JEE Mathematics Relations and Functions MCQs Set 10

Answer: (a) \([75,89]\)

Explanation:
1. \(f'(x)=6x^2-24=0\) gives \(x=2\) in \([1,3]\) (a minimum).
2. \(f(1)=85\), \(f(2)=75\), \(f(3)=89\), so the image is \([75,89]\).

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) many-one into

Explanation:
1. \(f(x)=1-\dfrac5{x^2+1}\) is even, so \(f(1)=f(-1)\): many-one.
2. Its range is \([-4,1)\ne R\), so \(f\) 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: (c) many-one and onto

Explanation:
1. \(f(1)=f(2)=f(3)=0\), so \(f\) is many-one.
2. A cubic polynomial takes every real value, so it is onto.

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) \(\dfrac12\log_{10}\dfrac{2+3x}{2-3x}\)

Explanation:
1. Let \(y=\dfrac23\cdot\dfrac{t-1}{t+1}\) with \(t=10^{2x}\). Then \(t=\dfrac{2+3y}{2-3y}\).
2. \(x=\dfrac12\log_{10}t=\dfrac12\log_{10}\dfrac{2+3y}{2-3y}\), so \(f^{-1}(x)=\dfrac12\log_{10}\dfrac{2+3x}{2-3x}\).

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) \(\dfrac x{x-1}\)

Explanation:
1. \(f(f(x))=\dfrac{x/(x-1)}{x/(x-1)-1}=x\), so \(f\circ f\) is the identity.
2. An odd number of compositions gives \(f\) itself: \(\dfrac x{x-1}\).

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: (b) a straight line passing through \(\left(\dfrac\pi2,-\sin^21\right)\) and parallel to the \(x\)-axis

Explanation:
1. \(\cos x\cos(x+2)=\dfrac12\left[\cos(2x+2)+\cos2\right]\) and \(\cos^2(x+1)=\dfrac12\left[1+\cos(2x+2)\right]\).
2. The difference is \(\dfrac{\cos2-1}2=-\sin^21\), a constant: a horizontal line through \(\left(\dfrac\pi2,-\sin^21\right)\).

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) \(106!\)

Explanation:
1. A bijection from \(A\) to itself is a permutation of its elements.
2. The number is \(106!\).

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 a constant function

Explanation:
1. \([x]\) is an integer, so \(\sin([x]\pi)=0\).
2. Hence \(f(x)=0\) for all \(x\), a constant function.

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) \(\dfrac x{1-x}\)

Explanation:
1. The series is geometric: \(f(x)=\dfrac x{1+x}\).
2. \(y=\dfrac x{1+x}\) gives \(x=\dfrac y{1-y}\), so \(f^{-1}(x)=\dfrac x{1-x}\).

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.

Answer: (a) \(f(d)=g(b)\)

Explanation:
1. \(f(g(x))=acx+ad+b\) and \(g(f(x))=acx+cb+d\).
2. Equality requires \(ad+b=cb+d\), i.e. \(f(d)=g(b)\).

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) \(R\) and \(S\) transitive \(\Rightarrow R\cup S\) is transitive

Explanation:
1. Take \(A=\{1,2,3\}\), \(R=\{(1,2)\}\), \(S=\{(2,3)\}\): both are transitive, but \(R\cup S=\{(1,2),(2,3)\}\) lacks \((1,3)\).
2. The other three statements are true.

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. For \(x>20\), \(f(x)=x-2>0\), so \(g(x)=|x-2-2|=x-4\).
2. \(g'(x)=1\).

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: (b) \(-1\)

Explanation:
1. \(g\left(\dfrac85\right)=\dfrac85\) and \(f\left(\dfrac85\right)=1\).
2. \(f\left(-\dfrac85\right)=-2\) and \(g(-2)=2\), so the value is \(1-2=-1\).

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: (b) not an equivalence relation but a partial order relation

Explanation:
1. \(A\subseteq A\) (reflexive) and \(A\subseteq B\), \(B\subseteq C\Rightarrow A\subseteq C\) (transitive).
2. It is anti-symmetric but not symmetric (\(A\subseteq B\) does not give \(B\subseteq A\)), so it is a partial order and not 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: (d) \(11\)

Explanation:
1. \(f(5)=5\).
2. \(g(5)=11\), so \((g\circ f)(5)=11\).

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) \(0\)

Explanation:
1. \(f\left(\dfrac{\sqrt\pi}2\right)=\dfrac\pi4\), \(g\left(\dfrac\pi4\right)=\tan\dfrac\pi4=1\).
2. \(h(1)=\log1=0\).

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) \(x\), \(\forall x\in R\)

Explanation:
1. For \(x<0\): \(f(x)=x^3+1<1\), so \(g(f(x))=(x^3)^{1/3}=x\).
2. For \(x\ge0\): \(f(x)=x^2+1\ge1\), so \(g(f(x))=\sqrt{x^2}=x\).

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.

Answer: (a) symmetric and transitive

Explanation:
1. \(\sqrt2\) is related to no element (\(\sqrt2+b\in Z\) would need \(b=k-\sqrt2\notin A\)), so \(R\) is not reflexive (\(2\sqrt2\notin Z\)).
2. Symmetry is clear, and for integers \(a,b,c\) with \(a+b,b+c\in Z\) we get \(a+c\in Z\); so \(R\) is symmetric and 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: (b) \(a+b=0\)

Explanation:
1. \(f(f(x))=\dfrac{a\frac{ax+d}{cx+b}+d}{c\frac{ax+d}{cx+b}+b}=\dfrac{(a^2+cd)x+d(a+b)}{c(a+b)x+(cd+b^2)}\).
2. For this to equal \(x\) for all \(x\) we need \(a+b=0\) (then it becomes \(\dfrac{(a^2+cd)x}{cd+a^2}=x\)).

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.

Answer: (d) reflexive only

Explanation:
1. \(x-x+\sqrt5=\sqrt5\) is irrational: reflexive.
2. Not symmetric: \(x-y=\sqrt5-1\) gives \(2\sqrt5-1\) irrational but \(y-x+\sqrt5=1\) rational. Not transitive: \(x-y=\sqrt2\), \(y-z=1-\sqrt5-\sqrt2\) give irrational values, while \(x-z+\sqrt5=1\) is rational.

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) \(\begin{cases}x+1,&-2\le x\le-1\\-1-x,&-1\le x\le0\\x-1,&0<x\le2\end{cases}\)

Explanation:
1. \(g(x)=x+3\) on \([-2,-1]\) and \(g(x)=1-x\) on \([-1,2]\). \(f(u)=u-2\) for \(u\ge1\) and \(f(u)=-u\) for \(u<1\).
2. For \(-2\le x\le-1\): \(u=x+3\ge1\) gives \(x+1\). For \(-1\le x\le0\): \(u=1-x\ge1\) gives \(-1-x\). For \(0<x\le2\): \(u=1-x<1\) gives \(x-1\).

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: (a) \(x-2\)

Explanation:
1. For \(4<x<6\), \(2<\dfrac x2<3\) so \(\left[\dfrac x2\right]=2\).
2. \(f(x)=x+2\) and \(f^{-1}(x)=x-2\).

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.