The quadratic equation whose roots are the minimum value of `sin^(2)theta-sin theta+1/2` and `lim_(xto oo)sqrt((x+1)(x+2))-x` is

Options

A. (a)`3x^(2)-7x+3=0`
B. (b)`8x^(2)-14x+3=0`
C. (c)`x^(2)-7x+3=0`
D. (d)`2x^(2)-8x+3=0`

Answer

To find the quadratic equation whose roots are the minimum value of \( \sin^2 \theta - \sin \theta + \frac{1}{2} \) and \( \lim_{x \to \infty} \sqrt{(x+1)(x+2)} - x \), we will follow these steps:

### Step 1: Find the minimum value of \( \sin^2 \theta - \sin \theta + \frac{1}{2} \)

Let \( y = \sin^2 \theta - \sin \theta + \frac{1}{2} \).

1. Rewrite \( y \) in terms of \( a = \sin \theta \):
\[ y = a^2 - a + \frac{1}{2} \]

2. To find the minimum value, differentiate \( y \) with respect to \( a \):
\[ \frac{dy}{da} = 2a - 1 \]

3. Set the derivative equal to zero to find critical points:
\[ 2a - 1 = 0 \implies a = \frac{1}{2} \]

4. Substitute \( a = \frac{1}{2} \) back into the equation for \( y \):
\[ y = \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right) + \frac{1}{2} = \frac{1}{4} - \frac{1}{2} + \frac{1}{2} = \frac{1}{4} \]

Thus, the minimum value \( \alpha = \frac{1}{4} \).

### Step 2: Find the limit \( \lim_{x \to \infty} \sqrt{(x+1)(x+2)} - x \)

1. Rewrite the limit:
\[ \lim_{x \to \infty} \left( \sqrt{(x+1)(x+2)} - x \right) \]

2. Simplify the expression inside the limit:
\[ \sqrt{(x+1)(x+2)} = \sqrt{x^2 + 3x + 2} \]

3. Rationalize the expression:
\[ \lim_{x \to \infty} \frac{(\sqrt{(x+1)(x+2)} - x)(\sqrt{(x+1)(x+2)} + x)}{\sqrt{(x+1)(x+2)} + x} \]
This gives:
\[ = \lim_{x \to \infty} \frac{x^2 + 3x + 2 - x^2}{\sqrt{(x+1)(x+2)} + x} = \lim_{x \to \infty} \frac{3x + 2}{\sqrt{(x+1)(x+2)} + x} \]

4. As \( x \to \infty \), \( \sqrt{(x+1)(x+2)} \approx x \):
\[ = \lim_{x \to \infty} \frac{3x + 2}{2x} = \frac{3}{2} \]

Thus, \( \beta = \frac{3}{2} \).

### Step 3: Form the quadratic equation

The quadratic equation can be formed using the roots \( \alpha \) and \( \beta \):

1. The general form of the quadratic equation is:
\[ x^2 - (\alpha + \beta)x + \alpha \beta = 0 \]

2. Substitute \( \alpha = \frac{1}{4} \) and \( \beta = \frac{3}{2} \):
\[ \alpha + \beta = \frac{1}{4} + \frac{3}{2} = \frac{1}{4} + \frac{6}{4} = \frac{7}{4} \]
\[ \alpha \beta = \frac{1}{4} \cdot \frac{3}{2} = \frac{3}{8} \]

3. The quadratic equation becomes:
\[ x^2 - \frac{7}{4}x + \frac{3}{8} = 0 \]

4. To eliminate the fractions, multiply through by 8:
\[ 8x^2 - 14x + 3 = 0 \]

### Final Answer

The quadratic equation whose roots are the minimum value of \( \sin^2 \theta - \sin \theta + \frac{1}{2} \) and \( \lim_{x \to \infty} \sqrt{(x+1)(x+2)} - x \) is:
\[ 8x^2 - 14x + 3 = 0 \]