Question

For constant number ' $a$ ', consider the function $f(x)=a x+\cos 2 x+\sin x+\cos x$ on $R$ (the set of real numbers) such that $f(u)

Answer 1

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