Functions practice · Mixed difficulty

Option A (6) is correct because the value of $\lambda$ is found by constructing a polynomial that satisfies the given conditions and then evaluating it at x=6.

Option B (\{0, 1\}) is correct because the range of the function $f(x) = 2x - 2[x]$ is \{0, 1\}.

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 $x \in \left( { - \infty ,2} \right]$ 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 B (-1) is correct because by finding g(x) and h(x) in terms of f(x), the product f(x) * g(x) * h(x) simplifies to -1.

Option C is correct. The domain requires $\frac{{2 - \phi'(x)}}{{x + 1}} > 0$. After finding $\phi'(x)$ and simplifying, we get $\frac{{(x - 4)(x + 1)}}{{x + 1}} < 0$, which leads to $x < 4$ and $x \neq -1$.

Option D (1399) is correct because by evaluating the function for different ranges of n and summing the results, we arrive at the value 1399.

Option B ($2^{\mathrm{n}}-2$) is correct because it represents the number of surjective mappings from a set A with n elements to a set B with 2 elements, calculated using the inclusion-exclusion principle.

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 ($3f(x)$) is correct because the composition of the functions simplifies to $3 \log \left( {\frac{{1 + x}}{{1 - x}}} \right)$.

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 ($2^n-2$) is correct because it represents the total number of functions from A to B minus the functions that map all elements of A to a single element in B.

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 A (2) is correct because substituting the given function into the equation and simplifying leads to K = 2.

Option B is correct because the derivative of an even function is always an odd function.

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 D ($f(\frac{\pi}{2})=-1$) is correct because substituting $x = \frac{\pi}{2}$ into the simplified function yields -1.

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.

The correct answer is B ($(x - 5)^{1/3}$) because it is the inverse function obtained by solving for x in terms of y and then swapping x and y.

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 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 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.