Question:
The domain of \( f(x) = \cos^{-1}[\log_2(x^2+5x+8)] \) is:
The domain of \( f(x) = \cos^{-1}[\log_2(x^2+5x+8)] \) is:
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When solving domain problems for inverse trigonometric functions, always convert the interior logarithmic or algebraic expressions into strict boundaries before solving the quadratic inequalities.
Updated On: Jun 9, 2026
- \( [-4, -3] \)
- \( [-3, -2] \)
- \( [-2, -1] \)
- \( [-1, 2] \)
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The Correct Option is B
Solution and Explanation
Concept:
The function \(\cos^{-1}(u)\) is defined only when \( -1 \le u \le 1 \). Here, \( u = \log_2(x^2+5x+8) \).
Step 1: Set the inequality constraint.
\[ -1 \le \log_2(x^2+5x+8) \le 1 \]
Step 2: Transform into exponential form.
Since the base 2 is greater than 1, the inequality direction is preserved: \[ 2^{-1} \le x^2+5x+8 \le 2^1 \Rightarrow 0.5 \le x^2+5x+8 \le 2 \]
Step 3: Solve the inequalities.
1. \( x^2+5x+8 \le 2 \Rightarrow x^2+5x+6 \le 0 \Rightarrow (x+2)(x+3) \le 0 \). This yields \( x \in [-3, -2] \). 2. \( x^2+5x+8 \ge 0.5 \Rightarrow x^2+5x+7.5 \ge 0 \). The discriminant \( D = 25 - 4(7.5) = -5 \). Since \( D < 0 \) and the leading coefficient is positive, this part is always true for all real \( x \). The intersection is \([-3, -2]\). [-3, -2]
Step 1: Set the inequality constraint.
\[ -1 \le \log_2(x^2+5x+8) \le 1 \]
Step 2: Transform into exponential form.
Since the base 2 is greater than 1, the inequality direction is preserved: \[ 2^{-1} \le x^2+5x+8 \le 2^1 \Rightarrow 0.5 \le x^2+5x+8 \le 2 \]
Step 3: Solve the inequalities.
1. \( x^2+5x+8 \le 2 \Rightarrow x^2+5x+6 \le 0 \Rightarrow (x+2)(x+3) \le 0 \). This yields \( x \in [-3, -2] \). 2. \( x^2+5x+8 \ge 0.5 \Rightarrow x^2+5x+7.5 \ge 0 \). The discriminant \( D = 25 - 4(7.5) = -5 \). Since \( D < 0 \) and the leading coefficient is positive, this part is always true for all real \( x \). The intersection is \([-3, -2]\). [-3, -2]
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