JEE Mathematics Relations and Functions MCQs Set 09

Answer: (b) not symmetric

Explanation:
1. \((2,3)\in R\) but \((3,2)\notin R\), so \(R\) is not symmetric.
2. It is not reflexive (no \((1,1)\)), not transitive (\((4,2),(2,3)\in R\) but \((4,3)\notin R\)) and not a function (the element \(2\) has two images).

Teacher's Note:
1. For a relation check reflexive, symmetric and transitive separately, and use a small counterexample to disprove a property.
2. For a function decide one-one (no two inputs share an output) and onto (range equals codomain) before finding any inverse or composition.

Answer: (c) symmetric

Explanation:
1. \(x^2+y^2=y^2+x^2\), so \((x,y)\in P\Rightarrow(y,x)\in P\): symmetric.
2. Not reflexive (\((x,x)\in P\) only if \(2x^2=1\)); not transitive (\((1,0),(0,1)\in P\) but \((1,1)\notin P\)); not anti-symmetric (\((0.6,0.8)\) and \((0.8,0.6)\)).

Teacher's Note:
1. For a relation check reflexive, symmetric and transitive separately, and use a small counterexample to disprove a property.
2. For a function decide one-one (no two inputs share an output) and onto (range equals codomain) before finding any inverse or composition.

Answer: (a) \(1\)

Explanation:
1. From (ii), \(\dfrac1n\sum f^{-1}(x_i)=\bar x\) where \(\bar x=\dfrac{\sum x_i}n\). Then (i) gives \(f(\bar x)=\bar x\).
2. \(\bar x^2+3\bar x-3=\bar x\Rightarrow\bar x^2+2\bar x-3=0\Rightarrow\bar x=1\) (as \(\bar x>0\)).

Teacher's Note:
1. For a relation check reflexive, symmetric and transitive separately, and use a small counterexample to disprove a property.
2. For a function decide one-one (no two inputs share an output) and onto (range equals codomain) before finding any inverse or composition.

Answer: (c) \([-1,\infty)\) and \([0,\infty)\)

Explanation:
1. \((f\circ g)(x)=|e^x-1|\) for \(x\in[-1,\infty)\) (and \(e^x>0\) lies in \(R^+\)).
2. It decreases on \([-1,0]\) from \(1-\dfrac1e\) to \(0\) and increases on \([0,\infty)\) to \(\infty\), so the range is \([0,\infty)\).

Teacher's Note:
1. For a relation check reflexive, symmetric and transitive separately, and use a small counterexample to disprove a property.
2. For a function decide one-one (no two inputs share an output) and onto (range equals codomain) before finding any inverse or composition.

Answer: (d) \((f\circ g)(9)=683\)

Explanation:
1. \(g(9)=85\) and \(f(85)=8\cdot85+3=683\).
2. \(g(-3)=0\), \(f(0)=-8\) so \((f\circ g)(-3)=-8\); \(f(0)=-8\), \(g(-8)=8\) so \((g\circ f)(0)=8\); \(f(6)=29\), \(g(29)=845\) so \((g\circ f)(6)=845\).

Teacher's Note:
1. For a relation check reflexive, symmetric and transitive separately, and use a small counterexample to disprove a property.
2. For a function decide one-one (no two inputs share an output) and onto (range equals codomain) before finding any inverse or composition.

Answer: (a) \(-1+\sqrt{x+1}\)

Explanation:
1. \(y=(x+1)^2-1\) gives \(x+1=\pm\sqrt{y+1}\).
2. Since \(x\ge-1\), \(x+1\ge0\), so \(x=-1+\sqrt{y+1}\).

Teacher's Note:
1. For a relation check reflexive, symmetric and transitive separately, and use a small counterexample to disprove a property.
2. For a function decide one-one (no two inputs share an output) and onto (range equals codomain) before finding any inverse or composition.

Answer: (c) \(\{(8,11),(10,13)\}\)

Explanation:
1. \(x=11\to y=8\), \(x=12\to y=9\notin\{8,10,12\}\), \(x=13\to y=10\), so \(R=\{(11,8),(13,10)\}\).
2. \(R^{-1}=\{(8,11),(10,13)\}\).

Teacher's Note:
1. For a relation check reflexive, symmetric and transitive separately, and use a small counterexample to disprove a property.
2. For a function decide one-one (no two inputs share an output) and onto (range equals codomain) before finding any inverse or composition.

Answer: (d) reflexive but neither symmetric nor transitive

Explanation:
1. \(x^2-4xy+3y^2=(x-y)(x-3y)\), so \(xRy\) iff \(x=y\) or \(x=3y\). Reflexive since \(x=x\).
2. \((3,1)\in R\) but \((1,3)\notin R\); \((9,3),(3,1)\in R\) but \((9,1)\notin R\). Not symmetric and not transitive.

Teacher's Note:
1. For a relation check reflexive, symmetric and transitive separately, and use a small counterexample to disprove a property.
2. For a function decide one-one (no two inputs share an output) and onto (range equals codomain) before finding any inverse or composition.

Answer: (a) \(360\)

Explanation:
1. A one-to-one function assigns distinct images to the \(4\) elements.
2. Number \(={}^6P_4=6\cdot5\cdot4\cdot3=360\).

Teacher's Note:
1. For a relation check reflexive, symmetric and transitive separately, and use a small counterexample to disprove a property.
2. For a function decide one-one (no two inputs share an output) and onto (range equals codomain) before finding any inverse or composition.

Answer: (d) Range of \(R=\{5\}\)

Explanation:
1. \(|1-4|=3\), \(|4-9|=5\) are allowed but \(|1-9|=8\) is not, so \(R=\{(1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2)\}\).
2. This is symmetric, so \(R^{-1}=R\); the domain and range are both \(\{1,2,3\}\). Hence 'range \(=\{5\}\)' is false.

Teacher's Note:
1. For a relation check reflexive, symmetric and transitive separately, and use a small counterexample to disprove a property.
2. For a function decide one-one (no two inputs share an output) and onto (range equals codomain) before finding any inverse or composition.

Answer: (c) \(5\)

Explanation:
1. Each equivalence relation corresponds to a partition of \(S\).
2. Partitions of a \(3\)-element set: \(1\) (all together) \(+3\) (one pair, one single) \(+1\) (all separate) \(=5\).

Teacher's Note:
1. For a relation check reflexive, symmetric and transitive separately, and use a small counterexample to disprove a property.
2. For a function decide one-one (no two inputs share an output) and onto (range equals codomain) before finding any inverse or composition.

Answer: (b) \(f\) is one-one into

Explanation:
1. \(f(x_1)=f(x_2)\) gives \((x_1-m)(x_2-n)=(x_2-m)(x_1-n)\), so \((m-n)(x_1-x_2)=0\) and \(x_1=x_2\): one-one.
2. The value \(1\) is never attained (it would need \(m=n\)), so \(f\) is not onto \(R\): it is into.

Teacher's Note:
1. For a relation check reflexive, symmetric and transitive separately, and use a small counterexample to disprove a property.
2. For a function decide one-one (no two inputs share an output) and onto (range equals codomain) before finding any inverse or composition.

Answer: (b) \(n^m\)

Explanation:
1. Each of the \(m\) elements of \(A\) can be sent to any of the \(n\) elements of \(B\).
2. Total \(=n\times n\times\dots\times n=n^m\).

Teacher's Note:
1. For a relation check reflexive, symmetric and transitive separately, and use a small counterexample to disprove a property.
2. For a function decide one-one (no two inputs share an output) and onto (range equals codomain) before finding any inverse or composition.

Answer: (d) reflexive and symmetric

Explanation:
1. \(1+a^2>0\) gives reflexive; \(1+ab=1+ba\) gives symmetric.
2. Not transitive: \((1,-0.5)\) and \((-0.5,-3)\) are in \(R\) (\(1-0.5>0\), \(1+1.5>0\)) but \(1+1\cdot(-3)=-2<0\).

Teacher's Note:
1. For a relation check reflexive, symmetric and transitive separately, and use a small counterexample to disprove a property.
2. For a function decide one-one (no two inputs share an output) and onto (range equals codomain) before finding any inverse or composition.

Answer: (a) \(2\)

Explanation:
1. Let \(f(x)=ax+b\). Then \(a(x+ax+b)+b=x+ax+b\) gives \(a^2=1\) and \(ab=0\).
2. So \(b=0\) and \(a=\pm1\): \(f(x)=x\) or \(f(x)=-x\), i.e. \(2\) functions.

Teacher's Note:
1. For a relation check reflexive, symmetric and transitive separately, and use a small counterexample to disprove a property.
2. For a function decide one-one (no two inputs share an output) and onto (range equals codomain) before finding any inverse or composition.

Answer: (c) \(f(y)<f(x)<f(z)\)

Explanation:
1. Check the six permutations; exactly one statement is true only for \(f(x)=2\), \(f(y)=1\), \(f(z)=3\) (then \(f(x)\ne2\) is false, \(f(y)=2\) is false, \(f(z)\ne1\) is true).
2. So \(f(y)=1<f(x)=2<f(z)=3\).

Teacher's Note:
1. For a relation check reflexive, symmetric and transitive separately, and use a small counterexample to disprove a property.
2. For a function decide one-one (no two inputs share an output) and onto (range equals codomain) before finding any inverse or composition.

Answer: (b) \([-1,3]\)

Explanation:
1. \(\sin x-\sqrt3\cos x=2\sin\left(x-\dfrac\pi3\right)\), with range \([-2,2]\).
2. Adding \(1\) gives \([-1,3]\).

Teacher's Note:
1. For a relation check reflexive, symmetric and transitive separately, and use a small counterexample to disprove a property.
2. For a function decide one-one (no two inputs share an output) and onto (range equals codomain) before finding any inverse or composition.

Answer: (a) \(R_1=\{u:-1\le u\le1\}\), \(R_2=\{v:-\infty<v\le0\}\)

Explanation:
1. \((f\circ g)(x)=\sin(\log_e|x|)\); \(\log_e|x|\) takes every real value, so the range is \([-1,1]\).
2. \((g\circ f)(x)=\log_e|\sin x|\); \(|\sin x|\in(0,1]\), so the range is \((-\infty,0]\).

Teacher's Note:
1. For a relation check reflexive, symmetric and transitive separately, and use a small counterexample to disprove a property.
2. For a function decide one-one (no two inputs share an output) and onto (range equals codomain) before finding any inverse or composition.

Answer: (b) an equivalence relation

Explanation:
1. Dividing by \(abcd\): \(\dfrac{b+c}{bc}=\dfrac{a+d}{ad}\), i.e. \(\dfrac1b+\dfrac1c=\dfrac1a+\dfrac1d\), i.e. \(\dfrac1b-\dfrac1a=\dfrac1d-\dfrac1c\).
2. Equality of the quantity \(\dfrac1b-\dfrac1a\) gives reflexive, symmetric and transitive: an equivalence relation.

Teacher's Note:
1. For a relation check reflexive, symmetric and transitive separately, and use a small counterexample to disprove a property.
2. For a function decide one-one (no two inputs share an output) and onto (range equals codomain) before finding any inverse or composition.

Answer: (c) \(2^n-2\)

Explanation:
1. Total functions \(=2^n\).
2. Remove the two constant functions (not onto): \(2^n-2\).

Teacher's Note:
1. For a relation check reflexive, symmetric and transitive separately, and use a small counterexample to disprove a property.
2. For a function decide one-one (no two inputs share an output) and onto (range equals codomain) before finding any inverse or composition.

Answer: (a) \(\{1,2,3\}\)

Explanation:
1. We need \(x-3\ge0\) and \(7-x\ge x-3\), so \(x\in\{3,4,5\}\).
2. \(f(3)={}^4P_0=1\), \(f(4)={}^3P_1=3\), \(f(5)={}^2P_2=2\), so the range is \(\{1,2,3\}\).

Teacher's Note:
1. For a relation check reflexive, symmetric and transitive separately, and use a small counterexample to disprove a property.
2. For a function decide one-one (no two inputs share an output) and onto (range equals codomain) before finding any inverse or composition.

Answer: (d) \(-\log_2(1-x)\)

Explanation:
1. \(y=1-2^{-x}\) gives \(2^{-x}=1-y\).
2. \(x=-\log_2(1-y)\), so \(f^{-1}(x)=-\log_2(1-x)\).

Teacher's Note:
1. For a relation check reflexive, symmetric and transitive separately, and use a small counterexample to disprove a property.
2. For a function decide one-one (no two inputs share an output) and onto (range equals codomain) before finding any inverse or composition.