JEE Mathematics Relations and Functions MCQs Set 14

Answer: (a) 18

Explanation:
1. \( p(x) = (a - a_1)x^2 + (b - b_1)x + (c - c_1) \) is a quadratic with the single root \( x = -1 \), so it has a repeated root: \( p(x) = k(x+1)^2 \) with \( k = a - a_1 \ne 0 \).
2. \( p(-2) = k(-1)^2 = k = 2 \).
3. \( p(2) = 2 \cdot 3^2 = 18 \). Answer: (a)

Teacher's Note:
1. A quadratic with only one real root must have that root repeated, so it is a constant times a perfect square.
2. Use the given value to find the constant.

Answer: (b) Statement I is true, Statement II is true; Statement II is a correct explanation of Statement I

Explanation:
1. On \( x \ge 1 \), f is strictly increasing with range \( [1,\infty) \), so it is a bijection onto \( [1, \infty) \). Solving \( y = (x-1)^2 + 1 \) with \( x \ge 1 \) gives \( x = 1 + \sqrt{y-1} \), so \( f^{-1}(x) = 1 + \sqrt{x-1} \). Statement II is true.
2. \( f(x) = f^{-1}(x) \) means \( f(x) = x \) (the graph of an increasing function meets its inverse on the line \( y = x \)): \( (x-1)^2 + 1 = x \Rightarrow (x-1)(x-2) = 0 \), so \( x = 1, 2 \). Statement I is true.
3. Statement II supplies the formula for \( f^{-1} \) used to determine the set in Statement I, so it is a correct explanation. Answer: (b)

Teacher's Note:
1. For an increasing function, f and its inverse meet only on the line \( y=x \).
2. Always state the domain and codomain before calling a function a bijection.

Answer: (c) S is an equivalence relation but R is not an equivalence relation

Explanation:
1. Relation R: reflexive (take \( w = 1 \)) but not symmetric, because \( 0 = w \cdot 1 \) with \( w = 0 \), yet \( 1 = w \cdot 0 \) has no solution. So R is not an equivalence relation.
2. Relation S: \( \frac{m}{n} S \frac{p}{q} \) means \( qm = pn \), i.e. \( \frac{m}{n} = \frac{p}{q} \) (equality of the fractions). Equality is reflexive, symmetric and transitive.
3. So S is an equivalence relation and R is not. Answer: (c)

Teacher's Note:
1. Relation S is just equality of rational numbers written in fraction form, so it is automatically an equivalence relation.
2. Check symmetry of R with the pair \( (0,1) \) and \( (1,0) \).

Answer: (b) \( B = C \)

Explanation:
1. Take any \( x \in B \). If \( x \in A \), then \( x \in A \cap B = A \cap C \), so \( x \in C \). If \( x \notin A \), then \( x \in A \cup B = A \cup C \) forces \( x \in C \).
2. So \( B \subseteq C \). By the same argument with B and C swapped, \( C \subseteq B \).
3. Hence \( B = C \). Answer: (b)

Teacher's Note:
1. Equal intersection and equal union with a common set A force the other two sets to be equal (cancellation law).
2. Split into the cases x in A and x not in A.

Answer: (c) f is one-one and onto \( \mathbf{R} \)

Explanation:
1. \( f'(x) = 3x^2 + 5 \gt 0 \) for all x, so f is strictly increasing and therefore one-one.
2. f is continuous with \( f(x) \to -\infty \) as \( x \to -\infty \) and \( f(x) \to \infty \) as \( x \to \infty \), so it takes every real value.
3. So f is one-one and onto. Answer: (c)

Teacher's Note:
1. A strictly monotonic function is one-one; a continuous function with limits \( \pm\infty \) at the ends is onto \( \mathbf{R} \).
2. Every odd-degree polynomial is onto \( \mathbf{R} \).

Answer: (a) T is an equivalence relation on R but S is not

Explanation:
1. S: \( (1, 2) \in S \) but \( (2, 1) \notin S \), so S is not symmetric (it is not even reflexive: \( (5,5) \notin S \)). So S is not an equivalence relation.
2. T: \( x - x = 0 \) is an integer (reflexive). If \( x - y \in Z \) then \( y - x \in Z \) (symmetric). If \( x - y \) and \( y - z \) are integers then \( x - z = (x-y) + (y-z) \) is an integer (transitive).
3. So T is an equivalence relation and S is not. Answer: (a)

Teacher's Note:
1. S is just a line segment, so it cannot be reflexive on all of R.
2. T is the relation 'differ by an integer' and is a standard equivalence relation.

Answer: (a) \( f^{-1}(y) = \frac{y-3}{4} \)

Explanation:
1. f is strictly increasing, so it is one-one. Its codomain Y is exactly its range, so it is onto. Hence f is invertible.
2. Put \( y = 4x + 3 \) and solve for x: \( x = \frac{y-3}{4} \).
3. So \( f^{-1}(y) = \frac{y-3}{4} \). Answer: (a)

Teacher's Note:
1. A function is invertible exactly when it is both one-one and onto.
2. To find the inverse, solve \( y=f(x) \) for x and swap the roles.

Answer: (a) reflexive, symmetric and not transitive

Explanation:
1. Reflexive: every word has a letter in common with itself.
2. Symmetric: if x shares a letter with y, then y shares it with x.
3. Not transitive: CAT and TOY share T, TOY and YOU share O and Y, but CAT and YOU share no letter.
4. So R is reflexive and symmetric but not transitive. Answer: (a)

Teacher's Note:
1. Transitivity is the property most often broken by relations like 'has something in common'.
2. One explicit counter-example (CAT, TOY, YOU) is enough to disprove transitivity.

Answer: (d) reflexive and transitive only

Explanation:
1. Reflexive: \( (3,3), (6,6), (9,9), (12,12) \in R \).
2. Not symmetric: \( (6, 12) \in R \) but \( (12, 6) \notin R \).
3. Transitive: the pairs with different entries are \( (6,12), (3,9), (3,12), (3,6) \). The only chains are \( (3,6),(6,12) \Rightarrow (3,12) \in R \); others hold trivially.
4. So R is reflexive and transitive only. Answer: (d)

Teacher's Note:
1. Check reflexivity first, then symmetry, then transitivity by looking at chains of pairs.
2. This relation is the partial order 'divides' restricted to some pairs.

Answer: (a) \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \)

Explanation:
1. For \( |x| \lt 1 \), \( \tan^{-1}\left(\frac{2x}{1-x^2}\right) = 2\tan^{-1}x \).
2. As \( x \in (-1, 1) \), \( \tan^{-1}x \in \left(-\frac{\pi}{4}, \frac{\pi}{4}\right) \), so \( f(x) = 2\tan^{-1}x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \).
3. f is strictly increasing (one-one) and its range is \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \), so f is onto this B. Answer: (a)

Teacher's Note:
1. The identity \( \tan^{-1}\frac{2x}{1-x^2}=2\tan^{-1}x \) holds only for \( |x|<1 \).
2. For onto, set the codomain equal to the range.

Answer: (d) \( -f(x) \)

Explanation:
1. Put \( x = y = 0 \): \( f(0) = f(0)^2 - f(a)^2 \), so \( 1 = 1 - f(a)^2 \) and \( f(a) = 0 \).
2. \( f(2a - x) = f(a - (x - a)) = f(a)f(x - a) - f(a - a)f(a + (x - a)) \), using the given equation with \( x \to a \) and \( y \to x - a \).
3. This is \( 0 \cdot f(x-a) - f(0)f(x) = -f(x) \). Answer: (d)

Teacher's Note:
1. Substituting 0 for x and y gives the key value \( f(a)=0 \).
2. Write \( 2a-x \) as a difference of the form \( p-q \) so that the equation can be applied.

Answer: (c) not symmetric

Explanation:
1. Not a function: 2 is related to both 4 and 3, i.e. \( (2,4), (2,3) \in R \).
2. Not transitive: \( (1,3), (3,1) \in R \) but \( (1,1) \notin R \).
3. Not symmetric: \( (2,3) \in R \) but \( (3,2) \notin R \). So statement (c) is true.
4. Not reflexive: \( (1,1), (2,2), (3,3), (4,4) \notin R \). Answer: (c)

Teacher's Note:
1. Check each property with one explicit pair from the list.
2. A relation is a function only if each first element appears in exactly one pair.

Answer: (b) \( f(2+x) = f(2-x) \)

Explanation:
1. Symmetry about the vertical line \( x = 2 \) means that points equally far to the left and right of 2 have the same function value.
2. The points at distance x from 2 are \( 2 + x \) and \( 2 - x \), so \( f(2 + x) = f(2 - x) \). Answer: (b)

Teacher's Note:
1. Symmetry about \( x=a \) is expressed as \( f(a+x)=f(a-x) \); symmetry about the y-axis is the special case \( a=0 \).
2. Option (d) would force \( f\equiv0 \).

Answer: (b) [2, 3)

Explanation:
1. \( \sin^{-1}(x-3) \) needs \( -1 \le x - 3 \le 1 \), i.e. \( 2 \le x \le 4 \).
2. The denominator needs \( 9 - x^2 \gt 0 \), i.e. \( -3 \lt x \lt 3 \).
3. Intersection: \( 2 \le x \lt 3 \), i.e. \( [2, 3) \). Answer: (b)

Teacher's Note:
1. The endpoint 3 is excluded because the denominator becomes zero there.
2. Intersect the conditions from the numerator and the denominator.

Answer: (d) \( \frac{7n(n+1)}{2} \)

Explanation:
1. Repeated use of the equation gives \( f(r) = r f(1) = 7r \) for every natural number r.
2. \( \sum_{r=1}^{n} f(r) = 7(1 + 2 + \dots + n) = \frac{7n(n+1)}{2} \). Answer: (d)

Teacher's Note:
1. Additive functions satisfy \( f(rx)=rf(x) \) for natural r.
2. The sum of the first n natural numbers is \( \frac{n(n+1)}{2} \).

Answer: (d) \( (-1, 0) \cup (1, 2) \cup (2, \infty) \)

Explanation:
1. \( x^3 - x \gt 0 \) means \( x(x-1)(x+1) \gt 0 \), so \( x \in (-1, 0) \cup (1, \infty) \).
2. \( 4 - x^2 \ne 0 \) means \( x \ne \pm 2 \), which removes \( x = 2 \) from \( (1, \infty) \).
3. Domain \( = (-1, 0) \cup (1, 2) \cup (2, \infty) \). Answer: (d)

Teacher's Note:
1. Combine the restrictions from the rational part and the logarithm.
2. The wavy-curve method gives the sign pattern of \( x(x-1)(x+1) \).

Answer: (b) an odd function

Explanation:
1. \( f(-x) = \log\left(-x + \sqrt{x^2+1}\right) \).
2. \( f(x) + f(-x) = \log\left[\left(\sqrt{x^2+1} + x\right)\left(\sqrt{x^2+1} - x\right)\right] = \log\left(x^2 + 1 - x^2\right) = \log 1 = 0 \).
3. So \( f(-x) = -f(x) \) and f is odd. Answer: (b)

Teacher's Note:
1. Test parity by computing \( f(x)+f(-x) \); a zero sum shows an odd function.
2. This function is the inverse hyperbolic sine.

Answer: (b) \( \pi \)

Explanation:
1. \( \sin^2\theta = \frac{1 - \cos 2\theta}{2} = \frac{1}{2} - \frac{1}{2}\cos 2\theta \).
2. \( \cos 2\theta \) has period \( \frac{2\pi}{2} = \pi \). Answer: (b)

Teacher's Note:
1. Reduce even powers of sine or cosine to a cosine of a multiple angle.
2. The period of \( \cos k\theta \) is \( \frac{2\pi}{k} \).

Answer: (a) [1, 9]

Explanation:
1. The inverse sine needs \( -1 \le \log_3\left(\frac{x}{3}\right) \le 1 \).
2. Raising 3 to each part: \( 3^{-1} \le \frac{x}{3} \le 3 \), i.e. \( 1 \le x \le 9 \).
3. The logarithm is defined for these x (they are positive). Domain \( = [1, 9] \). Answer: (a)

Teacher's Note:
1. Convert the logarithmic inequality into exponential form to remove the logarithm.
2. Check that the argument of the logarithm is positive in the result.

Answer: (b) \( \frac{\pi}{2} \)

Explanation:
1. \( \sin^4 x + \cos^4 x = 1 - 2\sin^2 x\cos^2 x = 1 - \frac{1}{2}\sin^2 2x \).
2. \( \sin^2 2x = \frac{1 - \cos 4x}{2} \), so \( f(x) = \frac{3}{4} + \frac{1}{4}\cos 4x \).
3. The period of \( \cos 4x \) is \( \frac{2\pi}{4} = \frac{\pi}{2} \). Answer: (b)

Teacher's Note:
1. Rewrite higher powers using double-angle identities until only a single cosine remains.
2. Check that no smaller positive number works as a period.

Answer: (a) [1, 4]

Explanation:
1. The square root needs \( \log_{10}\left(\frac{5x - x^2}{4}\right) \ge 0 \), i.e. \( \frac{5x - x^2}{4} \ge 1 \) (this also makes the argument of the logarithm positive).
2. \( 5x - x^2 \ge 4 \Rightarrow x^2 - 5x + 4 \le 0 \Rightarrow (x-1)(x-4) \le 0 \).
3. So \( 1 \le x \le 4 \). Answer: (a)

Teacher's Note:
1. Work from the outside in: the root requires the log to be non-negative, and that already forces the log's argument to be positive.
2. Options (b) and (d) are not even valid intervals.