Question

The function $f(x) = \frac{\lambda \sin x + 6 \cos x}{2 \sin x + 3 \cos x}$ is increasing, if

• A. $\lambda \gt 2$
• B. $\lambda \lt 4$
• C. $\lambda \ge 4$
• D. $\lambda \gt 1$

Hint:

For any function of the form $f(x) = \frac{a \sin x + b \cos x}{c \sin x + d \cos x}$, the numerator of its derivative simplifies elegantly to a constant determinant value: $ad - bc$. For the function to be increasing, simply set $ad - bc \ge 0$. Here, $(\lambda)(3) - (6)(2) \ge 0 \implies 3\lambda - 12 \ge 0 \implies \lambda \ge 4$. This bypasses long differentiation entirely!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given a rational trigonometric function $f(x) = \frac{\lambda \sin x + 6 \cos x}{2 \sin x + 3 \cos x}$ and told that it is an increasing function. We need to determine the mathematical condition that the parameter $\lambda$ must satisfy.

Step 2: Key Formula or Approach:
For any differentiable function to be monotonically increasing, its first derivative must be non-negative, meaning $f'(x) \ge 0$. We will apply the quotient rule of differentiation: $$\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$$ Since the denominator $v^2 = (2 \sin x + 3 \cos x)^2$ is always positive where the function is defined, $f'(x) \ge 0$ simplifies to requiring the numerator of the derivative to be greater than or equal to zero.

Step 3: Detailed Explanation:
Let's differentiate $f(x)$ with respect to $x$: $$f'(x) = \frac{(\lambda \cos x - 6 \sin x)(2 \sin x + 3 \cos x) - (\lambda \sin x + 6 \cos x)(2 \cos x - 3 \sin x)}{(2 \sin x + 3 \cos x)^2}$$ Expanding the terms in the numerator: $$\text{Numerator} = (2\lambda \sin x \cos x + 3\lambda \cos^2 x - 12 \sin^2 x - 18 \sin x \cos x) - (2\lambda \sin x \cos x - 3\lambda \sin^2 x + 12 \cos^2 x - 18 \sin x \cos x)$$ Notice that the terms $2\lambda \sin x \cos x$ and $-18 \sin x \cos x$ appear in both groups and completely cancel out: $$\text{Numerator} = 3\lambda \cos^2 x - 12 \sin^2 x + 3\lambda \sin^2 x - 12 \cos^2 x$$ Group the matching coefficients together: $$\text{Numerator} = 3\lambda(\sin^2 x + \cos^2 x) - 12(\sin^2 x + \cos^2 x)$$ Using the fundamental identity $\sin^2 x + \cos^2 x = 1$: $$\text{Numerator} = 3\lambda(1) - 12(1) = 3\lambda - 12$$ For the function to be increasing, we require: $$3\lambda - 12 \ge 0 \implies 3\lambda \ge 12 \implies \lambda \ge 4$$

Step 4: Final Answer:
The function is increasing if $\lambda \ge 4$, which corresponds to option (C).