NIMCET 2019 Question Paper with solutions

CONCEPT:
${\mathit{Z}}^{+}$ is the set of all positive integers $\left(1,2,3,...\right)$
If quadratic equation ${\mathit{a}\mathit{x}}^2+\mathit{b}\mathit{x}+\mathit{c}=0$ , has real roots then its discriminant $\left({\mathit{b}}^2-4\mathit{a}\mathit{c}\right)$ should be positive.
CALCULATIONS:
Given equation is $\left[{\tan }^2\mathit{x}\right]-\tan \mathit{x}-\mathit{a}=0$ .
$\therefore \mathit{D}={\mathit{b}}^2-4\mathit{a}\mathit{c}={\left(-1\right)}^2-4\left(1\right)\left(-\mathit{a}\right)=1+4\mathit{a}$
So D must be an odd perfect square -
$\Rightarrow \sqrt{1+4\mathit{a}}=2\mathit{\lambda }+1$
$\Rightarrow 1+4\mathit{a}=4{\mathit{\lambda }}^2+1+4\mathit{\lambda }$
$\Rightarrow \mathit{a}=\mathit{\lambda }\left(\mathit{\lambda }+1\right)$
So, For different values of $\mathit{\lambda }(1,2,3,4,5,6,7,8,9$ but $\mathit{\lambda }$ can not be 0
as $\left\{\mathit{a}\in {\mathit{Z}}^{+}:\mathit{a}\le 100\right\})$ we will get different values of 'a'.
So, $\mathit{a}=2,6,12,20,30,42,56,72,90$ .
Hence there are 9 values of 'a'.