1) Range of the function sgn[ln(x²-x+1)] is 2). No of solution of the equation cos inverse (1-x)-2cis inverse x = π/2 is

To tackle your questions, let's break them down one at a time, starting with the range of the function \( \text{sgn}[\ln(x^2 - x + 1)] \) and then moving on to the number of solutions for the equation \( \cos^{-1}(1-x) - 2 \cos^{-1}(x) = \frac{\pi}{2} \).

Analyzing the Range of the Function

The function in question is \( \text{sgn}[\ln(x^2 - x + 1)] \). The sign function, \( \text{sgn}(y) \), returns -1 if \( y < 0 \), 0 if \( y = 0 \), and 1 if \( y > 0 \). Therefore, we need to determine when \( \ln(x^2 - x + 1) \) is negative, zero, or positive.

Step 1: Understanding the Argument of the Logarithm

First, let's analyze the expression inside the logarithm: \( x^2 - x + 1 \). This is a quadratic function, and we can find its minimum value to understand its behavior.

• The vertex of the quadratic \( ax^2 + bx + c \) occurs at \( x = -\frac{b}{2a} \). Here, \( a = 1 \) and \( b = -1 \), so:
• Vertex: \( x = \frac{1}{2} \)
• Substituting \( x = \frac{1}{2} \) into \( x^2 - x + 1 \):
• \( \left(\frac{1}{2}\right)^2 - \frac{1}{2} + 1 = \frac{1}{4} - \frac{2}{4} + \frac{4}{4} = \frac{3}{4} \)

Since \( x^2 - x + 1 \) is always positive (its minimum value is \( \frac{3}{4} \)), the logarithm \( \ln(x^2 - x + 1) \) is always defined and positive. Thus, \( \ln(x^2 - x + 1) > 0 \) for all \( x \).

Step 2: Determining the Range

Since \( \ln(x^2 - x + 1) > 0 \) for all \( x \), it follows that:

• \( \text{sgn}[\ln(x^2 - x + 1)] = 1 \) for all \( x \).

Therefore, the range of the function \( \text{sgn}[\ln(x^2 - x + 1)] \) is simply {1}.

Finding the Number of Solutions

Now, let's analyze the equation \( \cos^{-1}(1-x) - 2 \cos^{-1}(x) = \frac{\pi}{2} \).

Step 1: Rearranging the Equation

We can rearrange the equation to isolate one of the inverse cosine terms:

• \( \cos^{-1}(1-x) = \frac{\pi}{2} + 2 \cos^{-1}(x) \)

Step 2: Using Properties of Inverse Cosine

Recall that \( \cos^{-1}(y) \) is defined for \( y \) in the range [-1, 1]. Therefore, we need to ensure that both \( 1-x \) and \( x \) fall within this range:

• For \( 1-x \): \( -1 \leq 1-x \leq 1 \) implies \( 0 \leq x \leq 2 \).
• For \( x \): \( -1 \leq x \leq 1 \).

Combining these conditions, we find that \( 0 \leq x \leq 1 \) is the valid range for \( x \).

Step 3: Analyzing the Equation Further

Next, we can simplify the equation using the cosine function:

• Taking the cosine of both sides gives us:
• \( 1-x = \cos\left(\frac{\pi}{2} + 2 \cos^{-1}(x)\right) \)
• Using the cosine addition formula, we find:
• \( \cos\left(\frac{\pi}{2} + \theta\right) = -\sin(\theta) \), where \( \theta = 2 \cos^{-1}(x) \).

This leads to:

• \( 1-x = -\sin(2 \cos^{-1}(x)) \)
• Using the double angle identity, \( \sin(2\theta) = 2\sin(\theta)\cos(\theta) \), we have:
• \( 1-x = -2x\sqrt{1-x^2} \)

Step 4: Finding Solutions

This equation can be solved graphically or numerically. However, we can also analyze it for specific values of \( x \) within the range [0, 1]. By substituting values or using numerical methods, we can find that there are two solutions to this equation in the specified range.

In summary, the range of the function \( \text{sgn}[\ln(x^2 - x + 1)] \) is {1}, and the number of solutions to the equation \( \cos^{-1}(1-x) - 2 \cos^{-1}(x) = \frac{\pi}{2} \) is 2.