Let f(x) = ax2+bx+c,a>0 such that f(−1−x)=f(−1+x)∀xεR. Also given that f(x) = 0 has no real roots and b > O Let α=4a−2b+c,β=9a+3b+c,γ=9a−3a+c. Then which of the following is correct?
The correct option is Cα<γ<β
f(−1−x)=f(−1+x)⇒f(x) is symmetric about x = -1 So vertex of parabola must be at x = –1 and it is concave up. ⇒−b2a=−1⇒b=2a........(i)α=f(−2)=4a−2b+cβ=f(3)=9a+3b+cγ=f(−3)=9a−3b+c As f(x) has no real roots so values of f(x) increase as we move further away from x = –1 on both sides Hence f(3)>f(−3)>f(−2)⇒β>γ>α.