DS-Maths

[sanyalpritish]

If x>3?

a)x>0
b)Sq root(x^3-9x+4)>2

[kmittal82]

1 is clearly not enough on its own

2)

Sq both sides

x^3 - 9x + 4 > 4

x(x^2 - 9) > 0

x(x-3)(x+3) > 0

the above holds true for all x>3, and but also for say x = -1, so not sufficient

Combining 1 and 2, we know that x > 0, and for x(x-3)(x+3) > 0 to hold true with that precondition, x must be greater than 3

hence (C)

OA please?

[Rahul@gurome]

If x>3?

a)x>0
b)Sq root(x^3-9x+4)>2

I think the question should be Is x > 3?
In that case,

Statement 1: x > 0 => x may or may not be grater than 3.

Not sufficient.

Statement 2: sqrt(x^3 - 9x + 4) > 2 => (x^3 - 9x + 4) > 4 => (x^3 - 9x) > 0 => x(x + 3)(x - 3) > 0
Implies, either -3 < x < 0 or x > 3

Not sufficient.


1 & 2 Together: x > 3

Both statements together are sufficient.

[sanyalpritish]

When U take option B

we get
1. x>0
2.x>-3
3.x>3

Now as Option A gives x>0 hence x>-3 is eliminated.

But combing still does not tell us that x>3 because Option a and b give us just that x>0

So how come option C is the answer

[Rahul@gurome]

When U take option B

we get
1. x>0
2.x>-3
3.x>3

Now as Option A gives x>0 hence x>-3 is eliminated.

But combing still does not tell us that x>3 because Option a and b give us just that x>0

So how come option C is the answer

When we take statement B, we get either -3 < x < 0 or x > 3. To understand this consider the following (try putting some values like -4, -1, 2, 5 etc),
1. x(x + 3)(x - 3) < 0 for x < -3
2. x(x + 3)(x - 3) > 0 for -3 < x <0
3. x(x + 3)(x - 3) < 0 for 0 < x < 3
4. x(x + 3)(x - 3) > 0 for x > 3