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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. $- 4 < y < 1$

edited | 743 views
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 (12.3k 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?
+1
+2

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

$\quad y < 1$

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 (9.5k 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 (3.1k points)
Ans =>
B) - 2 < y < 1
by Boss (41.9k points)
-2 gives not determined so y cannot be -3 , -4 as they will include -2 in there range, now from a and b .

putting y=1 gives 1. as stated it should be less than 1 so B is the correct answer
by Boss (16.1k points)

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

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

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

$\implies \dfrac{y-1}{y+2}<0\:; y + 2 \neq 0$

$\implies \dfrac{y-1}{y+2}<0\:; y \neq -2$

Here, $\dfrac{\text{Numerator}}{\text{Denominator}}<0 \implies \left\{\begin{matrix} \text{Numerator}> 0\:\: \textbf{and}\:\: \text{Denominator} < 0 \\ \text{Numerator}< 0\:\: \textbf{and}\:\: \text{Denominator} >0 \end{matrix}\right.$

$\textbf{Case(i):}\:\text{Numerator}> 0\:\: \textbf{and}\:\: \text{Denominator} < 0$

$\implies y-1>0$ and $y+2<0$

$\implies y>1$ and $y<-2$

Here, nothing is common.

So, we can't express in the inequality form.

$\textbf{Case(ii):}\:\text{Numerator}< 0\:\: \textbf{and}\:\: \text{Denominator} > 0$

$\implies y-1<0$ and $y+2>0$

$\implies y<1$ and $y>-2$

Here, ${\color{Red}{\text{Red}}}$ color shaded area is common.

$\therefore -2<y<1$

So, the correct answer is $(B).$

by Veteran (59.3k points)