Application of Derivatives: Tricky Questions Solved (5)

Question

Find the maximum value of $f(x) = \sin x + \cos x$ on $[0, 2\pi]$.

Solution — Step by Step

The maximum value is $\sqrt{2}$.

Why This Works

For a smooth function on a closed interval, the maximum is either at a critical point (where $f' = 0$) or at an endpoint. We’ve checked both and confirmed the maximum is $\sqrt{2}$ at $x = \pi/4$.

The second derivative test classifies critical points: $f'' < 0$ means concave down (local max), $f'' > 0$ means concave up (local min). When $f'' = 0$, the test is inconclusive and we go to higher derivatives or the first-derivative sign chart.

Alternative Method

Rewrite $\sin x + \cos x = \sqrt{2}\sin(x + \pi/4)$. Maximum of $\sin$ is $1$, so the maximum of the expression is $\sqrt{2}$, achieved when $x + \pi/4 = \pi/2$, i.e. $x = \pi/4$. Two-line solution.

Common Mistake

Forgetting to check endpoints. CBSE board questions on closed intervals always require checking $f(a)$ and $f(b)$. JEE Main 2024 had a $4$-marker where the endpoint won, surprising students who had only used the second-derivative test.