Functions practice · Medium

Option C (0 < x < 1) is correct because it represents the intersection of the domains of f(x) and g(x), excluding points where either function is undefined.

Option A (I, II) is correct because polynomial functions have non-negative integer exponents for the variable x, which is true for both I and II. Function III has a fractional exponent (2/3), so it's not a polynomial.

Option B ($(2,3)$) is correct because it satisfies the inequality $\log_{1/2}(x^2 - 5x + 7) > 0$.

Option B $x \in \left( { - \infty ,2} \right]$ is correct because the inequality holds true for x less than or equal to 2.

Option C ($1 - {x^3}$) is correct because it is derived by swapping $x$ and $y$ and solving for $y$ after setting $y = f(x)$.

Option A ($f^{-1}(x)=f(x)$) is correct because the inverse of the function is equal to the function itself.

Option A ([0,1]) is correct because the domain of arccos(x) is [-1,1], so -1 <= 2x-1 <= 1 which implies 0 <= x <= 1.

Option B (Onto but not one – one) is correct because the function fog is not one-to-one as f(2) = f(1) = 1, but it is onto as every element in the codomain has a pre-image in the domain.

Option B (Statement -1 is True, Statement -2 True; Statement - 2 is NOT a correct explanation for Statement - 1) is correct because both statements are true, but the periodicity of f(x) does not explain why f'(c) = 0 in the given interval.

Option A (X=(-2, ∞), Y=(- ∞, log_3 4)) is correct because it satisfies the condition for the function to be one-to-one and onto, based on the given equation.

The correct answer is A ($x f^{\prime}(x)-2024 f(x)=0$) because by using the given functional equation and differentiation, we can derive the relationship between f(x) and its derivative.

Option B ($\frac{1}{9}$) is correct because by using the given functional equation and the Intermediate Value Theorem, we can deduce that $f(9) = \frac{1}{9}$.

Option B (8) is correct because it is the only solution that satisfies the given equation when considering both positive and negative cases of the absolute value.

Option B ($- \infty < x \le \frac{{e - 1}}{e}$) is correct because it represents the domain where the function is defined, considering the logarithm and square root.

Option B ($[0,\sqrt 3 ]$) is the correct range because it considers the domain restriction and the behavior of the tangent function within that domain.

Option B is correct. By applying componendo-dividendo and substituting the expression for x, we can express f(2x) in terms of f(x).

Option C ($\frac{{1-4 x+4 x^2}}{{1+4 x-4 x^2}}$) is correct because it is the result of substituting g(x) into f(x) and simplifying the expression.

Option B ($\frac{1}{9}$) is correct because by using the given functional equation and the Intermediate Value Theorem, we can deduce that $f(9) = \frac{1}{9}$.

Option B ($\{ 1,\;4,\;7 \}$) is correct because these are the values of n for which f(n) > 2 based on the given piecewise function.

Option C (R-[-1,0)) is correct because the range of the function is all real numbers except for the interval [-1, 0).

Option A ($\frac{3 f(x)+1}{f(x)+3}$) is correct because substituting f(x) and simplifying results in f(2x).

Option D ($x^{2}-5 x+6$) is correct because by substituting $x-1$ into the given function, we can find the expression for $f(x)$.

Option A ($fog(x) = 27x^2 - 18x + 5$) is correct because it is the result of substituting g(x) into f(x).

Option C is correct. By recursively applying the given formula $f(n)=2 \sum\limits_{{r=1}}^{n-1} f(r)$ and simplifying, we can find that $\sum\limits_{{r=1}}^m f(r) = 3^{m-1}$.

Option C (Neither even nor odd) is correct because the greatest integer function does not satisfy the conditions for even or odd functions.

Option D ($-f(x)$) is correct because by substituting specific values into the functional equation and using the given condition f(0)=1, we can derive that f(2a-x) = -f(x).

Option A (5) is correct because calculating the number of strictly increasing functions and onto functions, and then subtracting, results in a number with a last digit of 5.

Option B ($f$ is one-one into) is correct because the function is injective (one-one) but not surjective (onto) due to a restriction in its range.

Option D (None of these) is correct because by substituting the given function into the expression and simplifying using trigonometric identities, the result is 0.

Option D ($\mathrm{{P}} \in(-\infty, 4]$) is correct because it represents the range of P for which the given trigonometric equation has real solutions.