IPMAT/Number System•8mos
bheem0001

I did not understand this silly sum

1

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Comments (4)

A
aloo_samosa
SSCBS
•8mos

Same why cant the value of constant be more than 1

2
S
synicdx
IIM Indore (AIR 27)
•8mos
Correct

See the equation:

|x|² − 2|x| + |a − 2| = 0

Think of |x| as just a normal number y (because absolute value is always non-negative). So set y = |x|. Then the equation becomes:

y² − 2y + |a − 2| = 0.

Now rearrange it:

|a − 2| = 2y − y².

Left side is an absolute value, so it can’t be negative. That means 2y − y² must be ≥ 0.

So check small integer values of y.

If y = 0, right side = 0. Then |a − 2| = 0 → a = 2. Also y = |x| = 0 → x = 0. One solution (1 value of a for each of 1 value of x).

If y = 1, right side = 1. Then |a − 2| = 1 → a = 1 or 3. And |x| = 1 → x = 1 or −1. That gives 4 solutions (2 values of a for each of 2 values of x).

If y = 2, right side = 0. Then |a − 2| = 0 → a = 2. And |x| = 2 → x = 2 or −2. That gives 2 solutions (1 value of a for each of 2 values of x).

If y = 3, right side becomes negative, which is impossible. So we stop.

Total solutions = 1 + 4 + 2 = 7.

1
B
bheem0001
OP•8mos

Yes I also have the same doubt why can't the coefficient be more than 1. is there any logic /rule behind it?

1
S
synicdx
IIM Indore (AIR 27)
•8mos

Let y = |x| ≥ 0. Then the equation becomes

|a − 2| = 2y − y².

Now rewrite the right side:

2y − y² = 1 − (y − 1)².

Since (y − 1)² is always ≥ 0, we get

1 − (y − 1)² ≤ 1.

So the maximum possible value of 2y − y² is 1.

That means |a − 2| cannot be greater than 1.

There is no special rule about the constant - it comes directly from the algebra.

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