Functions practice · Mixed difficulty

Option B is correct because $f(-x) = f(x)$ for the function $f(x)=x\left(\frac{{a^{x}-1}}{{a^{x}+1}}\right)$.

Option C ($[0, 1]$) is correct because the identity $\sin^{-1}(x) + \cos^{-1}(x) = \frac{{\pi}}{{2}}$ holds for all $x$ in the interval [-1, 1], but since we have $\sqrt{x}$, $x$ must be non-negative, and $\sqrt{x}$ must be in [-1, 1], which means $x$ is in [0, 1].

Option D ($x \in(-\infty,-b) \cup(b, \infty)$) is correct because it represents all values of x whose absolute value is greater than b.

The correct answer is C (I is false, II is true) because if p < q, then the square root of a negative number results in a complex number, making statement I false and statement II true.

The correct answer is B ((-1,0]) because the range of the function $f(x)=\frac{{e^x-e^{|x|}}}{{e^x+e^{|x|}}}$ is $(-1,0]$, found by considering cases when $x \geq 0$ and $x < 0$.

The correct answer is D ([0,1]) because it is the intersection of the domains of all three square root functions.

Option B ($\{(x, y): y=|x|, x, y \in \mathbf{R}\}$) is correct because for each x in R, there is a unique y in R such that y = |x|.

Option C (A bijection) is correct because the function is both one-to-one and onto.

Option C ($\frac{{x^{2}-y^{2}}}{{4}}$) is correct because by substituting $u = x+ay$ and $v = x-ay$ and solving for x and y, then substituting into the original equation, we get $f(x, y) = \frac{{x^{2}-y^{2}}}{{4}}$.

Option C ($2^{8}$) is correct because the total number of reflexive relations on a set with $n$ elements is $2^{n(n-1)}$, and for a set with 4 elements, this is $2^{4(4-1)} = 2^{12}$. However, the explanation provided in the original answer is incorrect. The correct number of reflexive relations is $2^{n^2 - n} = 2^{4^2 - 4} = 2^{12}$. The provided answer $2^4$ is incorrect.

Option D ([0,2]) is correct because it represents the range of the composite function $g(\sqrt{{2}}f(x)+3)$ considering the range of $f(x)$ and the properties of the logarithm function.

Option A ($2^{n^2}$) is correct because a relation on A is a subset of A x A, and if A has n elements, A x A has $n^2$ elements, so there are $2^{n^2}$ possible subsets.

Option B ($\left\{3,-3\right\}, \phi$) is correct because $f^{-1}(83)$ yields $\pm 3$, and $f^{-1}(-2)$ is undefined in real numbers.

Option A (1) is correct because by evaluating the composite functions (gof)(-5/3) and (fog)(-5/3) and subtracting, the result is 1.

Option A (Injective but not surjective) is correct because the function is one-to-one (injective) but does not cover all possible values in its range (not surjective).

Option C is correct. The function is one-to-one because different natural numbers will always produce different outputs. It is not onto because not every natural number is in the range of f.

Option C ($g(f(S)) = g(S)$) is correct because it is NOT true. g(f(S)) = [-4, 4] and g(S) = [-2, 2], so they are not equal.

Option B ($f(x)=x(\frac{{a^x-1}}{{a^x+1}})$) is correct because $f(-x) = f(x)$, which is the definition of an even function.

Option B ($\left( {\pi ,2\;\pi } \right) \cup \left\{ {\pi /2} \right\}$) is correct because it identifies the intervals where [sin x] is either 1 or -1, which are the only values for which csc^-1[sin x] is defined.

Option B is correct because it represents the inverse function $f^{-1}(x)$ obtained by solving for x in terms of y and swapping x and y.

Option B is correct because solving the equation for y involves algebraic manipulation and recognizing the hyperbolic sine function.

Option B ([2,3]) is correct because the domain requires the argument of the logarithm to be positive and the logarithm itself to be non-negative.

Option B ($2 \log |x|$) is correct because the domain of $\log x^2$ includes negative values, which is accounted for by using the absolute value.

Option D (None of the above) is correct because the relation R, defined by $x^2 + y^2 = 9$, includes the pairs (0, 3), (0, -3), (3, 0), and (-3, 0). The domain and range are both {-3, 0, 3}, so all the statements are true, making "None of the above" the correct answer.

Option C (5/3) is correct because by substituting x = 1 and x = -1/2 into the given equation, we obtain a system of two equations that can be solved for f(2).

Option B is correct because the condition $x^2 + ax + 1 > 0$ for all $x$ implies $a^2 - 4 < 0$, leading to 3 integers in the range of 'a'.

Option B (1) is correct because the equation simplifies to a single solution at x=0.

Option A (0) is correct because the integrand is an odd function, and the integral of an odd function over a symmetric interval is zero.

Option B ($n^{{m}}$) is correct because for each of the m elements in X, there are n possible images in Y, leading to $n^m$ possible functions.

Option A ([{1,\sqrt 2 }]) is correct because it represents the range of the secant function given the restricted domain of its argument.