Question

Let $f(x) = 2x^3 + ax^2 + bx - 3\cos^2 x$ then $f(x)$ is an increasing function for all $x \in R$ if

• A. $a^2 - 6b - 18 > 0$
• B. $a^2 - 6b + 18 < 0$
• C. $a^2 - 3b - 6 < 0$
• D. $a > 0, b > 0$

Answer 1

Answer: (B)

B

Answer 2

For a function $f(x)$ to be an increasing function for all $x \in R$, its derivative $f'(x)$ must be greater than or equal to zero for all $x \in R$.

Given the function $f(x) = 2x^3 + ax^2 + bx - 3\cos^2 x$.

First, let's find the derivative $f'(x)$: $f'(x) = \frac{d}{dx}(2x^3) + \frac{d}{dx}(ax^2) + \frac{d}{dx}(bx) - \frac{d}{dx}(3\cos^2 x)$

1. $\frac{d}{dx}(2x^3) = 6x^2$
2. $\frac{d}{dx}(ax^2) = 2ax$
3. $\frac{d}{dx}(bx) = b$
4. For $-3\cos^2 x$, we use the chain rule: $\frac{d}{dx}(-3\cos^2 x) = -3 \cdot 2\cos x \cdot (-\sin x)$ $= 6\sin x \cos x$ Using the identity $\sin(2x) = 2\sin x \cos x$, we get: $= 3(2\sin x \cos x) = 3\sin(2x)$

So, $f'(x) = 6x^2 + 2ax + b + 3\sin(2x)$.

For $f(x)$ to be an increasing function for all $x \in R$, we must have $f'(x) \ge 0$ for all $x \in R$. $6x^2 + 2ax + b + 3\sin(2x) \ge 0$ for all $x \in R$.

Let's rearrange the inequality: $6x^2 + 2ax + b \ge -3\sin(2x)$.

For this inequality to hold for all $x \in R$, the minimum value of the quadratic expression on the left-hand side must be greater than or equal to the maximum value of the trigonometric expression on the right-hand side.

1. Minimum value of the quadratic $P(x) = 6x^2 + 2ax + b$: This is a parabola opening upwards ($A=6 > 0$). Its minimum value occurs at $x = -\frac{2a}{2(6)} = -\frac{a}{6}$. The minimum value is $P\left(-\frac{a}{6}\right) = 6\left(-\frac{a}{6}\right)^2 + 2a\left(-\frac{a}{6}\right) + b$ $= 6\left(\frac{a^2}{36}\right) - \frac{2a^2}{6} + b$ $= \frac{a^2}{6} - \frac{a^2}{3} + b$ $= \frac{a^2 - 2a^2}{6} + b$ $= -\frac{a^2}{6} + b$ Alternatively, using the formula for the minimum value of $Ax^2+Bx+C$ which is $\frac{4AC-B^2}{4A}$: $P_{min} = \frac{4(6)(b) - (2a)^2}{4(6)} = \frac{24b - 4a^2}{24} = b - \frac{a^2}{6}$.

2. Maximum value of $-3\sin(2x)$: We know that $-1 \le \sin(2x) \le 1$. Multiplying by $-3$ (and reversing the inequality signs): $-3 \le -3\sin(2x) \le 3$. So, the maximum value of $-3\sin(2x)$ is $3$.

For $6x^2 + 2ax + b \ge -3\sin(2x)$ to hold for all $x$, we must have: $\min(6x^2 + 2ax + b) \ge \max(-3\sin(2x))$ $b - \frac{a^2}{6} \ge 3$

Multiply by 6 to clear the denominator: $6b - a^2 \ge 18$

Rearrange the terms to match the options: $-a^2 + 6b - 18 \ge 0$ $a^2 - 6b + 18 \le 0$

Comparing this with the given options: A) $a^2 - 6b - 18 > 0$ B) $a^2 - 6b + 18 < 0$ C) $a^2 - 3b - 6 < 0$ D) $a > 0, b > 0$

Our derived condition is $a^2 - 6b + 18 \le 0$. Option B is $a^2 - 6b + 18 < 0$. Since $f'(x) \ge 0$ is the condition for an increasing function, $a^2 - 6b + 18 \le 0$ is the correct condition. If the options contain a strict inequality, it usually implies that the function is strictly increasing, or that the problem expects a slightly stronger condition. In competitive exams, if $\le$ is derived and $<$ is an option, it's often the intended answer, implying that equality might lead to points where $f'(x)=0$ over an interval, which is not strictly increasing. However, for "increasing function", $f'(x) \ge 0$ is the standard definition. Given the options, $a^2 - 6b + 18 < 0$ is the closest match and represents a sufficient condition for $f'(x) > 0$ everywhere, which implies $f(x)$ is strictly increasing, and thus also increasing.