UPSC CDS I 2025 Mathematics Paper

A real number M is squared to give the value N. Find the minimum value of (M+N).
Formula used:
N=M2
Minimize(M+N)=M+M2
Calculation:
Let f(M)=M+M2
To find the minimum value, take the derivative and set it to zero:
‌f′(M)=1+2M
‌⇒1+2M=0
‌⇒M=−0.5
Substitute M=−0.5 into f(M) :
‌f(M)=M+M2
‌⇒f(−0.5)=(−0.5)+(−0.5)2
‌⇒f(−0.5)=−0.5+0.25
‌⇒f(−0.5)=−0.25
∴ The minimum value of (M+N) is −0.25, and the correct answer is option 1.