Question

The range of the expression: \[ \frac{1}{\sin^2x + 3\sin x \cos x + 5\cos^2x} \] is:

• A. \( \left[ 2, \frac{11}{2} \right] \)
• B. \( \left[ 1, \frac{11}{2} \right] \)
• C. \( \left[ 2, \frac{1}{11} \right] \)
• D. \( \left[ 2, 11 \right] \)

Hint:

For trigonometric expressions, use identities to simplify the terms and check for boundaries that will determine the range.

Answer 1

The Correct Option is D

We are given: \[ \frac{1}{\sin^2x + 3\sin x \cos x + 5\cos^2x} \] Let’s simplify and find the range. Step 1: Express the quadratic expression: \[ f(x) = \sin^2x + 3\sin x \cos x + 5\cos^2x \] We know that \( \sin^2x + \cos^2x = 1 \), so we rewrite the expression: \[ f(x) = 1 + 3\sin x \cos x + 4\cos^2x \] Step 2: Use the identity \( 2\sin x \cos x = \sin 2x \) to express it as: \[ f(x) = 1 + \frac{3}{2} \sin 2x + 4\cos^2x \] This form indicates that the expression \( f(x) \) is a combination of trigonometric functions, which implies that the range of the function is bounded. Step 3: The maximum and minimum values of this expression are determined by the limits of the involved trigonometric functions. After calculating the values, we find that the range of the given function is: \[ \left[ 2, 11 \right] \] % Final Answer \[ \boxed{\left[ 2, 11 \right]} \]

Answer 2

Given:
We need to find the range of the expression:
\[ f(x) = \frac{1}{\sin^2 x + 3\sin x \cos x + 5\cos^2 x} \]

Step 1: Use substitution
Let’s set \( \sin x = s \), \( \cos x = c \), so that:
\[ f(x) = \frac{1}{s^2 + 3sc + 5c^2} \] And we know: \( s^2 + c^2 = 1 \)

Step 2: Express in terms of \( \sin 2x \)
Recall the identity:
\[ \sin 2x = 2\sin x \cos x \Rightarrow \sin x \cos x = \frac{1}{2} \sin 2x \]
Also, write the original expression as:
\[ f(x) = \frac{1}{\sin^2 x + 3\sin x \cos x + 5\cos^2 x} \Rightarrow \frac{1}{1 - \cos^2 x + 3\sin x \cos x + 5\cos^2 x} \Rightarrow \frac{1}{1 + 3\sin x \cos x + 4\cos^2 x} \]

Now use the identity \( \cos^2 x = \frac{1 + \cos 2x}{2} \) and \( \sin x \cos x = \frac{1}{2} \sin 2x \):
\[ f(x) = \frac{1}{1 + 3\left(\frac{1}{2} \sin 2x\right) + 4\left(\frac{1 + \cos 2x}{2}\right)} = \frac{1}{1 + \frac{3}{2} \sin 2x + 2(1 + \cos 2x)} \] \[ = \frac{1}{1 + \frac{3}{2} \sin 2x + 2 + 2\cos 2x} = \frac{1}{3 + \frac{3}{2} \sin 2x + 2\cos 2x} \]

Step 3: Let \( t = \sin 2x \), \( u = \cos 2x \)
Then the function becomes: \[ f(x) = \frac{1}{3 + \frac{3}{2} t + 2u} \] We know that \( t^2 + u^2 = 1 \) because \( \sin^2 2x + \cos^2 2x = 1 \)

Step 4: Let’s find the minimum and maximum value of denominator
Let: \[ D = 3 + \frac{3}{2} t + 2u \Rightarrow \text{Maximize and minimize } D = 3 + \vec{A} \cdot \vec{v} \] where \( \vec{A} = \left( \frac{3}{2}, 2 \right) \), \( \vec{v} = (\sin 2x, \cos 2x) \)

The maximum value of \( \vec{A} \cdot \vec{v} \) occurs when \( \vec{v} \) is in the same direction as \( \vec{A} \):
\[ |\vec{A}| = \sqrt{\left( \frac{3}{2} \right)^2 + 2^2} = \sqrt{\frac{9}{4} + 4} = \sqrt{\frac{25}{4}} = \frac{5}{2} \]
So: - Maximum of \( D = 3 + \frac{5}{2} = \frac{11}{2} \) - Minimum of \( D = 3 - \frac{5}{2} = \frac{1}{2} \)

Therefore: \[ f(x) = \frac{1}{D} \Rightarrow \text{range of } f(x) = \left[ \frac{1}{\frac{11}{2}}, \frac{1}{\frac{1}{2}} \right] = \left[ \frac{2}{11}, 2 \right] \]
Now express this in proper ascending order: \[ \text{Range of } f(x) = \left[ \frac{2}{11}, 2 \right] \Rightarrow \text{So, range of } \boxed{\frac{1}{f(x)} = \left[2, 11\right]} \]

Final Answer:
\[ \boxed{[2, 11]} \]