Increasing & decreasing practice · Mixed difficulty

Option A ([-1,1]) is correct because the function is decreasing when its derivative is negative, which leads to the condition a² - 1 <= 0.

Option B is correct. The function decreases in $(-1, \infty)$ because its derivative is negative in that interval.

The correct answer is that g(x) is a continuous function. The function min(x, x²) is continuous because both x and x² are continuous, and the minimum of two continuous functions is also continuous.

Option C ($(0, \infty)$) is correct because the inequality $1+\log x < x$ holds for all $x$ greater than $0$, which is the domain of the logarithm.

Option C ((-2, 6)) is correct because the function is decreasing when its derivative is negative, which occurs in the interval (-2, 6).

Option D ($f^{\prime \prime}(x)$ attains the value 2 for some $x \in(1,3)$) is correct because by applying Rolle's theorem twice to the function $g(x) = f(x) - x^2$, we can show that there exists a point in (1,3) where the second derivative of f(x) is 2.

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.

Option D (Decreasing in $(-1, \infty)$ and increasing in $(-\infty,-1)$) is correct because the derivative is negative on $(-1, \infty)$ and positive on $(-\infty,-1)$.

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 B is correct because the function's derivative is positive in the interval (π/4, 3π/8), indicating an increasing function.

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.

Option D (No value of $a$) is correct because the condition for the function to be decreasing for all x leads to a contradiction, implying no such value of 'a' exists.

Option A (12) is correct because the function is decreasing in the interval (-2, -1) only when the derivative is negative and the roots of the derivative equation are -2 and -1.

The correct answer is B ($\frac{{\tan \beta}}{{\tan \alpha}} > \frac{{\alpha}}{{\beta}}$) because the function $f(x) = x \tan x$ is increasing on $(0, \pi/2)$.

Option A is correct because the function's derivative is non-negative in the interval [-1/2, 1], indicating an increasing function.