Set up the expression
We need the maximum and minimum values of
16 ( ( sec − 1 x ) 2 + ( cosec − 1 x ) 2 ) . 16\left(\left(\sec^{-1}x\right)^2+\left(\cosec^{-1}x\right)^2\right). 16 ( ( sec − 1 x ) 2 + ( cosec − 1 x ) 2 ) . Let
α = sec − 1 x , β = cosec − 1 x . \alpha = \sec^{-1}x, \qquad \beta = \cosec^{-1}x. α = sec − 1 x , β = cosec − 1 x . Then
sec α = x , csc β = x . \sec \alpha = x, \qquad \csc \beta = x. sec α = x , csc β = x . So,
cos α = 1 x , sin β = 1 x . \cos \alpha = \frac{1}{x}, \qquad \sin \beta = \frac{1}{x}. cos α = x 1 , sin β = x 1 . Hence,
cos α = sin β = cos ( π 2 − β ) . \cos \alpha = \sin \beta = \cos\left(\frac{\pi}{2}-\beta\right). cos α = sin β = cos ( 2 π − β ) . Thus,
α = π 2 − β \alpha = \frac{\pi}{2}-\beta α = 2 π − β using principal values.
Therefore,
α + β = π 2 . \alpha + \beta = \frac{\pi}{2}. α + β = 2 π . So the given expression becomes
16 ( α 2 + β 2 ) , where α + β = π 2 . 16(\alpha^2+\beta^2), \qquad \text{where } \alpha+\beta=\frac{\pi}{2}. 16 ( α 2 + β 2 ) , where α + β = 2 π .
Principal value ranges
Using standard principal values:
sec − 1 x ∈ [ 0 , π ] ∖ { π 2 } \sec^{-1}x \in [0,\pi] \setminus \left\{\frac{\pi}{2}\right\} sec − 1 x ∈ [ 0 , π ] ∖ { 2 π } cosec − 1 x ∈ [ − π 2 , 0 ) ∪ ( 0 , π 2 ] \cosec^{-1}x \in \left[-\frac{\pi}{2},0\right) \cup \left(0,\frac{\pi}{2}\right] cosec − 1 x ∈ [ − 2 π , 0 ) ∪ ( 0 , 2 π ]
For common real x x x ∣ x ∣ ≥ 1 |x|\ge 1 ∣ x ∣ ≥ 1
Now check the two cases.
Case 1: x ≥ 1 x\ge 1 x ≥ 1
Then
α = sec − 1 x ∈ [ 0 , π 2 ) , β = cosec − 1 x ∈ ( 0 , π 2 ] . \alpha = \sec^{-1}x \in [0,\tfrac{\pi}{2}),
\qquad
\beta = \cosec^{-1}x \in (0,\tfrac{\pi}{2}]. α = sec − 1 x ∈ [ 0 , 2 π ) , β = cosec − 1 x ∈ ( 0 , 2 π ] . So
α , β ≥ 0 , α + β = π 2 . \alpha,\beta \ge 0, \qquad \alpha+\beta=\frac{\pi}{2}. α , β ≥ 0 , α + β = 2 π . Case 2: x ≤ − 1 x\le -1 x ≤ − 1
Then
α = sec − 1 x ∈ ( π 2 , π ] , β = cosec − 1 x ∈ [ − π 2 , 0 ) . \alpha = \sec^{-1}x \in (\tfrac{\pi}{2},\pi],
\qquad
\beta = \cosec^{-1}x \in \left[-\tfrac{\pi}{2},0\right). α = sec − 1 x ∈ ( 2 π , π ] , β = cosec − 1 x ∈ [ − 2 π , 0 ) . Still,
α + β = π 2 . \alpha+\beta=\frac{\pi}{2}. α + β = 2 π . So in all cases, the expression reduces to finding extrema of
16 ( α 2 + β 2 ) 16(\alpha^2+\beta^2) 16 ( α 2 + β 2 ) subject to
α + β = π 2 . \alpha+\beta=\frac{\pi}{2}. α + β = 2 π .
Find the minimum
Using
α 2 + β 2 ≥ ( α + β ) 2 2 , \alpha^2+\beta^2 \ge \frac{(\alpha+\beta)^2}{2}, α 2 + β 2 ≥ 2 ( α + β ) 2 , with equality when α = β \alpha=\beta α = β
Thus,
α 2 + β 2 ≥ ( π 2 ) 2 2 = π 2 8 . \alpha^2+\beta^2 \ge \frac{\left(\frac{\pi}{2}\right)^2}{2} = \frac{\pi^2}{8}. α 2 + β 2 ≥ 2 ( 2 π ) 2 = 8 π 2 . Hence,
16 ( α 2 + β 2 ) ≥ 16 ⋅ π 2 8 = 2 π 2 . 16(\alpha^2+\beta^2) \ge 16\cdot \frac{\pi^2}{8} = 2\pi^2. 16 ( α 2 + β 2 ) ≥ 16 ⋅ 8 π 2 = 2 π 2 . Equality occurs when
α = β = π 4 . \alpha=\beta=\frac{\pi}{4}. α = β = 4 π . This is possible since then
x = sec π 4 = 2 = csc π 4 . x=\sec\frac{\pi}{4}=\sqrt{2}=
\csc\frac{\pi}{4}. x = sec 4 π = 2 = csc 4 π . So the minimum value is
2 π 2 . 2\pi^2. 2 π 2 .
Find the maximum
Since
β = π 2 − α , \beta = \frac{\pi}{2}-\alpha, β = 2 π − α , we get
α 2 + β 2 = α 2 + ( π 2 − α ) 2 . \alpha^2+\beta^2 = \alpha^2+\left(\frac{\pi}{2}-\alpha\right)^2. α 2 + β 2 = α 2 + ( 2 π − α ) 2 . This is a convex quadratic, so its maximum over the allowed interval occurs at an endpoint.
For principal values, endpoints correspond to:
x = 1 x=1 x = 1 α = 0 , β = π 2 \alpha=0,\ \beta=\frac{\pi}{2} α = 0 , β = 2 π x = − 1 x=-1 x = − 1 α = π , β = − π 2 \alpha=\pi,\ \beta=-\frac{\pi}{2} α = π , β = − 2 π
Now compute:
At x = 1 x=1 x = 1
α 2 + β 2 = 0 2 + ( π 2 ) 2 = π 2 4 \alpha^2+\beta^2 = 0^2+\left(\frac{\pi}{2}\right)^2 = \frac{\pi^2}{4} α 2 + β 2 = 0 2 + ( 2 π ) 2 = 4 π 2 So value is
16 ⋅ π 2 4 = 4 π 2 . 16\cdot \frac{\pi^2}{4} = 4\pi^2. 16 ⋅ 4 π 2 = 4 π 2 . At x = − 1 x=-1 x = − 1
α 2 + β 2 = π 2 + ( − π 2 ) 2 = π 2 + π 2 4 = 5 π 2 4 \alpha^2+\beta^2 = \pi^2+\left(-\frac{\pi}{2}\right)^2
= \pi^2+\frac{\pi^2}{4} = \frac{5\pi^2}{4} α 2 + β 2 = π 2 + ( − 2 π ) 2 = π 2 + 4 π 2 = 4 5 π 2 So value is
16 ⋅ 5 π 2 4 = 20 π 2 . 16\cdot \frac{5\pi^2}{4} = 20\pi^2. 16 ⋅ 4 5 π 2 = 20 π 2 . Thus the maximum value is
20 π 2 . 20\pi^2. 20 π 2 .
Required sum
Therefore,
maximum + minimum = 20 π 2 + 2 π 2 = 22 π 2 . \text{maximum} + \text{minimum} = 20\pi^2 + 2\pi^2 = 22\pi^2. maximum + minimum = 20 π 2 + 2 π 2 = 22 π 2 . So the correct option is
22 π 2 . \boxed{22\pi^2}. 22 π 2 . That is Option C .
Comparison with stored answer
Stored correct answer: C
Derived answer: C
They agree.