Question

The set of all values of a for which the function $f\left(x\right)=(a^2=-3a+2)({\cos }^{2}x/4-{\sin }^{2}x/4)+(a-1)x+\sin 1$ does not process critical points is

• A. $[1,\infty )$
• B. $(0,1)\cup (1,4)$
• C. $(-2,4)$
• D. $(1,3)\cup (3,5)$

Answer 1

The correct option is B $(0,1)\cup (1,4)$

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