CAT Exam Model Paper 3 with solutions for free online practice

Given, $4{\mathit{x}}^2-16\mathit{x}+\frac{\mathit{\lambda }}{4}=0$
⇒ $16{\mathit{x}}^2-64\mathit{x}+\mathit{\lambda }=0$...(i)
It is given that α and β are the roots of Eq. (i)
$\therefore \mathit{\alpha }=\frac{-\left(-64\right)+\sqrt{{\left(-64\right)}^2-4\times 16\times \mathit{\lambda }}}{2\times 16}$
and $\mathit{\beta }=\frac{-\left(-64\right)-\sqrt{{\left(-64\right)}^2-4\times 16\times \mathit{\lambda }}}{2\times 16}$
⇒$\mathit{\alpha }=\frac{64+8\sqrt{64-\mathit{\lambda }}}{32}$ and $\mathit{\beta }=\frac{64-8\sqrt{64-\mathit{\lambda }}}{32}$
⇒$\mathit{\alpha }=\frac{8\left(8+\sqrt{64-\mathit{\lambda }}\right)}{32}$ and
$\mathit{\beta }=\frac{8\left(8-\sqrt{64-\mathit{\lambda }}\right)}{32}$
⇒$\mathit{\alpha }=\frac{8+\sqrt{64-\mathit{\lambda }}}{4}$ and $\mathit{\beta }=\frac{8-\sqrt{64-\mathit{\lambda }}}{4}$
Since, 1< α < 2 and 2 < β < 3
∴ and
⇒ and
⇒ and
⇒ and
On squaring each term, we get
16 < 64 − λ < 0 and 0 < 64 − λ < 16
⇒ 16 − 64 < − λ < 0 − 64 and 0 − 64 < − λ < 16 − 64
⇒ − 48 < − λ < − 64 and − 64 < − λ < − 48
⇒ 48 > λ > 64 and 64 > λ > 48 [∵48 > λ > 64 not possible]
Hence, λ can take 15 values.