Question

The function \[ y=\frac{\sin x+2\cos x}{3\sin x+4\cos x} \] is

• A. decreases for all \(x\in\mathbb R\)
• B. increases for all \(x\in\mathbb R\)
• C. decreases only for \(x\gt 0\)
• D. increases only for \(x\gt 0\)

Hint:

For functions of the form \[ \frac{a\sin x+b\cos x} {c\sin x+d\cos x}, \] the derivative often simplifies to a constant multiple of \[ \frac1{(c\sin x+d\cos x)^2}. \]

Answer 1

The Correct Option is A

Concept: For a rational function \[ y=\frac{u}{v}, \] \[ y'=\frac{u'v-uv'}{v^2} \] The sign of \(y'\) determines whether the function is increasing or decreasing.

Step 1: Differentiate the function. Let \[ u=\sin x+2\cos x \] and \[ v=3\sin x+4\cos x \] Then \[ u'=\cos x-2\sin x \] and \[ v'=3\cos x-4\sin x \] Therefore, \[ y' = \frac{(\cos x-2\sin x)(3\sin x+4\cos x) -(\sin x+2\cos x)(3\cos x-4\sin x)} {(3\sin x+4\cos x)^2} \]

Step 2: Simplify the numerator. \[\begin{aligned} N &=(3\sin x\cos x+4\cos^2x-6\sin^2x-8\sin x\cos x)\\ &\quad-(3\sin x\cos x-4\sin^2x+6\cos^2x-8\sin x\cos x) \end{aligned}\] \[\begin{aligned} N &=4\cos^2x-6\sin^2x+4\sin^2x-6\cos^2x\\ &=-2(\sin^2x+\cos^2x) \end{aligned}\] \[\begin{aligned} N=-2 \end{aligned}\]

Step 3: Determine the sign of \(y'\). \[ (3\sin x+4\cos x)^2\gt 0 \] whenever the function is defined. Hence, \[ y' = \frac{-2}{(3\sin x+4\cos x)^2} \lt 0 \] for all \(x\) in its domain. Therefore the function is decreasing. \[\begin{aligned} \boxed{\text{Decreases for all }x} \end{aligned}\] Hence, option \(\mathbf{(A)}\) is correct.