If the function \(f(x) = \displaystyle\frac{k\sin x + 2\cos x}{\sin x + \cos x}\) is increasing for all values of x, then
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(k \gt 2\)
f is increasing when \(f'(x) \gt 0\). By the quotient rule:
\( \displaystyle f'(x) = \frac{(k\cos x - 2\sin x)(\sin x + \cos x) - (k\sin x + 2\cos x)(\cos x - \sin x)}{(\sin x + \cos x)^2} \)
Expanding the two products in the numerator:
\((k\cos x - 2\sin x)(\sin x + \cos x) = k\sin x\cos x + k\cos^2 x - 2\sin^2 x - 2\sin x\cos x\)
\((k\sin x + 2\cos x)(\cos x - \sin x) = k\sin x\cos x - k\sin^2 x + 2\cos^2 x - 2\sin x\cos x\)
Subtracting, the \(\sin x\cos x\) terms cancel:
\(\text{numerator} = k\cos^2 x + k\sin^2 x - 2\sin^2 x - 2\cos^2 x\)
\( = k(\sin^2 x + \cos^2 x) - 2(\sin^2 x + \cos^2 x) = k - 2\)
So
\( \displaystyle f'(x) = \frac{k-2}{(\sin x + \cos x)^2} \)
The denominator is a square, positive wherever f is defined. So \(f'(x) \gt 0\) for all x exactly when
\(k - 2 \gt 0,\quad k \gt 2\)