Question 16

The given function is defined as a $3 \times 3$ determinant:

Observe that:

• At $\theta = \pi/6$, the first row is identical to the second row. Hence $f(\pi/6) = 0$.
• At $\theta = \pi/3$, the first row is identical to the third row. Hence $f(\pi/3) = 0$.

Since trigonometric functions $\sin\theta$, $\cos\theta$, and $\tan\theta$ are continuous and differentiable on $[\pi/6, \pi/3]$, the function $f(\theta)$ is continuous on $[\pi/6, \pi/3]$ and differentiable on $(\pi/6, \pi/3)$.

By Rolle's Theorem, since $f(\pi/6) = f(\pi/3) = 0$, there exists at least one $\theta \in (\pi/6, \pi/3)$ such that $f'(\theta) = 0$. Thus Statement (I) is TRUE.

Now expand the determinant along the first row:

where $c_1, c_2, c_3$ are the cofactors. Since $\tan\theta$ is not a linear combination of $\sin\theta$ and $\cos\theta$ and the cofactors are non-zero, $f(\theta)$ is not identically constant on $[\pi/6, \pi/3]$.

Since $f(\theta)$ is non-constant, its derivative $f'(\theta)$ cannot be zero everywhere on $(\pi/6, \pi/3)$. Therefore, there exists $\theta \in (\pi/6, \pi/3)$ such that $f'(\theta) \neq 0$. Thus Statement (II) is TRUE.

Both I and II are TRUE.