Functions practice · Medium

Option A is the correct answer. The inverse function $g(x)$ is found by solving for $x$ in terms of $y$ and then swapping $x$ and $y$, resulting in $g(x) = \frac{1}{2} \log _2\left(\frac{2+x}{2-x}\right)$.

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 B (-3) is correct because the slope of g(x) is found by differentiating f(g(x)) and using the given f'(x).

Option B ($(f \circ f)(x)=x$) is correct because applying the function composition twice results in the original input x, regardless of whether x is rational or irrational.

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 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 B $[1,3 / 2]$ is correct because the expression inside the square root must be non-negative, leading to the given interval.

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.

The correct answer is A ($x(\frac{a^x-1}{a^x+1})$) because substituting -x into the function results in the same function, indicating even symmetry.

Option C $0 < a < 10$ is correct because it represents the range of values that $a$ can take given the condition on $r$.

Option C (One-one and into) is correct because the function is one-to-one (injective) but not onto (surjective) since its range is not the entire real number set.

Option C ($f(x) = ax + (1 - a)x^2, a \in (0, 2)$) is correct because it satisfies the given conditions: f(0) = 0, f(1) = 1, and f'(x) > 0 for x in (0, 1).

Option C ($X = {R^ + },Y = R$) is correct because it defines a function that is one-to-one but not onto.

Option D (7) is correct because it involves solving a functional equation and a geometric series.

Option D (None of these) is correct because the function is neither one-to-one nor onto.

Option D (-1) is correct because by substituting -x for x and adding the equations, we find that x² = 1, and thus x² - 2 = -1.

Option A ($x$) is correct because when the function $f(x)$ is applied to itself, the result simplifies back to $x$.

Option A ($x, \forall x \in R$) is correct because when you compute $(g \circ f)(x)$ for both cases of $f(x)$ and $g(x)$, the result simplifies to $x$ for all real numbers.

The correct answer is A (6π) because the period of cos(kx) is 2π/k, and in this case, k = 1/3.

Option A (f⁻¹(x) = f(x)) is correct because the inverse of the function is equal to the function itself, indicating it's a self-inverse function.

Option C (($- 3, - 2) \cup ( - 2, - 1) \cup (1,\infty )$) is correct because it satisfies the conditions for the logarithm to be defined.

Option C is correct. The value inside the greatest integer function becomes greater than or equal to 1 when r >= 33. Therefore, the sum is equal to the number of terms where r >= 33, which is 98 - 33 + 1 = 66.

Option B ((-∞, 0)) is correct because the function is defined only when |x| > x, which occurs when x is negative.

Option B (-1) is correct because after evaluating the functions f(x), g(x), and h(x), their product simplifies to -1.

Option B ($x \in \left[ {1,\;2} \right]$) is correct because the function exists when the argument of the square root is non-negative and the argument of the inverse sine lies between -1 and 1.

Option A ($\frac{{x}}{{x-1}}$) is correct because the function composed with itself twice returns the original input, so composing it an odd number of times is equivalent to the original function.

Option D $\left[ {0,\;1} \right)$ is correct because the range of the function, excluding 1, makes it a surjection.