Question

If $f\left(x\right)=\sin 2x-8(a+1)\sin x+(4a^2+8a-14)x$ increases for all $x\in R$ and has no critical points for all $x\in R$, then which of the following values of a are possible?

• A. $(-\infty ,-\sqrt{5},-2)$
• B. $(1,\infty )$
• C. $(\sqrt{7},\infty )$
• D. None of these

Answer 1

The correct option is C $(\sqrt{7},\infty )$

$f\left(x\right)=\sin 2x-8(a+1)\sin x+(4a^2-8a-14)x$

$\Rightarrow f^{\prime }\left(x\right)=2\cos 2x-8(a+1)\cos x+(4a^2+8a-14)$

$-1\le \cos 2x.\cos x\le 1$

if $a\ge -1,$

$\Rightarrow -2-8(a+1)+8a+4a^2-14\le f^{\prime }\left(x\right)\le 2+8(a+1)+4a^2+8a+4$

$4a^2-24\le f^{\prime }\left(x\right)\le 4a^2+16a-4$

$4(a^2-6)\le f^{\prime }\left(x\right)\le 4(a^2+4a-1)$

$a>\sqrt{6}$

$0\ge 4(a^2-6)\le f^{\prime }\left(x\right)$

If $a<1,$

$\Rightarrow -2+8(a+1)+4a^2+8a-14\le f^{\prime }\left(x\right)\le 2-2(a+1)+4a^2+8a+4$

$4a^2+16a-8\le f^{\prime }\left(x\right)\le 4a^2-20$

$4(a^2+4a-2)\le f^{\prime }\left(x\right)\le 4(a^2-20)$

If $a\le (-2-\sqrt{6})=1$ $0<4(a^2+4a-2)\le f^{\prime }\left(x\right)$

$\therefore \forall a\in (-\infty ,-2-\sqrt{6})0(\sqrt{6},\infty ),$ $f^{\prime }\left(x\right)\ge 0$

$\Rightarrow f\left(x\right)$ increases $\forall x\in R\&$ has no critical points.

$\therefore$ at $(-\infty ,-\sqrt{5}-2)$ is not possible.

at $(1,\infty )$ is not possible

$\&$ $a\in (\sqrt{7},\infty )$ is possible.

Hence, the answer is option C.