CAT 2020 Exam Shift 2 Solved Paper

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Question : 58 of 76

Marks: +1, -0

In how many ways can a pair of integers

(x, a)

be chosen such that

x^{2}-2 x|+|a-2|=0 ?

4
5
6
7

Solution:

x^{2}-2|x|+|a-2|=0

|x|=\;{2 ± √{4-4(|a-2|)}}/{2}

|x|=1 ± √{1-|a-2|}

\;\text " If "\; a>2 ;|a-2|=a-2

|x|=1 ± √{1-(a-2)}

=1 ± √{3-a}

since

x

is integer

3-a ≥ 0

a ≤ 3

The possible values of a is

=3

Then

x=± 1

If

a=2,|x|=|1 ± 1|, ⇒ x=± 2,0

If

a<2,|a-2|=2-a

|x|=1 ± √{1-(2-a)}

|x|=1 ± √{a-1}

Since

{x}

is integer

a-1 ≥ 0 ⇒ a ≥ 1

∴

The possible values of

a

is 1
If

a=1,|x|=1 ⇒ x=± 1

∴

The possible pairs

=(-1,3),(1,3),(1,1),(-1,1),(2,2),(-2,2),(0,2)

i.e., 7

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