if the maximum value of f(x) =a-16+8x-x^2 is smaller than the minimum value of g(x) =x4-2ax+10-2a, then the number of - Maths - Complex Numbers and Quadratic Equations

Question

if the maximum value of f(x) =a-16+8x-x^2 is smaller than the
minimum value of g(x) =x4-2ax+10-2a, then the number of positive
integral value of a is

Brijendra Pal · Accepted Answer

Hi,
second equation must be x2-2ax+10-2a only then we can solve further so assuming the same and solving, since for a quadratic equation ax2+bx+c= 0 max/min value is = -D4aso for equation -x2+8x+a-16= 0 max value since a is negative= -64+4a-64-4= a and for x2-2ax+10-2a =0 minimum value = -4a2-40+8a 4=-a2-2a+10now according to the question, a<-a2-2a+10a2+3a-10<0 a2+5a-2a-10<0aa+5-2a+5<0a+5a-2<0a∈-5,2so positive integral value of a is  1 that is 1 itself
â€‹Regards

if the maximum value of f(x) =a-16+8x-x^2 is smaller than the minimum value of g(x) =x4-2ax+10-2a, then the number of positive integral value of a is