UPSC NDA I 2024 Mathematics Paper

Concept:If $f(x)$ has a local maximum at $x=0$, then $f'(0)=0$ and $f''(0)<0$. The condition for $y = f(x)+ax-b$ to have a relative maximum at $x=0$ is $y'(0)=0$ and $y''(0)<0$.Explanation:We have $y = f(x) + ax - b$.Differentiate once: $y'(x) = f'(x) + a$.At $x=0$: $y'(0) = f'(0) + a$.Since $f$ has a local maximum at $x=0$, $f'(0)=0$. Thus $y'(0) = a$.For $y$ to have a relative maximum at $x=0$, we need $y'(0)=0$, so $a=0$.Differentiate again: $y''(x) = f''(x)$.At $x=0$: $y''(0) = f''(0)$.Since $f$ has a local maximum, $f''(0)<0$. Hence $y''(0)<0$ automatically.The parameter $b$ does not appear in the derivatives, so it can be any real number.Therefore, the condition is $a=0$ and $b$ can be any value.Answer:For all $b$ and $a=0$.