Functions practice · Medium

Option A ($\left( \frac{{x-7}}{{3}} \right)^{{1 / 2}}$) is correct because $(f \circ g)(x) = 3x^2 + 7$, and its inverse is found by solving for $x$ in terms of $y$, resulting in $x = \sqrt{{\frac{{y-7}}{{3}}}}$.

Option A (one-one and onto) is correct because the function (f-g)(x) maps each real number to a unique real number, covering the entire real number set.

Option C ($\frac{{x + 7}}{{10}}$) is correct because it is the inverse function obtained by swapping x and y in the equation y = 10x - 7 and solving for y.

Option C (8100) is correct because the value is obtained by correctly solving the recurrence relation and substituting the value in the required expression.

Option D (500) is correct because the function is linear, and $f(100) = 100 * f(1) = 500$.

Option B ([-3,-2]) is correct because it represents the domain for which the function is defined, considering the logarithmic and inverse cosine functions.

Option A (2) is correct because after two iterations of the function, f(x+2) simplifies back to f(x), indicating a period of 2.

Option A ($\frac{1}{2}(\sqrt 3 - 1)$) is correct because the area under the curve is found by integrating the composite function gof(x) = cos(x) between the limits alpha and beta, which are the roots of the given quadratic equation.

Option B ($[ - \frac{\pi }{4},\\;\frac{\pi }{4}]$) is correct because the composite function simplifies to sin(2x), which is invertible in the given domain.

The correct answer is A (112). The value is obtained by substituting appropriate values of x and y into the given function and summing the results.

Option C (-4 < x ≤ 1) is correct because it satisfies both the square root condition (5-4x-x² ≥ 0) and the logarithm condition (x+4 > 0).

Option A is correct. To be invertible, the function must be a bijection. By rewriting f(x) as $2\sin(x + \pi/6) + 4$, we find the domain and range that make it a bijection.

Option B $((-\infty,-2) \cup[6, \infty))$ is correct because the expression inside the square root must be positive, leading to the given domain after considering the greatest integer function.

Option C ($(- \infty ,\, - 2] \cup \left[ { - \frac{4}{3},\,\infty } \right)$) is correct because it represents the domain where |g(x)| <= 1, ensuring the composite function fog(x) is defined.

Option B ($0 \le g(2) < 2$) is correct because it provides the correct bounds for the integral based on the given inequalities for f(t).

Option D ($f(-a-b)$) is correct because $\frac{f(-a)}{f(b)} = e^{a+b} = f(-a-b)$ when $f(x) = e^{-x}$.

Option A ($\frac{{x^2-y^2}}{{8}}$) is correct because it is obtained by substituting $x$ and $y$ in terms of $u$ and $v$ in the given functional equation.

Option B ($\frac{1}{2}[f(2 x)+f(2 y)]$ ) is correct because it is the simplified form of the product f(x+y) . f(x-y) using the given function definition.

Option A is correct because the domain is found by solving -1 <= 4/(3+2cos(x)) <= 1, which simplifies to cos(x) >= 1/2.

Option A ($A=\{3,7,8\}$) is correct because these are the first elements of the ordered pairs in $A \times B$.

Option C ($1 \le x < 4$) is correct because it represents the intersection of the domains of the arcsine function and the logarithmic function, ensuring both are defined.

Option C (-6) is correct because for f(x) to equal g(x) for all x, g(-3) must equal f(-3), which is -6.

Option C ([-2,4]) is correct because the range of the cosine function is [-1, 1], and this is used to find the range of the given function.

Option B (${x^2} - 2$ for all $x$ satisfying $\left| x \right| \ge 2$) is correct because the function f(x) is defined as x² - 2 only when |x| >= 2, which is the range of x + 1/x.

Option A (1) is correct because after simplifying f(x), it becomes a constant value of 5/4, and g(5/4) is given as 1, thus (g o f)(x) = 1.

Option A ($\frac{3}{2}$) is correct because finding the domain of the given function involves considering the domains of inverse cosine and inverse sine functions, and their intersection gives the interval $(\frac{1}{2}, 1]$, so $\alpha + \beta = \frac{1}{2} + 1 = \frac{3}{2}$.

Option A (5) is correct because the number of non-empty subsets is 15, which equals 3 * 5, therefore a = 5.

Option B ($(x-1)^3$) is correct because it satisfies the given second derivative, passes through the point (2,1), and has the tangent y=3x-5 at that point.

Option B (Transitive but not symmetric) is correct because if a is a brother of b, and b is a brother of c, then a is a brother of c (transitive), but if a is a brother of b, b is not necessarily a brother of a (b could be a sister).

Option C (4) is correct because by substituting the function $f(x)$ and simplifying the summation, we can find the value of $\alpha$ to be 4.