Increasing & decreasing practice · Mixed difficulty

Option A $((0, \infty))$ is correct because the derivative of the function is positive for all $x > 0$, indicating an increasing function.

Option A ($\left( { - \infty ,\,0} \right) \cup \left( {\frac{{14}}{{15}},\,\infty } \right)$) is correct because the function is increasing when its derivative is greater than or equal to zero, which occurs in the specified intervals.

Option C is correct because the derivative of the function, f'(x), is always positive, indicating that the function is increasing over the entire real line.

The correct answer is A ($2 \sin x+\tan x > 3 x$) because the second derivative of the function $f(x)=2 \sin x+\tan x-3 x$ is positive, implying that $f(x)$ is greater than 0 for $x \in (0, \pi/2)$. The original answer incorrectly includes equality.

Option A (always increases) is correct because the derivative of the function, f'(x) = sec²(x) - 1, is always non-negative, indicating that the function is always increasing.

Option D ($\mathrm{R}-(-\sqrt{3}, \sqrt{3})$) is correct because the derivative of the function must be greater than 0 for all x, which leads to the condition $a^2 \geq 3$.

Option A is correct because the derivative of the function, f'(x), is positive in the interval (0, 2), indicating that the function is increasing in that interval.

Option B ($\left( {1,\;3} \right)$) is correct because the function is decreasing when its derivative is negative, which occurs in this interval.

Option D ($1 < x < 2$) is correct because the derivative of the polynomial is negative in this interval, indicating that the function is decreasing.

Option B (Decreases in $\left( {0,\;\infty } \right)$) is correct because the derivative of the function is negative for all x in the real numbers, indicating a decreasing function.

Option D ( ∪[1, ∞)) is correct because it represents the intervals where the derivative of h(x) is greater than or equal to zero, indicating an increasing function.

Option A (An increasing function) is correct because the composition of two decreasing functions results in an increasing function.

Option A ($k > 1$) is correct because for the function to be monotonically increasing, its derivative must be greater than 0, implying $k > \cos x$ for all x, hence $k > 1$.

Option A is correct because the second derivative of the function is positive in the interval (-2/a, a/3), indicating that the first derivative is increasing in that interval.

Option D ($x > \frac{1}{2}$) is correct because the function $f(x) = 4x^4 - 2x + 1$ is increasing when its derivative $f'(x) = 16x^3 - 2$ is positive, which occurs for $x > \frac{1}{2}$.

Option A (always increasing) is correct because the derivative of the function is always non-negative.

Option D (None of these) is correct because the interval in which the function is decreasing does not match any of the given options after calculating the derivative and finding the interval.

Option A ($x > \frac{1}{e}$) is correct because the derivative of the function is positive when $x$ is greater than $\frac{1}{e}$.

Option D is correct because $g'(x) < 0$ for $x \in (0, \frac{3\pi}{4})$, indicating that $g(x)$ is decreasing in this interval.

Option D (None of the above) is correct because the function is neither strictly increasing nor onto.

Option C ($x \ge 0$) is correct because the function is increasing for all real numbers, and the inequality holds for $x \ge 0$.

Option A ($\pi$) is correct because finding the critical point of g(x) leads to $\alpha = 1/2$, and substituting this value into the expression yields $\tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3)$, which equals $\pi$.

Option A $((-\infty,-3) \cup(0, \infty))$ is correct because for $f(x)$ to be decreasing on $\mathbb{R}$, $f'(x) \leq 0$ for all $x$, which leads to the given interval for $m$.

Option A (Always increasing) is correct because the derivative of the function $f(x) = x + \cos x$ is always non-negative, indicating that the function is always increasing.

The correct answer is D ($g(x)$ increases on [0,1/2] and decreases on [1/2,1]) because the function g(x) is increasing when its derivative is positive and decreasing when its derivative is negative, and f''(x) < 0 indicates concavity.

Option C ((1,2)) is correct because the function f decreases where its derivative f'(x) is negative, which occurs in the interval (1, 2).

Option A (Increases on the interval $\left( {0,\;3a/4} \right)$) is correct because the derivative of the function is positive in that interval, indicating an increasing function.

Option A is correct because the given condition leads to a quadratic expression that must be positive for all x, resulting in the specified range for 'a'.

Option C is correct because the function's derivative is positive in the interval $(-\infty, 1/3)$, indicating an increasing function.

Option C ($\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right)-\frac{\pi}{4}$) is correct because it represents the sum of the absolute maximum and absolute minimum values of the function within the given interval.