Question

If the function \(f(x)=\frac{k \sin x+2\cos x}{\sin x+\cos x}\) is increasing for all values of x, then

• A. k < 1
• B. 1 < k
• C. k > 2
• D. k < 2

Answer 1

The Correct Option is C

To determine the condition on k such that the function \(f(x)=\frac{k \sin x+2\cos x}{\sin x+\cos x}\) is increasing for all values of x, we need to find the derivative \(f'(x)\) and analyze its sign. The function is increasing if \(f'(x) > 0\) for all \(x\).

Let’s compute the derivative using the quotient rule. Recall: if \(u=g(x)\) and \(v=h(x)\), then \(\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}\).

Here, \(u=k \sin x+2\cos x\) and \(v=\sin x+\cos x\). Therefore:

1. \(u'=\frac{d}{dx}(k \sin x+2 \cos x)=k \cos x-2 \sin x\)

2. \(v'=\frac{d}{dx}(\sin x+\cos x)=\cos x-\sin x\)

Apply the quotient rule:

\(f'(x)=\frac{(\sin x+\cos x)(k \cos x-2 \sin x)-(k \sin x+2 \cos x)(\cos x-\sin x)}{(\sin x+\cos x)^2}\)

Simplify the numerator:

\(=(k \cos x-2 \sin x)(\sin x+\cos x)-(k \sin x+2 \cos x)(\cos x-\sin x)\)

Distribute the terms:

\(=k \cos x \sin x+k \cos^2 x-2 \sin^2 x-2 \sin x \cos x-(k \sin x \cos x-k \sin^2 x+2 \cos^2 x-2 \cos x \sin x)\)

Combine like terms:

\(=k(\cos^2 x+\sin^2 x)-2(\sin^2 x+\cos^2 x)\)

Use the Pythagorean identity \(\sin^2 x+\cos^2 x=1\):

\(=k(1)-2(1)=(k-2)\)

For the function to be increasing:

\(f'(x) > 0 \Rightarrow k-2>0 \Rightarrow k > 2\)

Thus, the function \(f(x)\) is increasing for all \(x\) if and only if \(k > 2\).

Conclusion: k > 2