Functions practice · Easy

Option C ($\frac{1}{2}, 4$) is correct because it represents the interval of x values that satisfy the given logarithmic inequality after simplification and solving.

Option A (0) is correct because substituting x=3 into the derived function f(x) results in f(3) = 0.

Option C ($f$ is one-one and onto $R$) is correct because the function is strictly increasing and continuous over the real numbers.

Option C (2f(x)) is correct because by substituting (2x)/(1+x²) into f(x) and using logarithm properties, we can simplify the expression to 2f(x).

The correct answer is D (13) because it is the absolute value of the sum of f(3) and f'(1/4) after solving the given functional equation and differentiating.

Option D ($\left[\frac{1}{3}, 3\right]$) is correct because it represents the range of the function, found by analyzing the quadratic equation and its discriminant.

Option D is correct. Statement 1 is true because a bijective function is onto. Statement 2 is also true because a bijective function has an inverse, but Statement 2 doesn't explain Statement 1.

Option A ($A \Delta B=\{6561\}$) is false because the symmetric difference $A \Delta B$ should include all elements in A or B but not in both, which is {3, 27, 243, 6561}, not just {6561}.

Option A ([1,4]) is correct because the domain is found by solving the inequality -1 <= log2(x/2) <= 1.

Option B ($[2,16]$) is correct because the minimum value of the function is 9 - 7(1) = 2 and the maximum value is 9 - 7(-1) = 16.

Option A (0) is correct because the left-hand side is always non-negative, while the right-hand side is negative, resulting in no solutions.

Option A ($3 f(x)$) is correct because the composition $(f \circ g)(x)$ simplifies to $3 \log_e(\frac{{1+x}}{{1-x}})$, which is $3f(x)$.

Option D (2014) is correct because the function repeats every 3 iterations, and 2010 is a multiple of 3, so f²⁰¹⁰(2014) = 2014.

Option B (Onto but not one-one) is correct because the function takes on all real values but fails the horizontal line test, indicating it's onto but not one-to-one.

Option C (An odd function) is correct because f(-x) = -f(x) for all x in the domain, which is the definition of an odd function.

Option C ([-2,3)) is correct because the domain of g(x) is determined by the condition -2 <= [x] <= 2, which translates to -2 <= x < 3.

Option B (Only transitive) is correct because if x < y and y < z, then x < z, which satisfies the transitive property. It is neither reflexive nor symmetric.

The correct answer is 9 because there are 2 even numbers that are not fixed, leading to a derangement problem with 2 elements, and the number of ways to arrange them is 1, multiplied by the number of ways to arrange the other 2 even numbers, which is also 1. This gives 3C2 * 2! = 3 * 2 = 6. Then we need to consider the case where both even numbers are swapped, which is 1. So 6 + 3 = 9

Option C $\left[ {1/2,\;1} \right) \cup \left( {2,\;5/2} \right]$ is correct because it is the solution set of the given logarithmic inequality considering the domain.

Option A ($\{ 2 \}$) is correct because the inverse function of $f(x) = x^3$ evaluated at 8 is 2.

Option C ([-1, $\infty$)) is correct because the domain is determined by the range of the arcsin function, which requires the argument to be between -1 and 1.

Option A (26) is correct because it is the result of evaluating $\frac{g(9)}{g^{\prime}(9)}$ using the given function and its inverse.

Option B ($\pi$) is correct because the period of $\sin^2 x$ can be found by rewriting it as $\frac{{1 - \cos 2x}}{{2}}$, which has a period of $\pi$.

Option C (-1,-2) is correct because these are the x-values that satisfy the equation g{f(x)} = 8.

Option A (5.5) is correct because the sum $S$ simplifies to 5.5 by pairing terms using the property $\phi(x) + \phi(-x) = 1$.

The correct answer is B (2) because there are two solutions to the equation $4\{x\}=x+[x]$, which are $x=0$ and $x=5/3$, found by analyzing the properties of the fractional and integral parts of x.

Option B is correct because solving for x in terms of y and swapping x and y gives $f^{-1}(x) = \frac{5x+1}{2-x}$.

Option B (72) is correct because there are 3 choices for the first digit (5, 7, or 9) and then 4! ways to arrange the remaining 4 digits.

Option A ($f(x + 2) - 2f(x + 1) + f(x) = (a - 1)^2f(x)$) is correct because it simplifies to an identity using the properties of exponents.

Option B ((f o g)(9)=683) is correct because when x=9, g(9) = 85 and f(85) = 683.