Using the principal values of the inverse trigonometric functions the sum of the maximum and the minimum values of 16((sec^{-1} x) ^2 + (cosec^{-1} x)^ 2 ) is : (1) 24 2 (2) 18 2 (3) 31 2 (4) 22 2

Key Concepts You Need

1. Principal values of $\sec^{-1}x$ and $\csc^{-1}x$
   • For $|x|\ge 1$,
   • $\sec^{-1}x = \cos^{-1}(1/x)$ lies in $[0,\pi]$ but not $\pi/2$.
   • $\csc^{-1}x = \sin^{-1}(1/x)$ lies in $[-\pi/2,\pi/2]$ but not $0$.

2. Complementary relation
   For any $t$ with $-1\le t\le 1$,
   $\sin^{-1}t + \cos^{-1}t = \frac{\pi}{2}$
   Substituting $t = 1/x$ directly links the two inverse functions in our expression.

3. Optimisation of quadratic in one variable
   Once you express everything in a single variable (say $u$), the expression turns into a simple quadratic. You use ordinary calculus or the vertex-formula to find minima/maxima.

Chain of thought a student should follow

1. Rewrite $\sec^{-1}x$ and $\csc^{-1}x$ in terms of $\cos^{-1}(1/x)$ and $\sin^{-1}(1/x)$ so that they share a common argument $1/x$.
2. Notice the complementary angle property so the sum of the two angles is fixed ($\pi/2$).
   This means their squares’ sum becomes a quadratic function of just one angle.
3. Determine the allowed interval of that single angle from the principal value ranges.
4. Differentiate (or use the vertex) to locate the minimum; check end-points for the maximum.
5. Multiply by 16, then add the two extreme values to get the required answer.
6. Finally match it with the given MCQ options.

What is the problem in super-simple words?

Imagine you have two special buttons:

• Button-1: tells you the angle whose cosine is $1/x$ (this is called $\sec^{-1}x$).
• Button-2: tells you the angle whose sine is $1/x$ (this is called $\csc^{-1}x$).

You square both angles, add them, and then multiply by 16.
We want to find how small and how big this number can become, and then add those two extreme values together.

The big idea

Because the two angles always add up to $90^{\circ}$ (or $\pi/2$ in radians), you only have to play with one of them.
So the game becomes: “If one angle plus the other is always $90^{\circ}$, what is the smallest and biggest possible value of their squares’ sum?”

Step-by-step solution

1. Rewrite using $1/x$
   Let $t = \frac{1}{x}, \quad |t|\le 1 \; (\text{because } |x|\ge 1).$
   Then $\sec^{-1}x = \cos^{-1}t, \qquad \csc^{-1}x = \sin^{-1}t.$

2. Use complementary property

   $\sin^{-1}t + \cos^{-1}t = \frac{\pi}{2}.$

   Put $u = \cos^{-1}t \; \Rightarrow \; \sin^{-1}t = \frac{\pi}{2} - u.$

3. Form the expression in a single variable
   $S = 16\Big[\big(\sec^{-1}x\big)^2 + \big(\csc^{-1}x\big)^2\Big] = 16\left[u^{2} + \left(\frac{\pi}{2} - u\right)^{2}\right].$

   Expand the bracket:

   $u^{2} + \left(\frac{\pi}{2} - u\right)^{2} = u^{2} + \frac{\pi^{2}}{4} - \pi u + u^{2} = 2u^{2} - \pi u + \frac{\pi^{2}}{4}.$

   So

   $S = 16\left(2u^{2} - \pi u + \frac{\pi^{2}}{4}\right).$

4. Domain of $u$ (principal values)
   • $t = 1/x$ sweeps $[-1,1]$.
   • $u = \cos^{-1}t$ therefore sweeps $[0,\pi]$ but $u\neq \pi/2$ (because $t=0$ implies $x\to\infty$).
   Hence $u \in [0,\pi/2) \cup (\pi/2,\pi].$

5. Optimise the quadratic
   Let $f(u)=2u^{2}-\pi u + \frac{\pi^{2}}{4}.$

   Derivative:

   $f'(u) = 4u - \pi.$

   Set $f'(u)=0$:

   $4u-\pi=0 \; \Rightarrow \; u = \frac{\pi}{4}.$
   This lies in $[0,\pi/2)$, so it is a valid critical point.

   • Minimum

   = \frac{\pi^{2}}{8}.$$ • **Maximum** The quadratic opens upward (positive coefficient of $u^{2}$), so maxima must occur at an end-point. Check $u=0$ and $u=\pi$ (the ends of the allowed set): $$f(0)=\frac{\pi^{2}}{4}, \quad f(\pi)=2\pi^{2}-\pi^{2}+\frac{\pi^{2}}{4}=\frac{5\pi^{2}}{4}.$$ Clearly $$f_{\max}=\frac{5\pi^{2}}{4}.$$

6. Multiply by 16 and add
   $\text{Minimum value of }S = 16\times\frac{\pi^{2}}{8}=2\pi^{2}.$ $\text{Maximum value of }S = 16\times\frac{5\pi^{2}}{4}=20\pi^{2}.$ Required sum: $2\pi^{2}+20\pi^{2}=22\pi^{2}.$

7. Pick the correct option
   Option (4) gives $22\pi^{2}$, so that is the answer.