SAT1000 - P966

Calculus Level 3

If function f ( x ) = x 1 3 sin 2 x + a sin x f(x)=x-\dfrac{1}{3} \sin 2x + a \sin x is monotonic increasing for all x R x \in \mathbb R , what is the range of a a ?


Have a look at my problem set: SAT 1000 problems

[ 1 , 1 3 ] \left[-1,\ - \frac 13 \right] [ 1 , 1 ] [-1,\ 1] [ 1 , 1 3 ] \left[-1,\ \frac 13\right] [ 1 3 , 1 3 ] \left[-\frac 13,\ \frac 13\right]

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1 solution

Chew-Seong Cheong
Jul 17, 2020

For f ( x ) = x 1 3 sin 2 x + a sin x f(x) = x - \frac 13 \sin 2x + a \sin x to be monotonic increasing for all real x x , f ( x ) f'(x) must be non-negative or

f ( x ) 0 1 2 3 cos 2 x + a cos x 0 1 4 3 cos 2 x + 2 3 + a cos x 0 4 cos 2 x 3 a cos x 5 0 4 ( cos x 3 a 8 ) 2 9 a 2 16 5 0 max ( cos x 3 a 8 ) 2 = 1 + 3 a 8 2 4 1 + 3 a 8 2 9 a 2 16 5 0 when cos x = { 1 for a > 0 1 for a < 0 9 a 2 16 + 3 a + 4 9 a 2 16 5 3 a 1 a 1 3 a [ 1 3 , 1 3 ] \begin{aligned} f'(x) & \ge 0 \\ 1 - \frac 23 \cos 2x + a \cos x & \ge 0 \\ 1 - \frac 43 \cos^2 x + \frac 23 + a \cos x & \ge 0 \\ 4 \cos^2 x - 3a \cos x - 5 & \le 0 \\ 4 \blue{\left(\cos x - \frac {3a}8 \right)^2} - \frac {9a^2}{16} - 5 & \le 0 & \small \blue{\max \left(\cos x - \frac {3a}8 \right)^2 = \left| 1 + \frac {3|a|}8 \right|^2} \\ 4 \blue{\left|1+ \frac {3|a|}8 \right|^2} - \frac {9a^2}{16} - 5 & \le 0 & \small \blue{\text{when }\cos x = \begin{cases} -1 & \text{for }a > 0 \\ 1 & \text{for }a < 0 \end{cases}} \\ \frac {9a^2}{16} + 3|a| + 4 - \frac {9a^2}{16} & \le 5 \\ 3 |a| & \le 1 \\ |a| & \le \frac 13 \\ \implies a & \in \boxed{\left[-\frac 13,\ \frac 13 \right]} \end{aligned}

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