Q86 / 160 · TS EAMCET 2021 (06 Aug Shift 2) · Mathematics
The origin belongs to a region between the lines $x+2 y-5=0$ and $3 x-4 y+5=0$. The number of points in that region of the form $\left((\alpha-1)^2, \alpha\right), \alpha \in Z$ is
- AInfinite in number
- Correct answerB3
- C4
- D2
Solution
Given equations of lines are
From Eq. (i), origin $O$ and $P$ are on the same side.
$\begin{array}{ccc}
& & \left.(-5)(a-1)^2+2 a-5\right)>0 \\
\Rightarrow & & a^2+1-2 a+2 a-5 < 0 \\
\Rightarrow & & a^2 < 4 \\
\Rightarrow & & -2 < a < 2
\end{array}$
From Eq. (ii), we get
$\begin{array}{ll}
& \text { (5) }\left[3(a-1)^2-4 a+5\right]>0 \\
\Rightarrow & 3\left(a^2+1-2 a\right)-4 a+5>0 \\
\Rightarrow \quad & 3 a^2-6 a+3-4 a+5>0 \\
\Rightarrow \quad & 3 a^2-10 a+8>0 \\
\Rightarrow \quad & (a-2)(3 a-4)>0 \\
\Rightarrow \quad & a < \frac{4}{3} \text { or } a>2
\end{array}$
Now, Eqs. (iii) $\cap$ (iv),
$\begin{array}{rlrl}
& & a \in(-2,4 / 3) \\
\therefore & & a & =\{-1,0,1\}
\end{array}$
Hence, there are 3 points.
Hence, the correct answer is option: 'B'.