Question

The domain of the function 
\(f(x) = \sin^{-1}\left(\frac{x^2 - 3x + 2}{x^2 + 2x + 7}\right)\)
is :

• A. \([1,∞)\)
• B. \([−1,2]\)
• C. \([−1,∞)\)
• D. \((−∞,2]\)

Answer 1

The Correct Option is C

To find the domain of the given function \(f(x) = \sin^{-1}\left(\frac{x^2 - 3x + 2}{x^2 + 2x + 7}\right)\), we need to determine when the expression inside the inverse sine function is valid. The domain of \( \sin^{-1}(y) \) is valid for \( y \) values in the range \([-1, 1]\). Therefore, we need to establish the values of \( x \) for which:

-1 \leq \frac{x^2 - 3x + 2}{x^2 + 2x + 7} \leq 1

We will break this down into two inequalities:

1. \(\frac{x^2 - 3x + 2}{x^2 + 2x + 7} \geq -1\)
2. \(\frac{x^2 - 3x + 2}{x^2 + 2x + 7} \leq 1\)

**Step 1:** Solve the inequality \(\frac{x^2 - 3x + 2}{x^2 + 2x + 7} \geq -1\)

Rewriting, we get:

x^2 - 3x + 2 \geq - (x^2 + 2x + 7)

Simplifying gives:

2x^2 - x - 9 \geq 0

This is a quadratic inequality. Solving \(2x^2 - x - 9 = 0\) using the quadratic formula:

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

where a = 2, b = -1, c = -9.

Calculating the discriminant:

(-1)^2 - 4 \times 2 \times (-9) = 1 + 72 = 73

Thus, the roots are:

x = \frac{1 \pm \sqrt{73}}{4}

This gives valid intervals where \(2x^2 - x - 9 \geq 0\), which is outside the roots.

**Step 2:** Solve the inequality \(\frac{x^2 - 3x + 2}{x^2 + 2x + 7} \leq 1\)

Rewriting, we get:

x^2 - 3x + 2 \leq x^2 + 2x + 7

Simplifying gives:

-5x + 2 \leq 7

So, -5x \leq 5 or x \geq -1. This inequality can only hold when \(x\) is less than or equal to 2.

**Combine Both Results:**

Combining both inequalities, the domain of the function \(f(x)\) where \( x \) satisfies both inequalities is \(x \in [-1, \infty)\).

Thus, the correct answer is \([-1,∞)\).

Answer 2

\(f(x) = \sin^{-1}\left(\frac{x^2 - 3x + 2}{x^2 + 2x + 7}\right)\)
\(-1 \leq \frac{x^2 - 3x + 2}{x^2 + 2x + 7} \leq 1\)
\(\frac{x^2 - 3x + 2x}{2 + 2x + 7} \leq 1\)
\(x^2−3x+2≤x^2+2x+7\)
\(5x≥−5\)
\(x≥−1 …(i)\)
\(\frac{x^2 - 3x + 2}{x^2 + 2x + 7} \geq -1\)
\(x^2−3x+2≥−x^2−2x−7\)
\(2x^2−x+9≥0\)
\(x∈R …(ii)\)
\((i)∩(ii)\)
\(Domain ∈ [−1,∞)\)
So, the correct option is (C): \([−1,∞)\)