CAT Exam Model Paper 3 with solutions for free online practice

Given, $4x^{2} - 16x + \frac{\lambda}{4} = 0$ ⇒ $16x^{2} - 64x + \lambda = 0$...(i) It is given that α and β are the roots of Eq. (i) $\therefore \alpha = \frac{-(-64) + \sqrt{(-64)^{2} - 4 \times 16 \times \lambda}}{2 \times 16}$and $\beta = \frac{-(-64) - \sqrt{(-64)^{2} - 4 \times 16 \times \lambda}}{2 \times 16}$⇒$\alpha = \frac{64 + 8 \sqrt{64 - \lambda}}{32}$ and $\beta = \frac{64 - 8 \sqrt{64 - \lambda}}{32}$⇒$\alpha = \frac{8 (8 + \sqrt{64 - \lambda})}{32}$ and $\beta = \frac{8 (8 - \sqrt{64 - \lambda})}{32}$⇒$\alpha = \frac{8 + \sqrt{64 - \lambda}}{4}$ and $\beta = \frac{8 - \sqrt{64 - \lambda}}{4}$Since, 1< α < 2 and 2 < β < 3∴ $1 < \frac{8 + \sqrt{64 - \lambda}}{4} < 2$ and $2 < \frac{8 + \sqrt{64 - \lambda}}{4} < 3$⇒$4 < 8 + \sqrt{64 - \lambda} < 8$ and $8 < 8 - \sqrt{64 - \lambda} < 12$⇒$4 - 8 < \sqrt{64 - \lambda} < 8 - 8$ and $8 - 8 < \sqrt{64 - \lambda} < 12 - 8$⇒$-4 < \sqrt{64 - \lambda} < 0$ and $0 < \sqrt{64 - \lambda} < 4$On squaring each term, we get16 < 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.