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UPSC CDS 1 2022 Math Paper

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Question : 29 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: 👈: Video Solution

Given:

x^{2}=17x+y

and

y^{2}=x+17y, x\neq y

Formula Used:

a^{2}-b^{2}=(a-b)(a+b)

Calculation:

We have

x^{2}=17x+y\;\cdots-(i)

y^{2}=x+17y\;\cdots(\;\text{ ii)}\;)

On applying (i) - (ii), we get

x^{2}-y^{2}=16x-16y

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

\Rightarrow (x+y)=16

Now, On applying (i)

+

(ii), we get

x^{2}+y^{2}=18x+18y

\Rightarrow x^{2}+y^{2}=18(x+y)

\Rightarrow x^{2}+y^{2}=18 \times 16 \;\; [\;\text{ from (i)]}\;

\Rightarrow x^{2}+y^{2}=288 \;\;

Now, we have to find the value of

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

So, from equation (iv), we get

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

\therefore

The required value of

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

is 17 .

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