Question

The set of the values of the parameter 'a' for which the function $f(x) = 6ax - a\sin4x - 6x - \sin6x$ increases $\forall x \in R$ and has no critical points is

• A. [-1, 1]
• B. (-\infty, -6)
• C. (-\infty, -6]
• D. (6, \infty)

Answer 1

Answer: (D)

(6, \infty)

Answer 2

For the function $f(x)$ to increase $\forall x \in R$ and have no critical points, its derivative $f'(x)$ must be strictly positive for all $x \in R$, i.e., $f'(x) > 0$.

The derivative of $f(x)$ is: $f'(x) = \frac{d}{dx}(6ax - a\sin4x - 6x - \sin6x)$ $f'(x) = 6a - 4a\cos4x - 6 - 6\cos6x$ $f'(x) = (6a - 6) - 4a\cos4x - 6\cos6x$

We require $f'(x) > 0$, so: $(6a - 6) - 4a\cos4x - 6\cos6x > 0$ $6a - 6 > 4a\cos4x + 6\cos6x$

This inequality must hold for all $x \in R$. Thus, $6a - 6$ must be greater than the maximum value of the expression $4a\cos4x + 6\cos6x$.

Let $M = \max_{x \in R} (4a\cos4x + 6\cos6x)$. The condition becomes $6a - 6 > M$.

Case 1: $a > 0$. To maximize $4a\cos4x + 6\cos6x$, we can choose $x$ such that $\cos4x = 1$ and $\cos6x = 1$. This is possible when $x$ is a multiple of $\pi$. For example, at $x = \pi$, $\cos(4\pi) = 1$ and $\cos(6\pi) = 1$. So, $M = 4a(1) + 6(1) = 4a + 6$. The condition $6a - 6 > M$ becomes: $6a - 6 > 4a + 6$ $2a > 12$ $a > 6$.

Case 2: $a < 0$. Let $a = -b$ where $b > 0$. The expression becomes $-4b\cos4x + 6\cos6x$. To find the minimum value, we consider the extreme values of $\cos4x$ and $\cos6x$. The minimum value of $-4b\cos4x$ is $-4b(-1) = 4b$. The minimum value of $6\cos6x$ is $6(-1) = -6$. However, these minimums may not occur at the same $x$. Let's consider $f'(x) = (6a-6) - (4a\cos4x + 6\cos6x)$. If $a < 0$, then $6a-6 < -6$. The term $4a\cos4x$ is negative or zero. The term $6\cos6x$ ranges from -6 to 6. Consider $x = \pi$. $f'(\pi) = (6a-6) - 4a(1) - 6(1) = 6a - 6 - 4a - 6 = 2a - 12$. Since $a < 0$, $2a < 0$, so $2a - 12 < -12$. This means $f'(\pi) < 0$. Thus, for $a < 0$, the function is not strictly increasing.

Case 3: $a = 0$. $f'(x) = -6 - 6\cos6x$. The maximum value of $f'(x)$ is $-6 - 6(-1) = 0$. So $f'(x) \le 0$, and the function is not strictly increasing.

Therefore, the only values of $a$ for which $f'(x) > 0$ for all $x$ is $a > 6$. The set of values is $(6, \infty)$.