If |x^2 + y^2 + z^2|

Question

If  |x^2 + y^2 + z^2| < 4 , is  |x^2 + z^2| < 1  ?

(1)  |x^2 + y^2| < 1

(2)  |y^2 + z^2 | < 1

broall · Answer

$|x^2 + y^2 + z^2| < 4 \iff x^2 + y^2 + z^2 < 4$ $|x^2 + z^2| < 1 \iff x^2+z^2<1$ (1) $x^2+y^2 < 1 \implies x^2 < 1$ If $z^2 = 3$ then $x^2+z^2>1$. If $z^2 = 0$ then $x^2+z^2<1$. Insufficient. (2) $y^2+z^2 < 1 \implies z^2 < 1$ If $x^2 = 3$ then $x^2+z^2>1$. If $x^2 = 0$ then $x^2+z^2<1$. Insufficient. Combine (1) and (2) $x^2+y^2+y^2+z^2<2 \implies x^2+2y^2+z^2<2 \implies x^2+z^2<2$ We could have $x^2+z^2=1$ or $x^2+z^2<1$ or $x^2+z^2>1$. Insufficient. The answer is E.

Mo2men · Answer

If  |x^2 + y^2 + z^2| < 4 , is  |x^2 + z^2| < 1  ?

(1)  |x^2 + y^2| < 1

Let x^2 = y^2 =0 & z^2 =1/2........Answer to question is Yes

Let x^2 = 1/2 & y^2 =0 & z^2 =1/2........Answer to question is No

Insufficient

(2)  |y^2 + z^2 | < 1

Let x^2 = y^2 =0 & z^2 =1/2........Answer to question is Yes

Let x^2 = 1/2 & y^2 =0 & z^2 =1/2........Answer to question is No

Insufficient

Combining 1 & 2

Same examples above for Statementa 1 & 2 prove Insufficient

Answer: E

Dsgshanky · Answer

1. x^2 +y^2 <1, this implies x^2 + y^2 lies in [0,1), and x^2 lies in [0,1)
hence, z^2 < 4 - (x^2 +y^2), implies z^2 lies between [0,4)
also, since x^2 is always positive so x^2+z^2 lies between [0+x^2, 4+x^2) thus min is 0, and max is 5, insufficient.

2. y^2 +z^2 <1, this implies, y^2 +z^2 lies in [0,1) also, z^2 lies in [0,1)
given x^2 < 4- (y^2 +z^2), implies x^2 lies in [0,4)

for the same reason as 1, insufficient.

answer E.

stonecold · Answer

Waoooo, Some really really byzantine solutions above.

Let me try to put this in a laconic way => in such questions -> ALWAYS (and I mean ALWAYS) try and use test cases or else the question can be pretty painful.

Statement 1 =>
Test case 1 => 0,0,0 => |x^2+z^2|<1 is true
Test case 2 => 0.9,0,0.9 => |x^2+z^2|<1 is false

Statment 2 =>
Test case 1 => 0,0,0 => |x^2+z^2|<1 is true
Test case 2 => 0.9,0,0.9 => |x^2+z^2|<1 is false

Combining them =>

Test case 1 => 0,0,0 => |x^2+z^2|<1 is true
Test case 2 => 0.9,0,0.9 => |x^2+z^2|<1 is false

Hence E.

Notice how the test case is the same which makes the job 10x times easier

Madhavi1990 · Answer

St 1: |x sq + y sq| <1
|+ve| < 1 --> either it adds to 0; or tend to 0.9 but less than 1

--> but we don't know exact value of x or y

St 2: |y sq + z sq| < 1
|+ve| < 1 --> either it adds to 0; or tend to 0.9 but less than 1
--> but we don't know exact value of x or y

1) + 2) if we take max value |0.49+0.64+0.81|< 4
--> we still don't know the exact values of x,y,z square

so Ans E

bumpbot · Answer

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If |x^2 + y^2 + z^2| < 4, is |x^2 + z^2| < 1 ?

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