Sets, Relations and Functions Questions and Answers

Answer: E

\(f^{-1}(g(x))\) is \(x^{3} - 1\).

Since \(f(x) = x + 1\), its inverse undoes the addition: \(f^{-1}(y) = y - 1\). Put \(y = g(x) = x^{3}\).

\[ f^{-1}(g(x)) = x^{3} - 1 \]

\(x^{3} + 2\) applies f a second time instead of its inverse, and \(x - 1\) uses \(f^{-1}(x)\) without composing with g. The other options mix the two functions in other ways. A check at x = 2: \(g(2) = 8\) and \(f^{-1}(8) = 7 = 2^{3} - 1\).

Answer: D

The range is {-1, 0, 1}.

\(\cos x\) takes every value in \([-1, 1]\). The greatest integer function maps these values as follows.

\[ \begin{aligned} \cos x = 1 &\Rightarrow [\cos x] = 1 \\ 0 \le \cos x \lt 1 &\Rightarrow [\cos x] = 0 \\ -1 \le \cos x \lt 0 &\Rightarrow [\cos x] = -1 \end{aligned} \]

All three values occur (at x = 0, \(\frac{\pi}{2}\) and \(\pi\)), so the range is the set {-1, 0, 1}. The intervals (-1, 1) and [-1, 1] are wrong because \([\cos x]\) takes only integer values, and the two-element sets each miss one of the three.

Answer: C

The domain is [-1, 2).

The square root needs \(\frac{x + 1}{2 - x} \ge 0\), and the denominator cannot be zero.

The fraction is zero at x = -1 (allowed) and undefined at x = 2 (excluded). It is non-negative when numerator and denominator have the same sign, which happens for \(-1 \le x \lt 2\).

[-1, 2] wrongly includes x = 2, where the denominator is zero. [1, 2) misplaces the zero of the numerator, and the intervals ending at 3 do not match any sign change.

Answer: D

The range is \(\left[\frac{1}{81}, \frac{1}{9}\right]\).

Since \(-1 \le \sin x \le 1\), the exponent \(3 + \sin x\) runs from 2 to 4. The base \(\frac{1}{3}\) is less than 1, so the function decreases as the exponent grows.

\[ \begin{aligned} \min f &= \left(\tfrac{1}{3}\right)^{4} = \tfrac{1}{81} \\ \max f &= \left(\tfrac{1}{3}\right)^{2} = \tfrac{1}{9} \end{aligned} \]

A positive base raised to any power is positive, so the options with negative endpoints are impossible. \(\frac{1}{3}\) would need the exponent 1, which \(3 + \sin x\) never reaches.

Answer: D

The domain is \((-\infty, -5] \cup [3, \infty)\).

The square root needs a non-negative argument:

\[ x^{2} + 2x - 15 = (x + 5)(x - 3) \ge 0 \]

The product is non-negative outside the roots and zero at them, so \(x \le -5\) or \(x \ge 3\). Both endpoints are included because \(\sqrt{0} = 0\) is defined.

The options with an open bracket at \(-5\) or 3 wrongly exclude a root, and \([-5, 3]\) is exactly where the expression is negative or zero.

Check with a value: at \(x = 0\) the expression is −15, so 0 must be excluded, and only the options without the interval around 0 survive.

Answer: C

The sets are equal: A = B.

From A = B ∩ C, every element of A is in B, so A is a subset of B. From B = C ∩ A, every element of B is in A, so B is a subset of A. Two sets each contained in the other are equal.

'A is a proper subset of B' and 'A contains B' each give only one of the two inclusions, and A cannot sit inside the complement of B unless A is empty.

Answer: B

A - B equals \(A \cap \overline{B}\).

The difference A - B is the set of elements that are in A but not in B, that is, in A and in the complement of B. Here it is the multiples of 3 that are not multiples of 5, such as 3, 6, 9, 12, but not 15.

\(\overline{A} \cap B\) is B - A, the multiples of 5 that are not multiples of 3. \(\overline{A} \cap \overline{B}\) contains numbers in neither set, and \(\overline{A \cap B}\) is everything except the multiples of 15.

Answer: A

The function \(f(x)=\frac{1-x}{1+x}\) is its own inverse.

A function is its own inverse when f(f(x)) = x. Substituting:

\[ \begin{aligned} f(f(x)) &= \frac{1 - \frac{1-x}{1+x}}{1 + \frac{1-x}{1+x}} \\ &= \frac{(1+x) - (1-x)}{(1+x) + (1-x)} \\ &= \frac{2x}{2} = x \end{aligned} \]

5^{log x} and 2^x(x-1) do not return x when applied twice; the second is not even one-to-one. Since option A works, 'None of these' is wrong.

Answer: D

The one that is not a type of relation is Non-linear.

Relations on a set are classified by properties such as reflexive (every element related to itself), symmetric (a related to b implies b related to a) and transitive (a to b and b to c imply a to c). A relation with all three is an equivalence relation.

'Non-linear' describes the shape of a function or equation, as opposed to a linear one; it is not a property used to classify relations.

Answer: C

The function is neither odd nor even.

Test with f(-x):

\[ f(-x) = x^{2} - 4x + 4 \]

This is not equal to f(x) = x² + 4x + 4, so f is not even, and it is not equal to -f(x) = -x² - 4x - 4, so f is not odd. Since f(x) = (x + 2)², its graph is symmetric about x = -2, not about the y-axis. A polynomial of degree 2 is not periodic.