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If the point $(1, a)$ lies between the straight lines $x+y=1$ and $2(x+y)=3$ then a lies in interval
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Verified Answer
The correct answer is:
$\left(0, \frac{1}{2}\right)$
$\left(0, \frac{1}{2}\right)$
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 $\left(0, \frac{1}{2}\right)$
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