Question

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

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

Answer 1

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

$f\left(x\right)=(k^2-3x+2)\cos \frac{x}{2}+(k-1)x+\sin 1$
$f^{\prime }\left(x\right)=(k-1)(k-2)(-\frac{1}{2}\sin \frac{x}{2})+(k-1)$
$=(k-1)[1-\frac{k-2}{2}\sin \frac{x}{2}]$
Since f(x) does not possess critical points therefore f'(x) is not equal to zero.
i.e.,
$k\ne 1$ or $1-\frac{k-2}{2}\sin \frac{x}{2}=0$ does not posses a solution or $\sin \frac{x}{2}=\frac{2}{k-2}$ does not have a solution.
Hence we must
have $\left|\frac{2}{k-2}\right|>1$ as
$\left|\sin \frac{x}{2}\right|<1.$
Above implies that ${|k-2|}^2\le 4$
or $-2<(k-2)<2$
$\because x^2<a^2\Rightarrow (x^2-a^2)=-ive$ or $-a<x<a$
$\therefore 0<k<4$. Also $k\ne 1.$
$\therefore kϵ(0,1)\cup (1,4)$