If 3 < x < 7 and 4 > y > − 2, which of the following must be true?

\(3 < x < 7\)
\(-2 < y < 4\)

i.e.\( (x+y)_{max} < (7+4)\)
i.e. \((x+y)_{max} < (11)\)

i.e. \((x+y)_{min} > (3-2)\)
i.e. \((x+y)_{max} > (1)\)

i.e. \(1 < (x+y) < 11\)

\((2y-x)_{min} > 2*(-2)-7\)
i.e. \((2y-x)_{min} > -11\)

\((2x-y)_{min} > 2*(3)-4\)
i.e. \((2x-y)_{min} > 2\)

Testing the accuracy of given statements

I. \(x   + y > 0\) TRUE

II. \(2y −   x   > 0\) FALSE

III. \(2x − y > 1\) TRUE

Answer: Option D

Although this is not an out and out problem on max-min, we can use the concept of max-min to rule out statements and hence answer options.

In a max-min problem on inequalities, write down the given inequalities one below the other ensuring that the inequality signs are the same in both. Then, we take 4 extreme cases to find out the maximum and the minimum values of the given expression.

Applying this principle on the given inequalities will make it clearer. Let’s start with statement I; applying the max-min principle as outline above, our calculation can look like the one shown in the image.


It’s clear that the smallest value of the expression (x+y) is 1; therefore, x+y>0 is definitely true. Statement I has to be a part of our answer, therefore answer option B can be eliminated.

When we apply the max-min principle on statement II, the calculation looks like this:


It’s clear that the smallest value of 2y-x is -11. Therefore, 2y-x>0 is not always true. The answer options containing statement II viz., C and E can be eliminated.

When we evaluate statement III in a similar manner,


We find that the smallest value of the expression 2x – y is 2. Therefore, 2x – y > 1 is definitely true and hence answer option A can be eliminated. The only answer option left is D, which HAS TO be the correct answer.

The correct answer option is D.

Hope that helps!
Arvind

Please refer to the pic uploaded
IMO ans is D, that I and III only.

Posted from my mobile device

Solution:

Let’s analyze each statement.

I. x + y > 0

Since x > 3 and y > -2,

x + y > 3 + (-2)

x + y > 1

Since x + y > 1, x + y > 0. We see that I is true.

II. 2y - x > 0

If y = 1 and x = 4, we see that 2y - x = -2, which is not greater than 0. So II is not true.

III. 2x - y > 1

Since x > 3 and y < 4 (i.e., -y > -4),

2x - y > 2(3) - 4

2x - y > 2

Since 2x - y > 2, 2x - y > 1. We seet that III is true.

Answer: D

Given that \(3 < x < 7\) and \(4 > y > − 2\) and we need to find out which of the answer choices MUST be true

Let's solve it using two methods

Method 1: Substitution

Since it's a must be a true question so we need to prove the answer choices wrong

I. \(x   + y > 0\)
Now, the value of x is positive and minimum value of y can only be very close to -2. Even if we take that value then also x+y will be positive as minimum value of x is very close to 3
So, this is ALWAYS TRUE

II. \(2y −   x   > 0\)
Now, any negative value of y and a positive value of x will be sufficient to prove this one wrong
y = -1 and x = 4
=> 2y - x = 2*-1 - 4 < 0 => FALSE

III. \(2x − y > 1\)
We need to take minimum value of x and maximum value of y to try to prove this one wrong
=> x = 3.1, y = 3.9
=> 2x - y = 2*3.1 - 3.9 = 6.2 - 3.9 > 1 => ALWAYS TRUE

=> I and III are correct

So, Answer will be D

Method 2: Algebra

\(3 < x < 7\) and \(4 > y > − 2\)

I. \(x   + y > 0\)
Add (1) and (2) we get

3-2 < x + y < 7+4
=> 1 < x-y < 11
Clearly x+y will always be > 0 => TRUE

II. \(2y −   x   > 0\)
Multiply (1) by -1 we get
-7 < -x < 3 ...(4)

Multiple (2) by 2 we get
\(-4 < 2y < 8\) ..(5)

Adding (4) and (5) we get
-7-4 < 2y -x < 3+8
=> -11 < 2y-x < 11

Clearly 2y - x can be < 0 => FALSE

III. \(2x − y > 1\)
Multiply (1) by 2 we get
6 < 2x < 14 ...(6)

Multiple (2) by -1 we get
\(-4 < y < 2\) ..(7)

Adding (6) and (7) we get
6-4 < 2x-y < 14+2
=> 2 < 2x-y < 16
Clearly, 2x-y > 1 => TRUE

=> I and III are correct

So, Answer will be D
Hope it helps!

Watch the following video to learn the Basics of Inequalities