Question

The range of the real valued function \( f(x) = \cos^{-1} \left( \dfrac{3}{\sqrt{9x^2 - 12x + 22}} \right) \) is

• A. \( \left(0, \dfrac{\pi}{4} \right] \)
• B. \( \left[ \dfrac{\pi}{4}, \dfrac{\pi}{2} \right) \)
• C. \( [0, \pi] \)
• D. \( \left[ \dfrac{\pi}{6}, \dfrac{\pi}{2} \right) \)

Hint:

To find the range of a function involving inverse trigonometric expressions, first determine the domain of the inner expression and its bounds. Then use the monotonicity of the inverse trig function to deduce the final range.

Answer 1

The Correct Option is B

We are given: \[ f(x) = \cos^{-1} \left( \frac{3}{\sqrt{9x^2 - 12x + 22}} \right) \] Let us first simplify the expression under the square root. \[ 9x^2 - 12x + 22 = 9(x^2 - \frac{4}{3}x) + 22 \] Complete the square inside the bracket: \[ x^2 - \frac{4}{3}x = \left( x - \frac{2}{3} \right)^2 - \frac{4}{9} \] \[ 9(x^2 - \frac{4}{3}x) = 9\left[ \left( x - \frac{2}{3} \right)^2 - \frac{4}{9} \right] = 9 \left( x - \frac{2}{3} \right)^2 - 4 \] Now add 22: \[ 9x^2 - 12x + 22 = 9 \left( x - \frac{2}{3} \right)^2 + 18 \] Thus, \[ \sqrt{9x^2 - 12x + 22} = \sqrt{9 \left( x - \frac{2}{3} \right)^2 + 18} \] The minimum value of \( \left( x - \frac{2}{3} \right)^2 \) is 0, so the minimum value of the expression is: \[ \sqrt{18} = 3\sqrt{2} \] So the maximum value of the function inside the inverse cosine is: \[ \frac{3}{\sqrt{9x^2 - 12x + 22}} \leq \frac{3}{3\sqrt{2}} = \frac{1}{\sqrt{2}} \] And the minimum value approaches 0 (as \( \sqrt{9x^2 - 12x + 22} \to \infty \)), so the expression inside the inverse cosine lies in: \[ \left( 0, \frac{1}{\sqrt{2}} \right] \] Now consider the range of \( \cos^{-1}(x) \). If \( x \in (0, \frac{1}{\sqrt{2}}] \), then: \[ \cos^{-1}(x) \in \left[ \cos^{-1} \left( \frac{1}{\sqrt{2}} \right), \cos^{-1}(0) \right) = \left[ \frac{\pi}{4}, \frac{\pi}{2} \right) \]