Question

Let $\alpha +\beta =1,2{\alpha }^2+2{\beta }^2=1$ and f(x) be a continuous function such that f(x + 2) + f(x) = 2 $\forall \times ϵ[0,2]and(p+4)={\int }_0^{4}f\left(x\right)dx\&q=\frac{\alpha }{\beta }$ exactly one root of the equation $a{x}^2-bx+c=0$ is lying between p and q when a, b, $cϵN$ then

• A. $b^2-4ac\le 0$
• B. $c(a-b+c)>0$
• C. $b^2-4ac\ge 0$
• D. $c(a-b+c)<0$

Answer 1

Answer: (C), (D)

The correct options are
C $b^2-4ac\ge 0$
D $c(a-b+c)<0$

Given : $\alpha +\beta =1$

$2{\alpha }^2+2{\beta }^2=1(\alpha +\beta {)}^2={\alpha }^2+{\beta }^2+2\alpha \beta (1{)}^2=\frac{1}{2}+2\alpha \beta \frac{1}{4}=\left(\alpha \beta \right)(\alpha -\beta {)}^2=(\alpha +\beta {)}^2-4\alpha \beta 1-4\times \frac{1}{4}=0(\alpha =\beta )=\frac{1}{2}q=\frac{\alpha }{\beta }=1p+4={\int }_0^{4}f\left(x\right)dxp+4={\int }_0^{2}f\left(x\right)dx+{\int }_2^{4}f\left(x\right)dxp+4=2+2p=0$

One root of $a^2-bx+c=0$ lies in $(1,0)$

The $D\ge 0b^2-4ac\ge 0$ as roots are real

Also, $f\left(0\right)f\left(1\right)<0$

$\left(c\right)(a-b+c)<0$