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UPSC CDS 2 2023 Math Paper

Question Numbers: 37-38

Consider the following for the next items that follow:

Let

p = x4 - y2z2,q = y4 - z2x2,r = z4 - x2y2

Question : 38 of 100

Marks: +1, -0

What is x2 (px2 + qy2 + rz2) + qr - p2 equal to?

Solution:

Concept:

Substitute simple numeric values for

x

y

z

to compute

p

q

r

and then evaluate the given expression.

Explanation:

Let

x = 1

y = 1

, and

z = 2

Given

p = x^{4} - y^{2}z^{2}

, we get

p = 1^{4} - (1^{2} \times 2^{2}) = 1 - 4 = -3

Given

q = y^{4} - z^{2}x^{2}

, we get

q = 1^{4} - (2^{2} \times 1^{2}) = 1 - 4 = -3

Given

r = z^{4} - x^{2}y^{2}

, we get

r = 2^{4} - (1^{2} \times 1^{2}) = 16 - 1 = 15

Now substitute into the expression

x^{2}(p x^{2} + q y^{2} + r z^{2}) + q r - p^{2}

= 1^{2}\big[(-3)(1^{2}) + (-3)(1^{2}) + 15(2^{2})\big] + (-3)(15) - (-3)^{2}

= 1 \times [(-3) + (-3) + 60] - 45 - 9

= ( -3 -3 + 60 ) - 54 = 54 - 54 = 0

Thus the entire expression simplifies to

0

Answer:

0

(Option B).

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