Mathematics - Application of Derivative Question with Solution | TestHub

$f(x)=2x^3+a{x}^2+bx-3{\cos }^{2}x$

$\therefore f^{\prime }(x)=6x^2+2ax+b+3\sin 2x$

$\therefore f(x)$ is increasing for all $x\Rightarrow 6x^2+2ax+b+3\sin 2x>0$

Also, $6x^2+2ax+b+3\sin 2x\ge 6x^2+2ax+b-3$ as $\sin 2x\ge -1$

Hence $6x^2+2ax+b-3>0$

$\therefore 4a^2-4\cdot 6(b-3)<0\Rightarrow a^2-6b+18<0$