If \(\left(\lambda^2, \lambda+1\right), \lambda \in Z\) belongs to the region between the lines \(x+2…

If \(\left(\lambda^2, \lambda+1\right), \lambda \in Z\) belongs to the region between the lines \(x+2 y-5=0\) and \(3 x-y+1=0\) which includes the origin, then the possible number of such points is
  1. 4
  2. 3
  3. 2
  4. Infinite

Solution

Given equation of lines are \(\begin{aligned} & x+2 y-5=0 \quad \ldots (i) \\ & \text{and } 3 x-y+1=0 \quad \ldots (ii) \end{aligned}\) from Eq. (i), origin \(O\) and \(P\) are on the same side.
\(\begin{array}{lll} \Rightarrow & (-5)\left(\lambda^2+2 \lambda-3\right) > 0 \\ \Rightarrow & \lambda^2+2 \lambda-3 < 0 \\ \Rightarrow & \lambda^2+3 \lambda-\lambda-3 < 0 \\ \Rightarrow & \lambda(\lambda+3)-1(\lambda+3) < 0 \\ \Rightarrow & (\lambda+3)(\lambda-1) < 0] \end{array}\) \(\Rightarrow \quad \lambda \in(-3,1)\) For Eq. (ii), (\(\lambda)\left(\lambda-\frac{1}{3}\right) > 0\) \(\Rightarrow \quad \lambda \in(-\infty, 0) \cup\left(\frac{1}{3}, \infty\right)\) \(\therefore \quad \lambda=\{-2,-1\}\) Hence, there are 2 points.

Asked in: AP EAMCET 2019 (22 Apr Shift 1)

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