Explanation

(d)
$\because\ \ f(x) = \frac{\lambda\sin{x + 6\cos x}}{2\sin{x + 3\cos x}}$
On differentiating w.r.t. $x$, we get
$f^{'}(x) = \frac{\left\lbrack \begin{array}{r}
\left( 2\sin{x + 3\cos x} \right)\left( \lambda\cos{x - 6\sin x} \right) \\
- \left( \lambda\sin{x + 6\cos x} \right)(2\cos{x - 3\sin x})
\end{array} \right\rbrack}{\left( 2\sin{x + 3\cos x} \right)^{2}}$
The function is monotonic increasing, if $f^{'}(x) > 0$
$\Rightarrow 3\lambda\ \left( \sin^{2}x + \cos^{2}x \right) - 12\left( \sin^{2}x + \cos^{2}x \right) > 0$
$\Rightarrow$3$\lambda - 12 > 0\ \ \ \ \ (\because\sin^{2}x + \cos^{2}x = 1)$
$\Rightarrow \ \ \ \lambda > 4$