Question

Find the value of \( k \) if the function \( f(x) = \dfrac{k\sin x + 2\cos x}{\sin x + \cos x} \) is increasing for all \( x \).

• A. \( k<2 \)
• B. \( k = 2 \)
• C. \( k \geq 2 \)
• D. \( k \leq 2 \)

Hint:

For functions of the form \( \frac{a\sin x + b\cos x}{\sin x + \cos x} \), the derivative often simplifies using the identity \( \sin^2 x + \cos^2 x = 1 \). Always look for cancellations after expansion.

Answer 1

The Correct Option is C

Concept: A function is increasing if its derivative is non-negative for all \(x\) in its domain: \[ f'(x) \geq 0 \] For a function of the form \( \dfrac{N}{D} \), derivative is: \[ f'(x) = \frac{N'D - ND'}{D^2} \] Since \(D^2>0\), the sign depends only on the numerator.
Step 1: {Differentiate numerator and denominator.}
\[ N = k\sin x + 2\cos x, \quad D = \sin x + \cos x \] \[ N' = k\cos x - 2\sin x, \quad D' = \cos x - \sin x \]
Step 2: {Apply quotient rule.}
\[ f'(x) = \frac{(k\cos x - 2\sin x)(\sin x + \cos x) - (k\sin x + 2\cos x)(\cos x - \sin x)}{(\sin x + \cos x)^2} \]
Step 3: {Simplify numerator.}
Expanding and simplifying: \[ = k(\cos^2 x + \sin^2 x) - 2(\sin^2 x + \cos^2 x) \] \[ = k - 2 \]
Step 4: {Apply increasing condition.}
\[ f'(x) = \frac{k - 2}{(\sin x + \cos x)^2} \geq 0 \] Since denominator is always positive, \[ k - 2 \geq 0 \Rightarrow k \geq 2 \]