Increasing & decreasing practice · Medium

Option C (-1/18) is correct because the integral can be simplified using trigonometric identities and then evaluated.

Option B ($a \geq 1$) is correct because it ensures that the derivative of the function is always negative, making the function decreasing for all real values of x.

Option A $\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$ is incorrect. Option B $\left(0, \frac{\pi}{4}\right)$ is correct because g'(x) is negative in this interval, indicating that g(x) is decreasing.

Option C ($y > x-1$) is not true because for $x > 1$, $\log x < x-1$. The other options are true based on properties of logarithms and inequalities.

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 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 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 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 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 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 $m$.

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

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

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.