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

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

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