Question

For constant number ' $a$ ', consider the function $f(x)=ax+\cos 2x+\sin x+\cos x$ on $R$ (the set of real numbers) such that $f(u)<f(v)$ for $u<v$. If the range of ' $a$ ' for any real numbers $u,v$ is $\left[\frac{m}{n},\infty \right)$ where $m\&n$ are co-prime, then find the value of $(m-n)$.

Answer 1

Answer: 9

We have $f(x)=ax+\cos 2x+\sin x+\cos x\Rightarrow f^{\prime }(x)=a-2\sin 2x+\cos x-\sin x$ As $f^{\prime }(x)\ge 0$ for any real number $x\Rightarrow a\ge 2\sin 2x+\sin x-\cos x\dots \dots [\ast ]$
Let $t=\sin x-\cos x=\sqrt{2}\sin \left(t-\frac{\pi }{4}\right)\Rightarrow -\sqrt{2}\le t\le \sqrt{2}$,
[13th, 30-01-2011, P-2] so the inequality can be written as $a\ge -2t^2+t+2$
Let $g(t)=-2t^2+t+2=-2{\left(t-\frac{1}{4}\right)}^2+\frac{17}{8}$
then range of $g(t)$ for $-\sqrt{2}\le t\le \sqrt{2}$ is $g(-\sqrt{2})\le g(t)\le g\left(\frac{1}{4}\right)\Rightarrow -2-\sqrt{2}\le g(t)\le \frac{17}{8}$
So, the range of a can be found $a\ge {\max }_{|t|\le \sqrt{2}}\Rightarrow a\ge \frac{17}{8}\Rightarrow a\in \left[\frac{17}{8},\infty \right)$
Hence, $(m+n{)}_{\text{least }}=17+8=25$.