If the function f(x)=x^3-3(a-2) x^2+3 a x+7 , for some a R is increasing in (0,1] and decreasing in [0,5) , then a root of the equation f(x)-14 (x-1)^2 =0(x 1) is

Step-by-step solution

aligned & f(x)=x^3-3(a-2) x^2+3 a x+7 & f^ (x)=3 x^2-6(a-2) x+3 a aligned f^ (x) changes its behaving at x=1 aligned & f^ (1)=0 & 3 1^2-6(a-2) 1+3 a=0 & a=b f(x)=x^3-9 x^2+15 x+7 & f(x)-14 (x-1)^2 =0 (x^3-9 x^2+15 x+7 )-14 (x-1)^2 =0 & x^3-9 x^2+15 x-7=0 & (x-1)(x-1)(x-7) & x=1,1,7 aligned