Functions practice · Easy

Option C ($\{\4, -4\}, \phi$) is correct because when x=17, the inverse function yields 4 and -4, but when x=-3, the inverse function has no real solution.

Option C ($\sin x^2$) is correct because the composite mapping $f \circ g$ is defined as $f(g(x))$, which means we substitute $g(x) = x^2$ into $f(x) = \sin x$, resulting in $\sin(x^2)$.

Option A (3) is correct because the domain of f(x) is [0,3) and the range of g(x) is an empty set, so the union contains integers 0, 1, and 2, summing to 3.

Option C ($x$) is correct because substituting $f(x)$ into itself simplifies to $x$ due to the inverse relationship created by the exponents and the subtraction from $a$.

Option A ($f(x) = \frac{{1 - x}}{{1 + x}}$) is correct because applying the function to itself results in the original input x, meaning it's its own inverse.

Option D, $\log \left(x+\left(x^2+1\right)\right)$, is the correct inverse function because it satisfies $f(f^{-1}(x)) = x$.

Option C is correct because substituting g(x) into f(x) yields the given expression after simplification.

Option C, $\left( { - \infty ,1} \right) \cup \left( {2,3} \right)$, is correct because it represents the solution set obtained by solving the inequality after factoring.

Option A (-44) is correct because it is the result of evaluating the composite function (fog)(-1) by first finding g(-1) and then substituting that value into f(x).

The provided answer and solution steps are incorrect. The correct answer is (-∞, ∞). The range of arcsin is [-π/2, π/2] and the argument of arcsin must be between -1 and 1. Since e^x is always positive, 1/(1+e^x) is always between 0 and 1. Therefore, the domain is all real numbers.

The correct answer is A ($2a$) because the function $f$ has a period of $2a$, derived by substituting $x+a$ into the given equation and simplifying.

Option C ($-\infty < x < 1$) is correct because the domain is restricted by the logarithm, requiring $1 - a^{x-1} > 0$, which leads to $x < 1$.

The correct answer is D (8) because the given relation implies that the function has a period of 8, derived by substituting $x-2$ into the original equation and comparing.

Option D ($\frac{{7n\left( {n + 1} \right)}}{2}$) is correct because it represents the sum of the series where each term is a multiple of 7 based on the given functional equation.

The correct answer is D ($\left( {\left( {4n - 1} \right)\frac{\pi }{2},\left( {4n + 1} \right)\frac{\pi }{2}} \right),n \in Z$) because the domain is determined by the condition where cos(x) is positive.

Option A (even) is correct because f(-x) = f(x), which is the definition of an even function.

The correct answer is C (6/7) because substituting x=11 into the given function and simplifying yields 6/7.

Option B is correct because the domain is determined by the intersection of the domain of arcsin(3-x) and log(|x|-2), excluding points where the logarithm is zero.

Option C ($e^{\left(-\frac{a^2}{2 b^2}\right)}$) is correct because it represents the limit of the given expression as x approaches 0, evaluated using exponential and trigonometric limit properties.

Option C (f is both one-one and onto) is correct because the function is both injective (one-to-one) and surjective (onto) within the specified domain and range.

Option A (1) is correct. The period of the function is determined by recognizing that the fractional part of kx has a period of 1/k, and the overall period is the LCM of these periods.

Option B (27) is correct because the function satisfying the given condition is an exponential function, and f(3) can be calculated once the base is determined from f(2)=9.

Option B (An odd function) is correct because the derivative of an even function is always an odd function, and $f(x)$ is even.

Option B ($[\frac{1}{3}, 1]$) is correct because the minimum value of the function is achieved when sin(3x) = 1 and the maximum value when sin(3x) = -1.

Option D (0) is correct because the expression simplifies to zero using the cosine addition formula.

Option D ($f(x)$) is correct because when you compute $x^{3} f(\frac{{1}}{{x}})$, it simplifies back to the original function $f(x)$.

Option A ($\frac{{{{x^2} - {y^2}}}}{{8}}$) is correct because it is the result of substituting $x$ and $y$ in terms of $u$ and $v$ into the given function.

Option D ($f(y)$) is correct because solving for x in terms of y results in the same functional form as f(y).

Option C (One-one and into) is correct because the function f(x) = e^x is injective (one-one) but not surjective (onto) over the real numbers.

Option A (1) is correct because the graphs of the logarithmic and absolute value functions intersect at only one point.