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UPSC CDS 1 2023 Math Paper

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Question : 45 of 100

Marks: +1, -0

If

x=a+b+\frac{(a-b)^{2}}{4a+4b}

y=\frac{a+b}{4}+\frac{ab}{a+b}

then what is the value of

(x - a)^{2} - (y - b)^{2}

?

$a^{2}$
$b^{2}$
ab
$a^{2}b^{2}$

Solution:

Concept:

Substituting simple numeric values for

a

and

b

simplifies the expressions and allows quick evaluation.

Explanation:

Let

a = 0

and

b = 1

.

Calculate

x

:

x = a + b + \frac{(a-b)^2}{4a+4b}

.

Substitute values:

x = 0 + 1 + \frac{(0-1)^2}{0+4\cdot1} = 1 + \frac{1}{4} = \frac{5}{4}

.

Calculate

y

:

y = \frac{a+b}{4} + \frac{ab}{a+b}

.

Substitute:

y = \frac{0+1}{4} + \frac{0}{1} = \frac{1}{4}

.

Now evaluate

(x - a)^2 - (y - b)^2

:

(x - a)^2 = \left(\frac{5}{4} - 0\right)^2 = \frac{25}{16}

.

(y - b)^2 = \left(\frac{1}{4} - 1\right)^2 = \left(-\frac{3}{4}\right)^2 = \frac{9}{16}

.

Difference:

\frac{25}{16} - \frac{9}{16} = \frac{16}{16} = 1

.

Observe that

b^2 = 1^2 = 1

matches the result.

Answer:

b^2

(Option B).

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