Increasing & decreasing practice · Easy

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.

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 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.

Option A ($x < 0$) is correct because it satisfies the condition for the function to be decreasing when a < 0, derived from analyzing the sign of the derivative.

Option B is correct because $g(x) > 0$ for all $x$ which implies that the integral of $g(t)$ from $a$ to $x$ is an increasing function.

Option C (Decreasing on (0, 1) and increasing on (1, 2)) is correct because the sign of the derivative of $\phi(x)$ changes from negative to positive at $x=1$, indicating a decreasing function before 1 and an increasing function after 1.

Option A $\left(-\frac{\pi}{4}, \frac{\pi}{4}\right)$ is correct because the derivative of the function is positive in this interval, indicating that the function is strictly increasing.

Option A (Increases on R) is correct because the derivative of the function is always positive, indicating an increasing function over all real numbers.

The correct answer is B ($[2, \infty)$) because the derivative $f'(x)$ is non-negative for $x \geq 2$, indicating that $f(x)$ is increasing on that interval.

Option C ($f(x)$ is increasing in its domain) is correct because the derivative of the function is always positive in its domain.

Option C (Exactly one real root) is correct because the function is monotonic and of odd degree, guaranteeing it crosses the x-axis only once.