Let $$f:R \to A = \left\{ {y\,|\,0 \leqslant y < \frac{\pi }{2}} \right\}$$ be a function such that $$f\left( x \right) = {\tan ^{ - 1}}\left( {{x^2} + x + k} \right),$$ where $$k$$ is a constant. The value of $$k$$ for which $$f$$ is an onto function, is :
A.
1
B.
0
C.
$$\frac{1}{4}$$
D.
none of these
Answer :
$$\frac{1}{4}$$
Solution :
Clearly, $${x^2} + x + k \geqslant 0$$ for all $$x.$$
So, $$D = 1 - 4k \leqslant 0\,\,\, \Rightarrow k \geqslant \frac{1}{4}$$
Releted MCQ Question on Calculus >> Function
Releted Question 1
Let $$R$$ be the set of real numbers. If $$f:R \to R$$ is a function defined by $$f\left( x \right) = {x^2},$$ then $$f$$ is: