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Let f: Rvec[0,pi/2) be defined by f(x)=t...

Let `f: Rvec[0,pi/2)` be defined by `f(x)=tan^(-1)(x^2+x+a)dot` Then the set of values of `a` for which `f` is onto is `(0,oo)` (b) `[2,1]` (c) `[1/4,oo]` (d) none of these

A

`[0,oo)`

B

`[2,1]`

C

`[(1)/(4),oo)`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
C
Since co-domain`=[0,(pi)/(2))`
This is possible only when `x^(2) +x+a` is perfect square.
` :. 1-4a=0 " or " a=(1)/(4)`
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