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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
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Verified Answer
The correct answer is:
2
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.
\(\begin{aligned}
& x+2 y-5=0 \quad \ldots (i) \\
& \text{and } 3 x-y+1=0 \quad \ldots (ii)
\end{aligned}\)
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.
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