If the point $(1, a)$ lies between the straight lines $x+y=1$ and $2(x+y)=3$ then a lies in interval
- $\left(\frac{3}{2}, \infty\right)$
- $\left(1, \frac{3}{2}\right)$
- $(-\infty, 0)$
- $\left(0, \frac{1}{2}\right)$
Solution
Since, $(1, a)$ lies between $x+y=1$ and $2(x+y)=3$ $\therefore$ Put $x=1$ in $2(x+y)=3$. We get the range of $y$. Thus, $ 2(1+y)=3 \Rightarrow y=\frac{3}{2}-1=\frac{1}{2} $ Thus ' $a$ 'lies in $\left(0, \frac{1}{2}\right)$
Asked in: JEE Main 2012 (12 May Online)