GATE CS 2015 Set 2 QGA-3 — Maxima and Minima, Mean Value Theorem, Integration

We are given the function $f(x) = 1 - |x|$ defined on the interval $[-1, 1]$.

To find the maximum value of $f(x)$ and the point $x$ at which it occurs, we analyze the absolute value term $|x|$:

• For any real number $x$, the absolute value satisfies $|x| \ge 0$.
• Multiplying by $-1$ reverses the inequality: $-|x| \le 0$.
• Adding $1$ to both sides gives: $$f(x) = 1 - |x| \le 1$$

Thus, the maximum possible value of $f(x)$ is $1$.

This maximum value is achieved when the subtracted term $|x|$ is minimized, which occurs when: $$|x| = 0 \implies x = 0$$ Since $x = 0$ lies within the given interval $[-1, 1]$, the function attains its maximum value of $1$ at $x = 0$.

Therefore, the point of maximum is $0$ and the maximum value is $1$, which corresponds to option C.