If x is real, the function (x-a)(x-b)(x-c) will assume all real values, provided

Let y=(x-a)(x-b)(x-c)or y(x-c)=x2-(a+b)x+abor x2-(a+b+y)x+ab+cy=0∆=(a+b+y)2-4(ab+cy)=y2+2y(a+b-2c)+(a-b)2Since x is real and y assumes all real values, we must have ∆≥0 for all real values of y. The sign of a quadratic in y is same as of first term provided its discriminant B2-4AC<0This will be so if 4(a+b-2c)2-4(a-b)2<0or 4(a+b-2c+a-b)(a+b-2c-a+b)<0or 16(a-c)(b-c)<0 or 16(c-a)(c-b)=-ve∴ c lies between a and b i.e., ac>b   ...(ii)Hence from (i) and (ii) we observe that Option 4 is correct answer.