Mathematics - Application of Derivative Question with Solution | TestHub

$f(x)=\left(a^2=-3a+2\right)\left({\cos }^{2}x/4-{\sin }^{2}x/4\right)+(a-1)x+\sin 1$ $\Rightarrow \mathrm{f}(\mathrm{x})=(\mathrm{a}-1)(\mathrm{a}-2)\cos \mathrm{x}/2+(\mathrm{a}-1)\mathrm{x}+\sin 1$ $\Rightarrow {\mathrm{f}}^{\prime }(\mathrm{x})=-\frac{1}{2}(\mathrm{a}-1)(\mathrm{a}-2)\sin \frac{\mathrm{x}}{2}+(\mathrm{a}-1)$ $\Rightarrow {\mathrm{f}}^{\prime }(\mathrm{x})=(\mathrm{a}-1)\left[1-\frac{(\mathrm{a}-2)}{2}\sin \frac{\mathrm{x}}{2}\right]$ If $f(x)$ does not possess critical points, then $f^{\prime }(x)$ = 0 for any $x\in R$ $\Rightarrow (a-1)\left[1-\frac{(a-2)}{2}\sin \frac{x}{2}\right]$ = 0 for any $x\in R$ $\Rightarrow \mathrm{a}=1$ and $1-\left(\frac{\mathrm{a}-2}{2}\right)\sin \frac{\mathrm{x}}{2}=0$ must not have any solution in $\mathrm{R}$. $\Rightarrow \mathrm{a}=1$ and $\sin \frac{\mathrm{x}}{2}=\frac{2}{\mathrm{a}-2}$ is not solvable in $\mathrm{R}$. $\Rightarrow \mathrm{a}=1$ and $\left|\frac{2}{\mathrm{a}-2}\right|>1[$ For $\mathrm{a}=2,\mathrm{f}(\mathrm{x})=\mathrm{x}+\sin 1\therefore {\mathrm{f}}^{\prime }(\mathrm{x})=1$ $\Rightarrow \mathrm{a}=1$ and $0<\mathrm{a}<4\Rightarrow \mathrm{a}\in (0,1)\cup (1,4)$.