find the number of integral values of a in the interval [0,100] so that the range of the function f(x)= (x+a)/(x2-1) contains the interval [0,1]???

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fx=x+ax2-1=x+ax-1x+1When a=1fx=x+1x-1x+1fx cannot be 0 in this case as numerator becomes 0 at x=-1 but x=-1make denominator 0 too which makes function undefined.Hence a1 if we want range to have the interval 0,1For all other values of a, you will 0 in range at x=-ae.g.for a=2fx will be 0 at x=-1Set fx=1x+ax2-1=1x+a=x2-1a=x2-x-1x2-x-1-a=0Now a is integer in the interval 0,100a=1 is already discarded for a=0,2,3,4....,100 -1-a will always be negative.Roots of Ax2+Bx+C is always real when AC<0Here A=1 and C=-1-aAC<0Hence roots of x2-x-1-a=0 will be real.Hence 1 will also be part of range0,1 will be for all a0,100 except a=1Hence 100 is the answer

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