Question

Let $f:R\to \left[0.\frac{\pi }{2}\right)$ be defined by $f\left(x\right)=ta{n}^{-1}(x^2+x+a)$. Then the set of values of a for which f is onto is

• A. $[0,\infty )$
  
• B. $[\frac{1}{4},\infty )$
  
• C. $(-\infty ,\frac{1}{4}]$
  
• D. $\left\{\frac{1}{4}\right\}$

Answer 1

The correct option is B $[\frac{1}{4},\infty )$


We know for a function to be an onto function, Range should be equal to co-domain.
Here co-domain which is given is - $\left[0.\frac{\pi }{2}\right)$
To have the range $\left[0.\frac{\pi }{2}\right)$ domain should have all non zero values.
$x^2+x+a\ge 0$
this quadratic will be non zero if its discriminant is less than and equal to zero, Since it has coefficient of $x^2$ positive.
Discriminant $\Rightarrow 1-4a\le 0$
$\Rightarrow a\ge \frac{1}{4}$