Question

If \left| ax-1 \right| le 1 where a is a positive even integer, which of the following CANNOT be a value of x^2?
A) 0 B) 1/4 C) 1/2 D) 1 E) 4

Answer 1

sorry the equation is \[\left| ax-1 \right| \le 1\]

Answer 2

See solve for:
\[ax - 1 \le 0\]

From here,
find x in terms of a..

Are you able to do this?

Answer 3

sorry there will come 1 in place of 0 in the RHS..

Answer 4

Why 1?

Answer 5

Because it is given in the question...

Answer 6

yea but what do i do after that?

Answer 7

See the original equation is:

\[|ax - 1| \le 1\]

See from here we can form two equations:

We solve for first:

the first equation is:

\[ax - 1 \le 1\]

Solve this to find x..
try by adding 1 to both the sides..

Answer 8

after solving you get x ≤ 2/a and x ≤ 0, then what

Answer 9

No first one is right but second one is incorrect..
Show me how you make the second equation and how you solved it...

Answer 10

ax-1 ≤ -1
ax ≤ 0
x ≤ 0/a or 0 ?

Answer 11

Second equation will become:

\[-(ax - 1) \le 1\]

Now solve for x..

Answer 12

x ≥ 0/a or 0 then?

Answer 13

Remember one thing:

When we multiply both sides of an Inequality equation by (-1),

THEN SIGN OF INEQUALITY REVERSES...

Answer 14

Yes now you are right..

so what will be the square of x from here??

Answer 15

greater than 0?

Answer 16

So what is the value that cannot be possible..
Check from the choices given..

Answer 17

i would say A but the answer choice says E

Answer 18

See, find the square of x got in the first equation??

Answer 19

4/a^2?

Answer 20

Now combine the x^2:

\[0 \le x^2 \le \frac{4}{a^2}\]

Answer 21

Sorry I have ignored the equal sign there..

So from this:

x can take 0 to the values less than 4 because as given "a" is positive even integer:

So 4/a^2 will be definitely less than 4..

Agree??

Answer 22

oooh i get it thank you :D

Answer 23

Welcome dear..