Question

If $\alpha $ and $\beta $ are the roots of $4x^{2}-16x+t=0, \, \forall t>0$ such that $1 < \alpha < 2 < \beta < 3,$ then the number of integral values of $t$ are

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

Answer: (B)

$f\left(x\right)=4x^{2}-16x+t$
Opening upward parabola

$f\left(1\right)>0, \, f\left(2\right) < 0, \, f\left(3\right)>0$
$f\left(1\right)>0\Rightarrow 4-16+t>0\Rightarrow t>12$
$f\left(2\right) < 0\Rightarrow 16-32+t < 0\Rightarrow t < 16$
$f\left(3\right)>0\Rightarrow 36-48+t>0\Rightarrow t>12$
$\Rightarrow 12 < t < 16\Rightarrow t=13, \, 14, \, 15$
Hence, the number of integral values $\left(t = 13,14,15\right)$ is $3.$