For any two real numbers a and b, {(a-b)²}+{(b-a)²} is

Correct Answer is : positive only if a ≠ b

Correct Answer is : positive only if a ≠ b

Test Index

UPSC CDS I 2019 Math Paper

Question : 26 of 100

Marks: +1, -0

\sqrt{(a-b)^{2}}+\sqrt{(b-a)^{2}}

Solution:

Concept:

The square root of a square yields the absolute value.

Explanation:

For any real

x

\sqrt{x^{2}} = |x|

\sqrt{(a-b)^{2}} = |a-b|

Also

\sqrt{(b-a)^{2}} = |b-a| = |a-b|

(absolute value is symmetric).

Hence the sum is

|a-b| + |a-b| = 2|a-b|

|a-b| = 0

only when

a = b

|a-b| > 0

when

a \neq b

Thus the expression is zero if

a = b

, and positive if

a \neq b

Therefore it is positive only when

a \neq b

Answer:

Option C: positive only if a ≠ b.

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