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UPSC CDS 2 2023 Math Paper

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Question : 74 of 100

Marks: +1, -0

If

p=\frac{a^{2}}{\left(b-a\right)\left(c-a\right)},q=\frac{b^{2}}{\left(c-b\right)\left(a-b\right)},r=\frac{c^{2}}{\left(a-c\right)\left(b-c\right)}

(p+q+r)^{2}

equal to?

9
4
1
0

Solution:

Concept:

Use simple numeric values to evaluate the expression and find the required value.

Explanation:

Assume convenient values for

a

,

b

, and

c

that satisfy the conditions (all distinct).

Let

a=1

,

b=2

,

c=3

.

Compute

p

:

p = \frac{a^{2}}{\left(b-a\right)\left(c-a\right)} = \frac{1^{2}}{\left(2-1\right)\left(3-1\right)} = \frac{1}{2}

.

Compute

q

:

q = \frac{b^{2}}{\left(c-b\right)\left(a-b\right)} = \frac{2^{2}}{\left(3-2\right)\left(1-2\right)} = \frac{4}{\left(1\right)\left(-1\right)} = -4

.

Compute

r

:

r = \frac{c^{2}}{\left(a-c\right)\left(b-c\right)} = \frac{3^{2}}{\left(1-3\right)\left(2-3\right)} = \frac{9}{\left(-2\right)\left(-1\right)} = \frac{9}{2}

.

Now find

p+q+r

:

p+q+r = \frac{1}{2} + (-4) + \frac{9}{2} = \frac{1+9}{2} - 4 = \frac{10}{2} - 4 = 5 - 4 = 1

.

Then

(p+q+r)^{2} = (1)^{2} = 1

.

Answer:

1

(Option C).

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