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Let f:R rarr R, f(x)=x+log(e)(1+x^(2)). ...

Let `f:R rarr R, f(x)=x+log_(e)(1+x^(2))`. Then

A

f is injective

B

f is surjective

C

there is a point on the graph of y= f(x) where tangent is not parallel to any of the chords

D

inverser of f(x) exists.

Text Solution

Verified by Experts

The correct Answer is:
A, B, C, D
`f(x)=x+log_(e)(1+x^(2))`
`rArr" "f'(x)=1+(2x)/(1+x^(2))`
`=((x+1)^(2))/(1+x^(2))ge 0, AA x in R`
So, f(x) in injective and ata `x=-1`, tangent is not parallel to any of the chords.
`"Also "underset(xrarr0oo)(lim)f(x)`
`=underset(xrarr-oo)(lim)log_(e)(e^(2)(1+x^(2)))`
`=log_(e)(underset(xrarr-oo)(lim)(1+x^(2))/(e^(-x)))`
`=log_(e)(underset(xrarr-oo)(lim)(2x)/(e^(-x)))`
`=log_(e)(underset(xrarr-oo)(lim)(2)/(e^(-x)))`
`=-oo`
`underset(xrarroo)(lim)f(x)=oo`
Thus f(x) has range R, so f(x) is surjective.
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