For an integer $K$, if the point $P\left(K^2, K+1\right)$ and the origin $O(0,0)$ lie in the same region between the lines $x+2 y-5=0$ and $3 x-y+1=0$, then the possible number of such points $P$ is

• A

  4
• Correct answer
  B

  2
• C

  6
• D

  Infinitely many

Solution

Given, lines $x+2 y-5=0$ and $3 x-y+1=0$
lie in the same region of lines $x+2 y-5=0$ and $\begin{aligned} & 3 x-y+1=0 \\ & \left(k^2+2(k+1)-5\right) < 0 \text { and } 3 k^2-(2 k+1)+1>0 \\ & \Rightarrow k^2+2 k-3 < 0 \text { and } 3 k^2-k>0 \\ & (k+3)(k-1) < 0 \text { and } k(3 k-1)>0 \\ & k \in(-3,1) \text { and } k \in(-\infty, 0) \cup\left(\frac{1}{3}, \infty\right) \\ & \therefore \quad k \in(-3,0) \cup\left(\frac{1}{3}, 1\right) \end{aligned}$ Integral value of $k\{-2,-1\}$ $\therefore$ Possible number of points is 2 .

Hence, the correct answer is option: 'B'.