GATE2011 MN: GA-61

Question

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

• A. $- 2 < y < 2$
• B. $- 2 < y < 1$
• C. $- 3 < y < 1$
• D. $- 4 < y < 1$

Comments

(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

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Option A) is wrong here, for $y=1,1<1=f$

options C, D)  for $y=-2$ condition is false.

option B) is correct.

Answer 1

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: $B$

Comments

Can you please explain your inequality approach on the number line a bit more?

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Inequality related terms

https://www.mathsisfun.com/algebra/inequality-solving.html

These might be helpful

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https://www.youtube.com/watch?v=HMo47a52yt4

Answer 2

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

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

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

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

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

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 $(B).$                                                                                         

Answer 3

$\quad\dfrac{2y+1}{y+2} < 1$

  $\quad y < 1$

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.

Comments

y=-2

also satisfied the condition.

because -infinity<1.

Answer 4

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.