Question

Let $f:R\to R$ defined as $f(x)=\frac{x}{2}+\frac{\sin 2x}{4}-(2-c)\sin x-2cx$ is strictly increasing on $R$, then the largest integral value of $c$ is equal to

• A. 0
• B. 1
• C. -1
• D. -2

Answer 1

Answer: (C)

$\Theta f(x)=\frac{x}{2}+\frac{\sin 2x}{4}-(2-c)\sin x-2cx$
$f^{\prime }(x)=\frac{1}{2}+\frac{\cos 2x}{2}-(2-c)\cos x-2c$
$f^{\prime }(x)={\cos }^{2}x+(c-2)\cos x-2c$
$\Rightarrow f^{\prime }(x)=(\cos x+c)(\cos x-2)\ge 0$
$\Rightarrow \cos x+c\le 0$
$\Rightarrow c+1\le 0\Rightarrow c\le -1$