Let a,b,c be non-zero real number such that∫01 1+cos8⁡xax2+bx+cdx=∫02 1+cos8⁡xax2+bx+cdxThen the quadratic equation ax2+bx+c=0 has

Let f(β)<0f(x)=1+cos8⁡xax2+bx+cWe are given∫01 f(x)dx=∫02 f(x)dx∫01 f(x)dx=∫01 f(x)dx+∫12 f(x)dx⇒ ∫12 f(x)dx=0If f(x)>0(f(x)<0)∀x∈[1,2]then ∫12 f(x)dx>0∫12 f(x)dx<0But ∫12 f(x)dx=0∴f(x) is partly positive and partly negative on [1, 2].⇒ there exist α,β∈[1,2] such thatf(α)>0 and f(β)<0.As f is continuous on [1,2] there exists γ lying between α and β (and hence between 1 and 2) such that f(γ)=0⇒ 1+cos8⁡γaγ2+bγ+c=0⇒ aγ2+bγ+c=0 ∵1+cos8⁡γ≥1Thus, ax2+bx+c=0 has at least one root in [1,2].