UPSC CDS 1 2022 Math Paper

Given:
x2=17x+y and y2=x+17y,x≠y
Formula Used:
a2−b2=(a−b)(a+b)
Calculation:
We have
x2=17x+y‌⋯−(i)
y2=x+17y‌⋯(‌ ii) ‌
On applying (i) - (ii), we get
x2−y2=16x−16y
⇒(x−y)(x+y)=16(x−y)
⇒(x+y)=16
Now, On applying (i) + (ii), we get
x2+y2=18x+18y
⇒x2+y2=18(x+y)
⇒x2+y2=18×16‌‌[‌ from (i)] ‌
⇒x2+y2=288‌‌
Now, we have to find the value of √x2+y2+1
So, from equation (iv), we get
⇒√288+1=√289=17
∴ The required value of √x2+y2+1 is 17 .