If the point (1, a) lies between the straight lines x + y = 1 and 2(x+y) = 3 then a lies in interval

Question

If the point (1, a) lies between the straight lines x + y = 1 and 2(x+y) = 3 then a lies in interval (A) ((3/2),∞) (B) (1,(3/2)) (C) (-∞, 0) (D) (0,(1/2))

Tardigrade Admin · Accepted Answer

Since, (1, a) lies between x + y = 1 and 2(x+y) = 3 ∴ Put x = 1 in 2(x+y) = 3. We get the range of y. Thus, 2(1+y) = 3 ⇒ y = (3/2)-1 = (1/2) Thus âaâlies in (0, (1/2))

Q. If the point $(1, a)$ lies between the straight lines $x + y = 1$ and $2 (x + y) = 3$ then a lies in interval

$(\frac{3}{2}, \infty)$

$(1, \frac{3}{2})$

$(- \infty, 0)$

$(0, \frac{1}{2})$

Since, $(1, a)$ lies between $x + y = 1$
and $2 (x + y) = 3$
$∴$ 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 $(0, \frac{1}{2})$

Q. If the point (1,a) lies between the straight lines x+y=1 and 2(x+y)=3 then a lies in interval

(23​,∞)

(1,23​)

(−∞,0)

(0,21​)