CSEIT – GATE 1997 Subject – Engineering Mathematic/ Topic – Calulus What is the maximum value.

CSEIT – GATE 1997 Subject – Engineering Mathematic/ Topic – Calulus What is the maximum value.

In the GATE 1997 exam for Computer Science and Engineering (CSE), a calculus question was posed as follows:

Question: What is the maximum value of the function f(x)=2×2−2x+6f(x) = 2x^2 – 2x + 6f(x)=2x2−2x+6 in the interval $[0,2]$?

Solution:

1. Evaluate $f(x)$ at the endpoints of the interval:

   • At $x=0$: f(0)=2(0)2−2(0)+6=6f(0) = 2(0)^2 – 2(0) + 6 = 6f(0)=2(0)2−2(0)+6=6

   • At $x=2$: f(2)=2(2)2−2(2)+6=8f(2) = 2(2)^2 – 2(2) + 6 = 8f(2)=2(2)2−2(2)+6=8

2. Find the critical points within the interval by setting the derivative f′(x)f'(x)f′(x) to zero:

   • First, compute the derivative: f′(x)=ddx(2×2−2x+6)=4x−2f'(x) = \frac{d}{dx}(2x^2 – 2x + 6) = 4x – 2f′(x)=dxd​(2x2−2x+6)=4x−2

   • Set the derivative equal to zero to find critical points: $4x-2=0$ x=12x = \frac{1}{2}x=21​

3. Evaluate $f(x)$ at the critical point x=12x = \frac{1}{2}x=21​:

   f(12)=2(12)2−2(12)+6=2(14)−1+6=12−1+6=5.5f\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right)^2 – 2\left(\frac{1}{2}\right) + 6 = 2\left(\frac{1}{4}\right) – 1 + 6 = \frac{1}{2} – 1 + 6 = 5.5f(21​)=2(21​)2−2(21​)+6=2(41​)−1+6=21​−1+6=5.5

4. Determine the maximum value:

   Comparing the function values at the endpoints and the critical point:

   • $f(0)=6$
   • f(12)=5.5f\left(\frac{1}{2}\right) = 5.5f(21​)=5.5
   • $f(2)=8$

   The maximum value of $f(x)$ on the interval $[0,2]$ is 8, occurring at $x=2$.

Answer: The maximum value of the function f(x)=2×2−2x+6f(x) = 2x^2 – 2x + 6f(x)=2x2−2x+6 in the interval $[0,2]$ is 8.

This solution aligns with the analysis provided on GATE Overflow, where the function’s behavior within the given interval was examined to determine the maximum value.

CSEIT – GATE 1997 Subject – Engineering Mathematic/ Topic – Calulus What is the maximum value.

Engineering Mathematics Notes

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