If f x = x 2 + 3 x + 8 x 2 + 6 x + 29 , then (A) f(x) is one-one (B) f(x) is not continuous (C) f(x) is not one-one (D) domain of function is not the set of real numbers Share with your friends Share 5 Aarushi Mishra answered this fx=x2+3x+8x2+6x+29Note: x2+6x+29>0 for all x as its discriminant D=b2-4ac, D=36-116<0Hence its domain mut be all real number. Also it must be continuous for all x∈ℝNow let fx=kx2+3x+8x2+6x+29=kx2+3x+8=kx2+6kx+29k1-kx2+3-6kx+8-29k=0For real roots D≥03-6k2-41-k8-29k≥09+36k2-36k-429k2-37k+8≥09+36k2-36k-116k2+148k-32≥0-80k2+112k-23≥0Discriminant of this quadratic, D=-1122-4×-80×-23=12544-7360>0Hence for some value of k, D>0Hence for some k, there will exist two distinct values of x satisfying 1-kx2+3-6kx+8-29k=0Hence fx cannot be one-one -1 View Full Answer