Question:

<p><strong>Step 1:</strong> For <span>$f(x)$</span> to be one-one, we need <span>$f'(x) \geq 0$</span> for all <span>$x \in \mathbb{R}$</span>.</p><p><strong>Step 2:</strong> Calculate the derivative: <span>$f'(x) = 3x^2 - 6x - 4 + b\cos x - c\sin x$</span></p><p><strong>Step 3:</strong> For <span>$f'(x) \geq 0$</span> for all <span>$x \in \mathbb{R}$</span>, we need: <span>$3x^2 - 6x - 4 \geq c\sin x - b\cos x$</span> for all <span>$x \in \mathbb{R}$</span></p><p><strong>Step 4:</strong> The minimum value of <span>$3x^2 - 6x - 4$</span> is <span>$-7$</span> (at <span>$x = 1$</span>), and the maximum value of <span>$c\sin x - b\cos x$</span> is <span>$\sqrt{b^2 + c^2}$</span>.</p><p><strong>Step 5:</strong> Therefore, <span>$-7 \geq \sqrt{b^2 + c^2}$</span>, which gives <span>$b^2 + c^2 \leq 1$</span>.</p><p>∴ Answer is (a) <span>$\leq 1$</span>.</p>