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UPSC CDS 1 2022 Math Paper

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Question : 15 of 100

Marks: +1, -0

If

x^{2} = 17x + y

y^{2} = x + 17y

x \neq y

, then what is the value of

\sqrt{x^{2}+y^{2}+1}

?

17
19
23
27

Solution:

Concept:

Use algebraic manipulation of the given equations to find

x+y

and

x^{2}+y^{2}

.

Explanation:

Given:

x^{2} = 17x + y

and

y^{2} = x + 17y

, with

x \neq y

.

Subtract the second equation from the first:

x^{2} - y^{2} = (17x + y) - (x + 17y) = 16x - 16y

.

Factor left side:

(x - y)(x + y) = 16(x - y)

.

Since

x \neq y

,

x - y \neq 0

, so divide both sides by

x - y

:

x + y = 16

.

Now add the two original equations:

x^{2} + y^{2} = (17x + y) + (x + 17y) = 18x + 18y = 18(x + y)

.

Substitute

x + y = 16

:

x^{2} + y^{2} = 18 \times 16 = 288

.

We need

\sqrt{x^{2} + y^{2} + 1}

.

Plug in the value:

\sqrt{288 + 1} = \sqrt{289} = 17

.

Answer:

17

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