GATE CS 1997 Q4.1 — Maxima and Minima, Mean Value Theorem, Integration

To find the absolute maximum value of the continuous function $f(x) = 2x^2 - 2x + 6$ on the closed interval $[0, 2]$, we evaluate the function at its critical points within the interval and at the boundaries of the interval.

Step 1: Find the critical points
First, we find the derivative of $f(x)$ and equate it to zero: $$f'(x) = \frac{d}{dx}(2x^2 - 2x + 6) = 4x - 2$$ Setting $f'(x) = 0$: $$4x - 2 = 0 \implies x = 0.5$$ Since $x = 0.5$ lies within the interval $(0, 2)$, it is a valid critical point.

Step 2: Evaluate $f(x)$ at the critical point and the endpoints

• At the critical point $x = 0.5$: $$f(0.5) = 2(0.5)^2 - 2(0.5) + 6 = 0.5 - 1 + 6 = 5.5$$
• At the left endpoint $x = 0$: $$f(0) = 2(0)^2 - 2(0) + 6 = 6$$
• At the right endpoint $x = 2$: $$f(2) = 2(2)^2 - 2(2) + 6 = 8 - 4 + 6 = 10$$

Step 3: Compare the values
Comparing the evaluated values: $$\text{Minimum value} = 5.5 \quad (\text{at } x = 0.5)$$ $$\text{Maximum value} = 10 \quad (\text{at } x = 2)$$

Thus, the maximum value of the function in the given interval is 10, which corresponds to option B.