UPSC CDS I 2025 Mathematics Paper

Given:
‌(x+y+z)=k
‌(x+z−y)×(x−z+y)∝yz
Formula used:
Algebraic manipulation and properties of proportionality.
From (x+y+z)=k, we can derive:
‌x+z−y=(x+y+z)−2y=k−2y
‌x−z+y=(x+y+z)−2z=k−2z
Calculations:
Given that (x+z−y)×(x−z+y)∝yz
‌⇒(k−2y)×(k−2z)=C×yz‌ (where ‌C‌ is the constant of proportionality) ‌
‌⇒k2−2kz−2ky+4yz=C×yz
‌⇒k2−2k(z+y)+4yz=C×yz
‌⇒k2−2k(y+z)=C×yz−4yz
‌⇒k2−2k(y+z)=(C−4)yz
Now, let's rearrange to find what (y+z−x) is proportional to.
From (x+y+z)=k, we have y+z=k−x.
Substitute (y+z) in the equation k2−2k(y+z)=(C−4)yz :
‌⇒k2−2k(k−x)=(C−4)yz
‌⇒k2−2k2+2kx=(C−4)yz
‌⇒−k2+2kx=(C−4)yz
‌⇒k(2x−k)=(C−4)yz
We are interested in (y+z−x).
We know (y+z−x)=(k−x)−x=k−2x
From k(2x−k)=(C−4)yz, we can write:
‌⇒−k(k−2x)=(C−4)yz
‌⇒k−2x=‌

(C−4)yz

−k

‌⇒k−2x=‌

(4−C)yz

k

Since k,C are constants, ‌

(4−C)

k

is also a constant.
Therefore, ( k−2x ) is proportional to yz .
Since (y+z−x)=k−2x,
∴(y+z−x) is proportional to yz.