plz provide solution of question 122 in detail 122. Let f : R → A = y | 0 ≤ y < π 2 be a function such that f(x) = tan–1(x2 + x + k) where k is a constant. The value of k for which f is an onto function is: (a) 1 (b) 0 (c) 1 4 (d) none of these Share with your friends Share 0 Zuhaib Ul Zamann answered this Dear student, Range is [0,π/2)So we have tan-1(x2+x+k)∈[0,π/2)x2+x+k∈[0,∞)Hence min(x2+x+k)=0Now we know that min of quadratic ax2+bx+c occurs at x=-b2aHence -122-12+k=0k=12-14=14Regards 0 View Full Answer