The domain of the function
\(f(x) = \sin^{-1}\left(\frac{x^2 - 3x + 2}{x^2 + 2x + 7}\right)\)
is :
The domain of the function
\(f(x) = \sin^{-1}\left(\frac{x^2 - 3x + 2}{x^2 + 2x + 7}\right)\)
is :
\([1,∞)\)
\([−1,2]\)
\([−1,∞)\)
\((−∞,2]\)
The Correct Option is C
Approach Solution - 1
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:
- \(\frac{x^2 - 3x + 2}{x^2 + 2x + 7} \geq -1\)
- \(\frac{x^2 - 3x + 2}{x^2 + 2x + 7} \leq 1\)
**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 \([-1,∞)\).
Approach Solution -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): \([−1,∞)\)
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Concepts Used:
Types of Functions
Types of Functions
One to One Function
A function is said to be one to one function when f: A → B is One to One if for each element of A there is a distinct element of B.
Many to One Function
A function which maps two or more elements of A to the same element of set B is said to be many to one function. Two or more elements of A have the same image in B.
Onto Function
If there exists a function for which every element of set B there is (are) pre-image(s) in set A, it is Onto Function.
One – One and Onto Function
A function, f is One – One and Onto or Bijective if the function f is both One to One and Onto function.
Read More: Types of Functions