Trigonometry & Inverse Trigonometry
Extrema of a quadratic in sec⁻¹x
nta_pyq_2025_apr
Grade 12
Question:
Using the principal values of the inverse trigonometric functions, the sum of the maximum and the minimum values of $16\!\left((\sec^{-1}x)^2+(\operatorname{cosec}^{-1}x)^2\right)$ is:
$24\pi^2$
$22\pi^2$
$31\pi^2$
$18\pi^2$
Step-by-Step Solution
Key Concept: Use $\sec^{-1}x+\operatorname{cosec}^{-1}x=\tfrac{\pi}{2}$ to write the expression as a quadratic in $a=\sec^{-1}x\in[0,\pi]\setminus\{\tfrac{\pi}{2}\}$, then find its maximum at the endpoint $a=\pi$ and minimum at the vertex.
Let $a=\sec^{-1}x\in[0,\pi]\setminus\{\tfrac{\pi}{2}\}$; then $\operatorname{cosec}^{-1}x=\tfrac{\pi}{2}-a$.
$f(a)=16\!\left[a^2+\left(\tfrac{\pi}{2}-a\right)^2\right]=16\!\left[2a^2-\pi a+\tfrac{\pi^2}{4}\right]$.
**Maximum** at $a=\pi$: $f(\pi)=16\!\left[2\pi^2-\pi^2+\tfrac{\pi^2}{4}\right]=16\cdot\dfrac{5\pi^2}{4}=20\pi^2$.
**Minimum** at $a=\tfrac{\pi}{4}$: $f\!\left(\tfrac{\pi}{4}\right)=16\!\left[\tfrac{\pi^2}{8}-\tfrac{\pi^2}{4}+\tfrac{\pi^2}{4}\right]=16\cdot\dfrac{\pi^2}{8}=2\pi^2$.
**Sum** $=20\pi^2+2\pi^2=22\pi^2$.
Correct Answer: 2