Question

Number of integral values of $\lambda$ for which $x^2-2\lambda x<41-6\lambda \forall x\in (1,6],$ is

Answer 1

$f\left(x\right)=x^2-2\lambda x+6\lambda -41$
$f\left(x\right)<0\forall x\in (1,6]$
So, $f\left(1\right)\le 0......\left(1\right)$
$f\left(6\right)<0......\left(2\right)$
So, $\left(1\right)\cap \left(2\right)\to \lambda \in (-\frac{5}{6},10]$
$\lambda =0,1,.....,10$