Functions practice · Mixed difficulty

The correct answer is B (3) because solving the inequality results in x > 2.5, and the least integer satisfying this is 3.

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 $x < 1$.

Option C ([-2,-1]$\cup$[1,2]) is correct because it represents the domain for which the function is defined, considering the logarithmic and inverse sine functions.

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 (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 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 D ($a\left( {\frac{{{b^r} - 1}}{{b - 1}}} \right) + {b^r}x$) is correct because it represents the general form of the function after being composed with itself 'r' times, which involves a geometric series.

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

Option A (2) is correct because it's derived from using Leibniz's rule to differentiate the integral equation and then solving for k(1).

Option C ($3^{2x} + 1, 3^{x^2+1}$) is correct because substituting $\psi(x)$ into $\phi(x)$ gives $3^{2x} + 1$ and substituting $\phi(x)$ into $\psi(x)$ gives $3^{x^2+1}$.

Option D (None of these) is correct because none of the given equations hold true for the function $f(x) = |x-1|$.

Option B ($\frac{x}{\sqrt{1+3 x^2}}$) is correct because after applying the function three times, the expression simplifies to $\frac{x}{\sqrt{1+3 x^2}}$.

Option D ($1 + {\{ g(x)\} ^5}$) is correct because by differentiating $f(g(x)) = x$ and using the given $f'(x)$, we can find $g'(x)$.

Option A ($f$ is a relation from $A$ to $B$) is correct because all first elements of the ordered pairs in $f$ belong to $A$, and all second elements belong to $B$, satisfying the definition of a relation from $A$ to $B$. However, the element 2 in A maps to two different elements in B (9 and 11), so it's not a function.

Option A (0) is correct because after applying the column operation C3 -> C3 - (C1 + C2), the determinant becomes zero for all x.

Option C (3) is correct because after using the given functional equation and solving the resulting exponential equation, the value of a is found to be 3.