Match the principal values of ` cos^(-1) ( 8 x^(4) - 8 x^(2) + 1)` given in column I with the corresponding intervals of x given in column II . For which it holds .
`{:(,"Column I",,"Column II"),(A,4 cos^(-1) x,p.,0 le x le 1/sqrt2),(B,4 cos^(-1)x - 2 pi,q.,1/sqrt2 le x le 1),(C,2pi - 4 cos^(-1)x,r.,-1le x le - 1/sqrt2),(D,4pi - 4 cos^(-1) x,s.,-1/sqrt2 le x le 0):}`

Answer

To solve the problem of matching the principal values of \( \cos^{-1}(8x^4 - 8x^2 + 1) \) with the corresponding intervals of \( x \), we will analyze each option step by step.

### Step 1: Analyze the expression \( 8x^4 - 8x^2 + 1 \)

First, we rewrite the expression:
\[ 8x^4 - 8x^2 + 1 = 8(x^4 - x^2) + 1 \]
Let \( y = x^2 \). Then, we have:
\[ 8(y^2 - y) + 1 = 8y^2 - 8y + 1 \]
This is a quadratic in \( y \). The roots of the quadratic \( 8y^2 - 8y + 1 = 0 \) can be found using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 8 \cdot 1}}{2 \cdot 8} = \frac{8 \pm \sqrt{64 - 32}}{16} = \frac{8 \pm \sqrt{32}}{16} = \frac{8 \pm 4\sqrt{2}}{16} = \frac{2 \pm \sqrt{2}}{4} \]
Thus, the roots are:
\[ y_1 = \frac{2 + \sqrt{2}}{4}, \quad y_2 = \frac{2 - \sqrt{2}}{4} \]

### Step 2: Determine the range of \( 8x^4 - 8x^2 + 1 \)

The quadratic opens upwards (since the coefficient of \( y^2 \) is positive), and we can find the vertex to determine the maximum value. The vertex \( y \) is given by:
\[ y = -\frac{b}{2a} = \frac{8}{2 \cdot 8} = \frac{1}{2} \]
Calculating the value at the vertex:
\[ 8\left(\frac{1}{2}\right)^4 - 8\left(\frac{1}{2}\right)^2 + 1 = 8 \cdot \frac{1}{16} - 8 \cdot \frac{1}{4} + 1 = \frac{1}{2} - 2 + 1 = -\frac{1}{2} \]
Thus, the minimum value of \( 8x^4 - 8x^2 + 1 \) is \( -\frac{1}{2} \).

### Step 3: Finding intervals for \( \cos^{-1} \)

Since \( \cos^{-1}(x) \) is defined for \( x \in [-1, 1] \), we need to ensure that:
\[ -\frac{1}{2} \leq 8x^4 - 8x^2 + 1 \leq 1 \]

1. **For \( 8x^4 - 8x^2 + 1 \leq 1 \)**:
\[ 8x^4 - 8x^2 \leq 0 \implies 8x^2(x^2 - 1) \leq 0 \]
This gives \( x^2 \in [0, 1] \) or \( x \in [-1, 1] \).

2. **For \( 8x^4 - 8x^2 + 1 \geq -\frac{1}{2} \)**:
\[ 8x^4 - 8x^2 + \frac{3}{2} \geq 0 \]
Solving this quadratic inequality will give us the intervals for \( x \).

### Step 4: Match the values to the intervals

Now we will analyze each option in Column I:

- **(A)** \( 4 \cos^{-1} x \):
- \( 0 \leq 4 \cos^{-1} x \leq \pi \) leads to \( 0 \leq \cos^{-1} x \leq \frac{\pi}{4} \) which corresponds to \( x \in [\frac{1}{\sqrt{2}}, 1] \) (matches with q).

- **(B)** \( 4 \cos^{-1} x - 2\pi \):
- Rearranging gives \( 0 \leq 4 \cos^{-1} x \leq 3\pi \) which leads to \( x \in [-\frac{1}{\sqrt{2}}, 0] \) (matches with s).

- **(C)** \( 2\pi - 4 \cos^{-1} x \):
- Rearranging gives \( 0 \leq 2\pi - 4 \cos^{-1} x \leq \pi \) leading to \( x \in [0, \frac{1}{\sqrt{2}}] \) (matches with p).

- **(D)** \( 4\pi - 4 \cos^{-1} x \):
- Rearranging gives \( 0 \leq 4\pi - 4 \cos^{-1} x \leq \pi \) leading to \( x \in [-1, -\frac{1}{\sqrt{2}}] \) (matches with r).

### Final Matching

- A → q
- B → s
- C → p
- D → r