Question

For constant number $a,$ consider the function $f\left(x\right)=ax+\cos 2x+\sin x+\cos x$ on $R$ such that $f\left(u\right)<f\left(v\right)$ for $u<v.$ If the range of $a$ for any real number $x$ is $(\frac{m}{n},\infty )$ where $m,n\in N,$ then the minimum value of $(m+n)$ is

Answer 1

$f\left(x\right)=ax+\cos 2x+\sin x+\cos x$
$\Rightarrow f^{\prime }\left(x\right)=a-2\sin 2x+\cos x-\sin x$
It is given that $f\left(x\right)$ is strictly increasing function.
$\Rightarrow f^{\prime }\left(x\right)>0$ for any real number $x$
$\Rightarrow a>2\sin 2x+\sin x-\cos x\Rightarrow a>-2(\sin x-\cos x{)}^2+2+\sin x-\cos x$

Let $t=\sin x-\cos x=\sqrt{2}\sin (x-\frac{\pi }{4})$
$\Rightarrow -\sqrt{2}\le t\le \sqrt{2}$
So the inequality can be written as
$a>-2t^2+t+2$

Let $g\left(t\right)=-2t^2+t+2=-2{(t-\frac{1}{4})}^2+\frac{17}{8}$
The range of $g\left(t\right)$ for $-\sqrt{2}\le t\le \sqrt{2}$ is
$g(-\sqrt{2})\le g\left(t\right)\le g\left(\frac{1}{4}\right)$
$\Rightarrow -2-\sqrt{2}\le g\left(t\right)\le \frac{17}{8}$
So, the range of $a$ is
$a>max\left\{g\right(t){\}}_{|t|\le \sqrt{2}}$
$\Rightarrow a>\frac{17}{8}$
$\Rightarrow a\in (\frac{17}{8},\infty )$
Hence, least value of $m+n$ is $17+8=25$