Question

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

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

Answer 1

Here in as we need to find the parameter $a$ if $f(x) = \sin 2x - 8(a + 1)\sin x + (4{a^2} + 8a - 14)x$ increases for all $x \in R$ and has no critical points for all $x \in R$ which means $f'(x) \geqslant 0$ because the function increases but it is mentioned that there are no critical points for all $x \in R$ so we need to find $f'(x){\text{ > }}0$ only as if there is no critical point then $f'(x) \ne 0$ and simplify the functions by differentiating.

Complete step by step solution:
Since in the above equation it is given that $f(x) = \sin 2x - 8(a + 1)\sin x + (4{a^2} + 8a - 14)x$
So

$$\Rightarrow f'(x) = 2\cos 2x - 8(a + 1)\cos x + (4{a^2} + 8a - 14) \\\ {\text{ }} - 1 \leqslant \cos 2x.\cos x \leqslant 1 \\\$$

Now if $a \geqslant - 1$
Then we will simplify

$$\- 2 - 8(a + 1) + 8a + 4{a^2} - 14 \leqslant f'(x) \leqslant 2 + 8(a + 1) + 4{a^2} + 8a + 4 \\\ \Rightarrow 4{a^2} - 24 \leqslant f'(x) \leqslant 4{a^2} + 16a - 4 \\\ \Rightarrow 4({a^2} - 6) \leqslant f'(x) \leqslant 4({a^2} + 4a - 1) \\\ a{\text{ > }}\sqrt 6 \\\ 0 \geqslant 4({a^2} - 6) \leqslant f'(x) \\\$$

If $\;a \leqslant ( - 2 - \sqrt 6 ) = 1\;0 < 4(a2 + 4a - 2) \leqslant f\prime (x)$
Therefore $\forall a \in ( - \infty , - 2 - \sqrt 6 )0(\sqrt 6 ,\infty ),\;\;f\prime (x) \geqslant 0$
$\Rightarrow f(x)$ Increases $\forall x \in R$ and has no critical points
Now we will from the given options
Therefore at $( - \infty , - \sqrt 5 , - 2)\;$ is not possible
At $(1,\infty )$is not possible
$a \in (\sqrt 7 ,\infty )\;$It’s possible which means option $\;C$ is correct
Formula: We will find the value of $f'(x){\text{ > }}0$ only as if there is no critical point then $f'(x) \ne 0$ and simplify the functions by differentiation.

Additional Information:
The range of function is set of all the possible outcomes of the function for all possible values of its domain.

Note:
We can see that the problem which is given contains a lot of calculations so we need to perform each step carefully to avoiding confusion and calculation mistakes. Here in the above question it is mentioned that there are no critical points which means $f'(x) \ne 0$ so $f'(x){\text{ > }}0$. Keep in mind that a non-decreasing function can contain strictly increasing intervals and intervals where the function is constant. So the possible values of parameter $a$ for the above problem will be $(\sqrt 7 ,\infty )$