one and into

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detailed solution

Correct option is A

fx = xx2+cosx+1 If x≥0 then fx = x22+cosx +1 f'x = x - sinx ≥0 for all x≥0 if x<0 then fx = -x22 + cosx +1 ⇒f'x=-x -sinx >0 since x is negative therefore f'x >0 for all x ∈R ⇒ f is increasing , hence f is one-one and fx→∞ when x→∞ and fx→ -∞ when x→-∞ Range is = R , hence f is onto

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Similar Questions

Which of the following statements are incorrect?

$I. If f (x) and g (x) are one-one then f (x) + g (x) is also one-one$

$II. If f (x) and g (x) are one-one then f (x) \cdot g (x) is also one-one$

$III. If f (x) is odd then it is necessarily one-one.$

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