Functions practice · Easy

Option B ($\frac{{1}}{{g^{\prime}(2)}}$) is correct because it follows from the derivative of the inverse function theorem.

Option B (${x^2} + 3x + 1 = 0$) is correct because it satisfies the given conditions f(0) = 1, f(1) = 5, and f(2) = 11.

The correct answer is B (f(x) = x + 2) because this function is both injective (one-to-one) and surjective (onto) when mapping from Z to Z.

Option D (Neither one to one nor onto) is correct because the function f(z) = |z| maps multiple complex numbers to the same real number and only produces non-negative real numbers.

Option A (An even function) is correct because all the powers of x in the function are even, making f(x) = f(-x).

Option C ($\cos 2 \theta$) is correct because substituting $f(x)$ into itself simplifies to $\cos 2 \theta$ using trigonometric identities.

Option C ($(0,+\infty)$) is the correct domain because the argument of the logarithm, $x-[x]$, must be greater than 0, and $x-[x]$ represents the fractional part of x, which is always in the interval [0, 1). Since the log is undefined at 0, the domain is (0, +∞).

Option A (${ }^{m+n} C_k$) is correct because the given expression represents the coefficient of $x^k$ in the expansion of $(1+x)^m (1+x)^n = (1+x)^{m+n}$.

Option C (6) is correct because the number of bijective functions from a set A to a set B, where both have n elements, is n!.

Option D is correct because the function f(x) = x⁴ is neither one-to-one (many-to-one) nor onto as it doesn't cover all real numbers in its range.

Option A (${2^{100}} - 2$) is correct because the number of onto mappings from a set A with m elements to a set B with n elements is given by the formula and simplifies to 2¹⁰⁰ - 2 when m = 100 and n = 2.

Option C $- 4 < x \le 1$ is correct because it satisfies both the square root and logarithmic domain restrictions.

Option D (Many-one into) is correct because the function $f(x) = x^2 + x$ is many-to-one and its range is not the entire set of real numbers.

Option A (Neither one-one nor onto) is correct because the cosine function is periodic and its range is limited to [-1, 1], making it neither injective nor surjective when mapping from R to R.

Option A ($f[F(x)]$) is correct because $F[f(x)] = a^{\log_a x} = x$ and $f[F(x)] = \log_a a^x = x$, so $F[f(x)] = f[F(x)]$.

Option B (4) is correct because it is the value obtained by solving the functional equation for f(x) and then substituting x = 7.

Option B ( -1,1 .$\left\\{-1,1\\}\right.$) is correct because when $x > -2$, the function is 1, and when $x < -2$, the function is -1.

Option D (8) is correct because g(x) is a constant function, and since g(3) = 8, g(8) must also be 8.

Option A ($\frac{{\left( {1 - {x^4}} \right)\left( {2{x^4} + 3} \right)}}{{5{x^4}}}$) is correct because it is the result of solving the system of equations for f(x²) and then substituting x⁴.

Option D ($\left( {\frac{2}{5},\frac{1}{2}} \right) \cup \left( {\frac{3}{5},\frac{4}{5}} \right]$) is correct because it represents the range of the function f(x) considering the greatest integer function and the defined intervals.

Option D is correct because the domain requires x+3 > 0 and the denominator not equal to zero, leading to the interval (-3, infinity) excluding -1 and -2.

Option C ($R - \\{-3, 3\\}$) is correct because the argument of the logarithm, $|x^2 - 9|$, must be greater than 0, which means $x^2 - 9 \neq 0$, so $x \neq \pm 3$.

Option C ($3 f(x)$) is correct because substituting $g(x)$ into $f(x)$ and simplifying the logarithm results in $3f(x)$.

Option D (${x^2} + 5$) is correct because gof(x) = g(f(x)) = g(x²) = x² + 5.

Option A ([{- 2,\;0}] ∪ [{2,\;4}]) is correct because it represents the values of x for which the argument of the inverse sine function lies between -1 and 1.

The correct answer is D (-1) because when α = -1, f[f(x)] simplifies to x, satisfying the given condition.

Option A ([0, 1]) is correct because the range of the function is determined by completing the square and considering the non-negative values of the square root.

The correct answer is Not defined because the function is not bijective over the given interval, as $f(4) = 1$ which is not within the range $[4, \infty[$.

Option B ($\left[ {12,\;34} \right]$) is correct because it represents the range of the function $f(x)$ over the interval $[2, 3]$.

Option A ([1,2]) is correct because it represents the domain for which the function is defined.