JEE Mathematics Relations and Functions MCQs Set 11

Answer: (a) reflexive and symmetric

Explanation:
1. All of \( (0,0),(1,1),(2,2),(3,3) \) are in R, so R is reflexive.
2. \( (0,1)\leftrightarrow(1,0) \) and \( (0,3)\leftrightarrow(3,0) \) are both present, so R is symmetric.
3. \( (1,0),(0,3)\in R \) but \( (1,3)\notin R \), so R is not transitive.

Teacher's Note:
1. Test the three properties one at a time and look for a single counterexample to transitivity.

Answer: (d) reflexive, transitive but not symmetric

Explanation:
1. Every n divides itself, so R is reflexive.
2. If \( n\mid m \) and \( m\mid r \) then \( n\mid r \), so R is transitive.
3. \( 3\mid6 \) but \( 6\nmid3 \), so R is not symmetric.

Teacher's Note:
1. Divisibility is a partial order, not an equivalence.

Answer: (a) reflexive and transitive

Explanation:
1. x divides x, so R is reflexive.
2. \( (2,6)\in R \) but \( (6,2)\notin R \), so R is not symmetric.
3. If y is divisible by x and z by y, then z is divisible by x, so R is transitive.

Teacher's Note:
1. Divisibility is reflexive and transitive but not symmetric.

Answer: (c) an equivalence relation

Explanation:
1. \( |a-a|=0 \) is a multiple of 4, so R is reflexive.
2. \( |a-b|=|b-a| \), so R is symmetric.
3. If \( a-b \) and \( b-c \) are multiples of 4, so is \( a-c \); hence R is transitive and therefore an equivalence relation.

Teacher's Note:
1. Congruence modulo 4 is an equivalence relation.

Answer: (c) an equivalence relation

Explanation:
1. Every triangle is similar to itself (reflexive), similarity can be reversed (symmetric) and chains of similarity combine (transitive).
2. So R is an equivalence relation.

Teacher's Note:
1. Similarity, congruence and parallelism are standard equivalence relations.

Answer: (b) R is reflexive and transitive but not symmetric

Explanation:
1. \( (1,1),(2,2),(3,3),(4,4)\in R \), so R is reflexive.
2. \( (1,2)\in R \) but \( (2,1)\notin R \), so R is not symmetric.
3. Checking pairs such as \( (1,3),(3,2)\Rightarrow(1,2)\in R \) shows R is transitive.

Teacher's Note:
1. Make a quick list of all chains \( (a,b),(b,c) \) when checking transitivity.

Answer: (a) 1

Explanation:
1. Reflexive and symmetric forces \( (1,1),(2,2),(3,3),(2,1),(3,1) \) to be in R.
2. With these, \( (2,1),(1,3) \) require \( (2,3) \) for transitivity, so R is not transitive as long as \( (2,3),(3,2) \) are absent.
3. Adding \( (2,3) \) would force \( (3,2) \) by symmetry and make R transitive, so only one relation works.

Teacher's Note:
1. Build the smallest relation satisfying the conditions and see whether anything can be added.

Answer: (b) 2

Explanation:
1. An equivalence relation corresponds to a partition of A in which 1 and 2 lie in the same class.
2. The partitions are \( \{1,2\},\{3\} \) and \( \{1,2,3\} \), so there are 2 such relations.

Teacher's Note:
1. Count partitions instead of relations.

Answer: (a) equivalence relation

Explanation:
1. \( |a-a|=0 \) is even, and \( |a-b|=|b-a| \), so R is reflexive and symmetric.
2. The sum of two even differences \( a-b \) and \( b-c \) is the even difference \( a-c \), so R is transitive.

Teacher's Note:
1. Same parity of a and b is an equivalent description.

Answer: (a) equivalence relation

Explanation:
1. Equal distances from the origin form an equivalence relation: it is reflexive, symmetric and transitive.
2. Its classes are circles centred at the origin.

Teacher's Note:
1. Any relation of the form 'f(P)=f(Q)' is an equivalence relation.

Answer: (d) equivalence relation

Explanation:
1. A line is parallel to itself, parallelism is symmetric and transitive.
2. So R is an equivalence relation.

Teacher's Note:
1. Parallelism is reflexive when a line is considered parallel to itself.

Answer: (d) \( (-3,\infty)-\{-1,-2\} \)

Explanation:
1. The logarithm needs \( x+3>0 \), i.e. \( x>-3 \).
2. The denominator \( x^2+3x+2=(x+1)(x+2) \) must not vanish, so \( x\ne-1,-2 \).

Teacher's Note:
1. Combine the logarithm's domain with the zeros of the denominator.

Answer: (d) none of these

Explanation:
1. \( \sec^{-1} \) needs \( |\log x|\ge1 \), so \( x\ge e \) or \( 0<x\le\frac1e \).
2. The values in options (a), (b), (c) are not all in D (for instance \( e-1\approx1.72 \) and \( 1 \) are outside D, and (a) includes negative numbers).

Teacher's Note:
1. \( \sec^{-1}u \) is defined only for \( |u|\ge1 \).

Answer: (a) \( x\in\{0\} \)

Explanation:
1. We need \( 0\le\log(1-x^2)\le1 \), i.e. \( 1\le1-x^2\le e \).
2. \( 1\le1-x^2 \) gives \( x^2\le0 \), so \( x=0 \); and then \( 1-0=1\le e \) holds.

Teacher's Note:
1. The argument of f must lie in the domain \( [0,1] \) of f.

Answer: (c) \( \left(\frac14,1\right) \)

Explanation:
1. Work from the outside in: \( \log_{0.5}(\log_{0.25}x)>0 \) needs \( 0<\log_{0.25}x<1 \).
2. \( \log_{0.25}x>0 \) gives \( x<1 \), and \( \log_{0.25}x<1 \) gives \( x>\frac14 \).
3. So the domain is \( \left(\frac14,1\right) \).

Teacher's Note:
1. For a base less than 1 the logarithm is positive when its argument is less than 1.

Answer: (d) none of these

Explanation:
1. We need \( \tan x\ne1 \), i.e. \( x\ne n\pi+\frac\pi4 \), and \( \tan x \) itself must exist, i.e. \( x\ne n\pi+\frac\pi2 \).
2. Both sets must be excluded, which none of the first three options does.

Teacher's Note:
1. Exclude both the zeros of the denominator and the points where tan x is undefined.

Answer: (b) \( \bigcup_{n\in I}(2n\pi,(2n+1)\pi) \)

Explanation:
1. We need \( \frac1{\sin x}\ge1 \), which holds exactly when \( 0<\sin x\le1 \).
2. \( \sin x>0 \) for \( x\in(2n\pi,(2n+1)\pi) \).

Teacher's Note:
1. Positive sine implies \( \frac1{\sin x}\ge1 \).

Answer: (c) \( \left[-\frac{\sqrt3}2,0\right)\cup\left(0,\frac{\sqrt3}2\right] \)

Explanation:
1. \( \sin^{-1} \) needs \( [2-4x^2]\in\{-1,0,1\} \), i.e. \( -1\le2-4x^2<2 \).
2. \( 2-4x^2<2 \) gives \( x\ne0 \), and \( 2-4x^2\ge-1 \) gives \( x^2\le\frac34 \).
3. So \( x\in\left[-\frac{\sqrt3}2,0\right)\cup\left(0,\frac{\sqrt3}2\right] \).

Teacher's Note:
1. Greatest-integer values must themselves lie in \( [-1,1] \).

Answer: (d) \( -\infty<x<1 \)

Explanation:
1. \( 2^y=2-2^x \) must be positive, so \( 2^x<2 \).
2. Hence \( x<1 \), and the domain is \( (-\infty,1) \).

Teacher's Note:
1. A power of 2 is always positive.

Answer: (a) \( [1,2] \)

Explanation:
1. \( 2x-x^2\ge0 \) gives \( 0\le x\le2 \).
2. \( [x]\ne0 \) removes \( 0\le x<1 \).
3. The domain is \( [1,2] \).

Teacher's Note:
1. The greatest integer function is zero on [0,1).

Answer: (c) \( \left(1,\frac73\right] \)

Explanation:
1. \( f(x)=1+\frac1{x^2+x+1} \) and \( x^2+x+1\ge\frac34 \) with all larger values possible.
2. So \( 0<\frac1{x^2+x+1}\le\frac43 \) and the range is \( \left(1,\frac73\right] \).

Teacher's Note:
1. Write the fraction as 1 plus something and bound the denominator.

Answer: (d) \( \{0,1,2\} \)

Explanation:
1. For \( 0\le x<\sqrt2 \) (and \( x=1 \), \( x=2 \)): \( f=0 \).
2. For \( \sqrt2\le x<\sqrt3 \): \( [x^2]=2 \), \( [x]=1 \), so \( f=1 \).
3. For \( \sqrt3\le x<2 \): \( [x^2]=3 \), \( [x]=1 \), so \( f=2 \). The set is \( \{0,1,2\} \).

Teacher's Note:
1. Split the interval at the points where \( x^2 \) crosses an integer.

Answer: (d) \( (0,1] \)

Explanation:
1. The minimum of \( ax^2+2bx+1 \) is \( 1-\frac{b^2}{1+b^2}=\frac1{1+b^2} \).
2. As \( b^2 \) varies over \( [0,\infty) \), \( m(b) \) takes every value in \( (0,1] \).

Teacher's Note:
1. Use the vertex formula \( c-\frac{B^2}{4a} \).

Answer: (d) none of these

Explanation:
1. The base \( \sin x \) lies in \( (0,1) \), so \( \log_{\sin x}\tan x\ge0 \) means \( 0<\tan x\le1 \).
2. This gives \( 0<x\le\frac\pi4 \), which is not among the first three options.

Teacher's Note:
1. For a base between 0 and 1 the logarithm is non-negative when the argument is at most 1.

Answer: (c) \( \{0,-1\} \)

Explanation:
1. A function and its inverse meet on the line \( y=x \), so solve \( f(x)=x \).
2. \( x^2+2x=x \) gives \( x=0 \) or \( x=-1 \), both \( \ge-1 \).

Teacher's Note:
1. For an increasing function, \( f(x)=f^{-1}(x) \) exactly when \( f(x)=x \).

Answer: (b) neither one-one nor onto

Explanation:
1. \( f(2)=f\left(\frac12\right)=\frac25 \), so f is not one-one.
2. The values lie in \( \left[-\frac12,\frac12\right] \), so f is not onto R.

Teacher's Note:
1. Look for two inputs with the same output and for the maximum value.

Answer: (c) 24

Explanation:
1. Choose images for the three elements without repetition.
2. \( 4\times3\times2=24 \).

Teacher's Note:
1. Injective maps from a 3-set to a 4-set number \( ^4P_3 \).

Answer: (c) \( \{4,-4\},\phi \)

Explanation:
1. \( x^2+1=17 \) gives \( x=\pm4 \).
2. \( x^2+1=-3 \) has no real solution, so the pre-image of \( -3 \) is empty.

Teacher's Note:
1. A pre-image is the set of all x mapped to the given value.

Answer: (a) onto

Explanation:
1. \( f(1)=f(2)=1 \), so f is not one-one.
2. Every \( m\in N \) equals \( f(2m) \), so f is onto.

Teacher's Note:
1. Show onto by exhibiting a pre-image for a general m.

Answer: (c) one-one and onto

Explanation:
1. \( \frac{x_1-2}{x_1-3}=\frac{x_2-2}{x_2-3} \) simplifies to \( x_1=x_2 \), so f is one-one.
2. Solving \( y=\frac{x-2}{x-3} \) gives \( x=\frac{3y-2}{y-1} \), defined for every \( y\ne1 \), so f is onto B.

Teacher's Note:
1. Solve for x in terms of y to prove ontoness.

Answer: (a) 14

Explanation:
1. There are \( 2^4=16 \) functions from E to F.
2. Two of them are constant (not onto), so \( 16-2=14 \) are onto.

Teacher's Note:
1. Subtract the constant functions from all functions.

Answer: (a) injective

Explanation:
1. On \( \left[0,\frac\pi2\right] \), \( \frac x2\in\left[0,\frac\pi4\right] \) where sine is strictly increasing, so f is one-one.
2. Its range is \( \left[0,\frac1{\sqrt2}\right]\ne[0,\infty) \), so f is not onto.

Teacher's Note:
1. Compare the range with the stated codomain.