Question

If f(x)=2-x^2+x^3 and f(a)=2, then all possible values of a are : (a) 0&1 only (b) 0,1,&-1 (c) 0 only (d) 1 only (e) 6 only

Answer 1

f(a)=2 - a^2 + a^3
2 = a^3 - a^2 + 2
a^3 - a^2 = 0
a^3 = a^2
a = 1

Answer 2

ur answer is d

Answer 3

Nope, solving the equation f(a)=2 is the same as\[2-a^2 + a^3 = 2\]\[a^3 - a^2=0\]Most commonly people divide by a^2 at this point, however in doing so you actually lose one of your solutions. Best to factorise it into \[a^2(a-1)=0\] so now, we are looking when either \[a^2 = 0\]or \[a-1=0\]In which you end up with TWO solutions, being a=0 or a=1. So your answer is a).

Answer 4

ok lol i forgot about the a^2 = 0 thx