Question

For all $x\in R$, if $\lambda$ $x^2-9\lambda x+5\lambda +1>0$, then $\lambda$ lies in the interval

• A. $\left(-\frac{4}{61},0\right)$
• B. $\left[0,\frac{4}{61}\right]$
• C. $[0,\frac{4}{61})$
• D. None of these

Answer 1

Answer: (C)

We have, $\lambda x^2-9\lambda x+5\lambda +1$
$=\lambda \left(x^2-9x\right)+5\lambda +1$
$=\lambda \left(x^2-9x+\frac{81}{4}\right)+\left(5\lambda -\frac{81}{4}\lambda +1\right)$
$=\lambda {\left(x-\frac{9}{2}\right)}^2+\left(1-\frac{61\lambda }{4}\right)>0$ if $\lambda \ge 0$ and $1-\frac{61\lambda }{4}>0$
(For $\lambda =0$, the given expression is $1$)
i.e., if $\lambda \ge 0$ and $\lambda <\frac{4}{61}$ i.e., if $0\le \lambda <\frac{4}{61}$.