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If $\dfrac{(2y+1)}{(y+2)} < 1,$ then which of the following alternatives gives the CORRECT range of $y$ ?

1. $- 2 < y < 2$
2. $- 2 < y < 1$
3. $- 3 < y < 1$
4. $- 4 < y < 1$

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0
(2y+1)/(y+2)<1

sol.

2y+1<y+2

y<1  upper limit

and you cant put y=-2 because it makes it infinite

so range of y is -2<y<1

option B is the Right answer

In inequality we should not do cross multiplication as we do not know the sign of $y$ (we know that on multiplying with negative number inequality will change).

$\frac{2y+1}{y+2} < 1$

$\implies \frac{2y+1}{y+2} - 1 < 0$

$\implies \frac{y-1}{y+2} < 0$

We can draw the following number line Since, we want negative interval $y \in (-2,1).$

Correct Answer: $B$

by Boss (12.2k points)
edited by
0
Good approach.
0
U hav studied it from where ??
0
any book of JEE preparation(inequality problem)
+1
Can you please explain your inequality approach on the number line a bit more?

$\quad\dfrac{2y+1}{y+2} < 1$

$\quad y < 1$

1. For any value of $y <{- 2},$ Numerator become greater than Denominator.

2. But since in LHS, denominator cannot be zero

i.e., $y > {-2}.$

Therefore, ${-2} < y < 1$
Option B.

by Loyal (9.4k points)
+1
y=-2

also satisfied the condition.

because -infinity<1.
+1 vote

GIVEN  CONDITION =>            2y+1)/(y+2) < 1          then,    y<1

Solution_Approch: check options one by one

A. -2 < y < 2  => y=-1,0, 1 but y=1 it gives 1  (1<1 which is not correct) so option A Is wrong.

B. -2 < y < 1  => y=-1,0  condition satisfied so option B Is CORRECT.

SO you don't need to check option c & d.

by Active (3k points)