NIMCET 2018 Question Paper with solutions

Given that, α, β are the roots of equation ${\mathit{x}}^2$ - px + r = 0
∴ α + β = - $\frac{\mathrm{Coefficient}\mathrm{of}\mathrm{x}}{\mathrm{Coefficient}\mathrm{of}{\mathit{x}}^2}$ = p ... (i)
and αβ = $\frac{\mathrm{constant}}{\mathrm{Coefficient}\mathrm{of}{\mathit{x}}^2}$ = r
Also, $\frac{\mathit{\alpha }}{2}$ and 2β are roots of equation ${\mathit{x}}^2$ - qx + r = 0
∴ $\frac{\mathit{\alpha }}{2}+2\mathit{\beta }$ = - $\frac{\mathrm{Coefficient}\mathrm{of}\mathrm{x}}{\mathrm{Coefficient}\mathrm{of}{\mathit{x}}^2}$ = q
⇒ α + 4β = 2q ... (ii)
Now, subtract Eq.(i) from Eq. (ii), we get
3β = 2q - p
⇒ β = $\frac{2\mathit{q}-\mathit{p}}{3}$
Put the values of P in Eq. (i),
⇒ α = p - $\frac{2\mathit{q}-\mathit{p}}{3}$ = $\frac{3\mathit{p}-2\mathit{q}+\mathit{p}}{3}$ = $\frac{4\mathit{p}-2\mathit{q}}{3}$
= $\frac{2}{3}$ (2p - q)
∴ r = αβ
⇒ r = $\frac{2}{3}$ (2p - q) $\frac{\left(2\mathit{q}-\mathit{p}\right)}{3}$
= $\frac{2}{9}$ (2p - q) (2q - p)