GATE CS 2012 Q9 — Maxima and Minima, Mean Value Theorem, Integration

To find the local minima of the function $f(x) = \sin(x)$ on the closed interval $I = \left[\frac{\pi}{4}, \frac{7\pi}{4}\right]$, we must examine both the interior critical points and the boundary points of the interval.

1. Interior Critical Points

We find the critical points by setting the first derivative to zero:

Within the interior interval $\left(\frac{\pi}{4}, \frac{7\pi}{4}\right)$, the solutions are:

Using the second derivative test, $f''(x) = -\sin(x)$:

• At $x = \frac{\pi}{2}$: $f''\left(\frac{\pi}{2}\right) = -1 < 0$, which indicates a local maximum.
• At $x = \frac{3\pi}{2}$: $f''\left(\frac{3\pi}{2}\right) = 1 > 0$, which indicates a local minimum.

2. Boundary Points

For a closed interval $[a, b]$, a boundary point $a$ is a local minimum if the function increases as we move into the interval (i.e., $f'(a) > 0$). Similarly, the boundary point $b$ is a local minimum if the function decreases as we approach it from the left (i.e., $f'(b) < 0$).

• At the left boundary $x = \frac{\pi}{4}$:
  $f'\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} > 0$.
  Since the derivative is positive, the function values increase as $x$ moves to the right of $\frac{\pi}{4}$. Thus, $x = \frac{\pi}{4}$ is a local minimum.
• At the right boundary $x = \frac{7\pi}{4}$:
  $f'\left(\frac{7\pi}{4}\right) = \cos\left(\frac{7\pi}{4}\right) = \frac{1}{\sqrt{2}} > 0$.
  Since the derivative is positive, the function values increase as we approach $\frac{7\pi}{4}$ from the left. Thus, $x = \frac{7\pi}{4}$ is a local maximum.

Conclusion

The function has exactly two local minima in the given interval, located at: