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NCERT Exemplar · Class 11 Mathematics Relations and Functions

42 questions · 42 still being checked

EXERCISE 2.3 1–10 (part 1 of 4)

  1. Exercise 1

    Let A={−1,2,3}\displaystyle \mathrm{A}=\{-1,2,3\} and B={1,3}\displaystyle \mathrm{B}=\{1,3\}. Determine
    (i)
    A×B\displaystyle \mathrm{A} \times \mathrm{B}
    (ii)
    B×A\displaystyle \mathrm{B} \times \mathrm{A}
    (iii)
    B×B\displaystyle \mathrm{B} \times \mathrm{B}
    (iv)
    A×A\displaystyle \mathrm{A} \times \mathrm{A}

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    NCERT’s answer
    (i)
    \(\displaystyle \{(-1, 1), (-1, 3), (2, 1), (2, 3), (3, 1), (3, 3)\}\)
    (ii)
    \(\displaystyle \{(1,-1),(1,2),(1,3),(3,-1),(3,2),(3,3)\}\)
    (iii)
    \(\displaystyle \{(1,1),(1,3),(3,1),(3,3)\}\)
    (iv)
    \(\displaystyle \{(-1, -1), (-1, 2), (-1, 3), (2, -1), (2, 2), (2, 3), (3, -1), (3, 2), (3, 3)\}\)
    Pair each element of the first set with each element of the second.\[n(A\times B)=3\cdot 2=6,\quad n(B\times B)=2\cdot 2=4,\quad n(A\times A)=3\cdot 3=9 \] \[A\times B=\{(-1,1),(-1,3),(2,1),(2,3),(3,1),(3,3)\} \] \[B\times A=\{(1,-1),(1,2),(1,3),(3,-1),(3,2),(3,3)\} \] \[B\times B=\{(1,1),(1,3),(3,1),(3,3)\} \] \[A\times A=\{(-1,-1),(-1,2),(-1,3),(2,-1),(2,2),(2,3),(3,-1),(3,2),(3,3)\} \]Answer: the four products are the sets of $\displaystyle 6$, $\displaystyle 6$, $\displaystyle 4$ and $\displaystyle 9$ ordered pairs listed above.

  2. Exercise 2

    If P={x:x<3,x∈N},Q={x:x≤2,x∈W}\displaystyle \mathrm{P}=\{x: x<3, x \in \mathbf{N}\}, \mathrm{Q}=\{x: x \leq 2, x \in \mathbf{W}\}. Find (P∪Q)×(P∩Q)\displaystyle (\mathrm{P} \cup \mathrm{Q}) \times(\mathrm{P} \cap \mathrm{Q}), where W\displaystyle \mathbf{W} is the set of whole numbers.

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    NCERT’s answer
    \(\displaystyle \{(0, 1), (0, 2), (1, 1), (1, 2), (2, 1), (2, 2)\}\)
    \(\displaystyle \mathbf N\) starts at $\displaystyle 1$ and \(\displaystyle \mathbf W\) starts at 0.\[P=\{1,2\},\qquad Q=\{0,1,2\} \] \[P\cup Q=\{0,1,2\},\qquad P\cap Q=\{1,2\} \] \[(P\cup Q)\times(P\cap Q)=\{(0,1),(0,2),(1,1),(1,2),(2,1),(2,2)\} \] \[n=3\cdot 2=6 \]Answer: \(\displaystyle \{(0,1),(0,2),(1,1),(1,2),(2,1),(2,2)\}\)

  3. Exercise 3

    If A={x:x∈W,x<2}B={x:x∈N,1<x<5}C={3,5}\displaystyle \mathrm{A}=\{x: x \in \mathbf{W}, x<2\} \quad \mathrm{B}=\{x: x \in \mathbf{N}, 1<x<5\} \quad \mathrm{C}=\{3,5\} find
    (i)
    A×(B∩C)\displaystyle \mathrm{A} \times(\mathrm{B} \cap \mathrm{C})
    (ii)
    A×(B∪C)\displaystyle \mathrm{A} \times(\mathrm{B} \cup \mathrm{C})

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    NCERT’s answer
    (i)
    \(\displaystyle \{(0, 3), (1, 3)\}\)
    (ii)
    \(\displaystyle \{(0, 2), (0, 3), (0, 4), (0, 5), (1, 2), (1, 3), (1, 4), (1,5)\}\)
    \[A=\{0,1\},\qquad B=\{2,3,4\},\qquad C=\{3,5\} \]
    \[B\cap C=\{3\},\qquad B\cup C=\{2,3,4,5\} \]
    (i)
    \[A\times(B\cap C)=\{(0,3),(1,3)\} \]
    (ii)
    \[A\times(B\cup C)=\{(0,2),(0,3),(0,4),(0,5),(1,2),(1,3),(1,4),(1,5)\} \]
    \[n=2\cdot 4=8 \]
    Answer: (i) \(\displaystyle \{(0,3),(1,3)\}\); (ii) the $\displaystyle 8$ pairs \(\displaystyle (a,b)\) with \(\displaystyle a\in\{0,1\}\), \(\displaystyle b\in\{2,3,4,5\}\).

  4. Exercise 4

    In each of the following cases, find a\displaystyle a and b\displaystyle b.
    (i)
    (2a+b,a−b)=(8,3)\displaystyle (2 a+b, a-b)=(8,3)
    (ii)
    (a4,a−2b)=(0,6+b)\displaystyle \left(\frac{a}{4}, a-2 b\right)=(0,6+b)

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    NCERT’s answer
    (i)
    \(\displaystyle a=\frac{11}{3}\) and \(\displaystyle b=\frac{2}{3}\)
    (ii)
    \(\displaystyle a=0\) and \(\displaystyle b=-2\)
    Equal ordered pairs have equal components.
    (i)
    \[2a+b=8,\qquad a-b=3 \]
    \[3a=11\ \Rightarrow\ a=\tfrac{11}{3} \]
    \[b=a-3=\tfrac{2}{3} \]
    \[2\cdot\tfrac{11}{3}+\tfrac{2}{3}=8 \]
    (ii)
    \[\tfrac{a}{4}=0\ \Rightarrow\ a=0 \]
    \[a-2b=6+b\ \Rightarrow\ -2b=6+b \]
    \[-3b=6\ \Rightarrow\ b=-2 \]
    \[a-2b=4=6+b \]
    Answer: (i) \(\displaystyle a=\tfrac{11}{3},\ b=\tfrac{2}{3}\); (ii) \(\displaystyle a=0,\ b=-2\).

  5. Exercise 5

    Given A={1,2,3,4,5},S={(x,y):x∈ A,y∈ A}\displaystyle \mathrm{A}=\{1,2,3,4,5\}, \mathrm{S}=\{(x, \mathrm{y}): x \in \mathrm{~A}, y \in \mathrm{~A}\}. Find the ordered pairs which satisfy the conditions given below:
    (i)
    x+y=5\displaystyle x+y=5
    (ii)
    x+y<5\displaystyle x+y<5
    (iii)
    x+y>8\displaystyle x+y>8

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    NCERT’s answer
    (i)
    \(\displaystyle \{(1,4),(2,3),(3,2),(4,1)\}\)
    (ii)
    \(\displaystyle \{(1,1),(1,2),(1,3),(2,1),(2,2),(3,1)\}\)
    (iii)
    \(\displaystyle \{(4,5),(5,4),(5,5)\}\)
    Pairs \(\displaystyle (x,y)\) of \(\displaystyle A\times A\) by their sum, \(\displaystyle 2\le x+y\le 10\).
    (i)
    \[x+y=5:\quad \{(1,4),(2,3),(3,2),(4,1)\} \]
    (ii)
    \[x+y<5\ \Rightarrow\ x+y\in\{2,3,4\} \]
    \[\{(1,1),(1,2),(1,3),(2,1),(2,2),(3,1)\} \]
    (iii)
    \[x+y>8\ \Rightarrow\ x+y\in\{9,10\} \]
    \[\{(4,5),(5,4),(5,5)\} \]
    Answer: (i) $\displaystyle 4$ pairs; (ii) $\displaystyle 6$ pairs; (iii) $\displaystyle 3$ pairs, as listed above.

  6. Exercise 6

    Given R={(x,y):x,y∈W,x2+y2=25}\displaystyle \mathrm{R}=\left\{(x, y): x, y \in \mathbf{W}, x^2+y^2=25\right\}. Find the domain and Range of R.

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    NCERT’s answer
    Domain of \(\displaystyle \mathrm{R}=\{0,3,4,5\}=\) Range of R
    Whole numbers, so \(\displaystyle x^2\le 25\) forces \(\displaystyle x\in\{0,1,2,3,4,5\}\); test each.\[x=0:\ y^2=25\Rightarrow y=5 \] \[x=1:\ y^2=24,\qquad x=2:\ y^2=21 \quad(\text{not squares}) \] \[x=3:\ y^2=16\Rightarrow y=4 \] \[x=4:\ y^2=9\Rightarrow y=3 \] \[x=5:\ y^2=0\Rightarrow y=0 \] \[R=\{(0,5),(3,4),(4,3),(5,0)\} \] \[\text{Domain}=\{0,3,4,5\},\qquad \text{Range}=\{5,4,3,0\} \]Answer: Domain \(\displaystyle =\{0,3,4,5\}\), Range \(\displaystyle =\{0,3,4,5\}\).

  7. Exercise 7

    If R1={(x,y)∣y=2x+7\displaystyle \mathrm{R}_1=\{(x, y) \mid y=2 x+7, where x∈R\displaystyle x \in \mathbf{R} and −5≤x≤5}\displaystyle -5 \leq x \leq 5\} is a relation. Then find the domain and Range of R1\displaystyle \mathrm{R}_1.

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    NCERT’s answer
    Domain of \(\displaystyle \mathrm{R}_1=[-5,5]\) and Range of \(\displaystyle \mathrm{R}_1=[-3,17]\)
    \[-5\le x\le 5 \ \Rightarrow\ \text{Domain}=[-5,5] \] \[-10\le 2x\le 10 \] \[-3\le 2x+7\le 17 \] \[x=-5\Rightarrow y=-3,\qquad x=5\Rightarrow y=17 \]The line is continuous, so \(\displaystyle y\) takes every value between these ends.NCERT_Solution_Class11_Maths_Exemplar_Ch2_Ex2-3_Q7Answer: Domain \(\displaystyle =[-5,5]\), Range \(\displaystyle =[-3,17]\).

  8. Exercise 8

    If R2={(x,y)∣x\displaystyle \mathrm{R}_2=\left\{(x, y) \mid x\right. and y\displaystyle y are integers and x2+y2=64}\displaystyle \left.x^2+y^2=64\right\} is a relation. Then find R2\displaystyle \mathrm{R}_2.

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    NCERT’s answer
    \(\displaystyle \mathrm{R}_2=\{(0,8),(8,0)(0,-8),(-8,0)\}\)
    Since \(\displaystyle y^2 \ge 0\), \(\displaystyle x^2 \le 64\), so \(\displaystyle x\) is an integer with \(\displaystyle |x|\le 8\).\[x=0:\ y^2=64 \Rightarrow y=\pm 8 \] \[x=\pm 8:\ y^2=0 \Rightarrow y=0 \] \[x=\pm1,\pm2,\dots,\pm7:\ y^2=64-x^2 = 63,\,60,\,55,\,48,\,39,\,28,\,15 \]None of these is a perfect square, so no other integer pair works.Answer: \(\displaystyle R_2=\{(0,8),(0,-8),(8,0),(-8,0)\}\)

  9. Exercise 9

    If R3={(x,∣x∣)∣x\displaystyle \mathrm{R}_3=\{(x,|x|) \mid x is a real number }\displaystyle \} is a relation. Then find domain and range of R3\displaystyle \mathrm{R}_3.

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    NCERT’s answer
    Domain of \(\displaystyle \mathrm{R}_3=\mathbf{R}\) and range of \(\displaystyle \mathrm{R}_3=\mathbf{R}^{+} \cup\{0\}\)
    \(\displaystyle |x|\) is defined for every real \(\displaystyle x\):\[\text{Domain}=\mathbb{R} \]\(\displaystyle |x|\ge 0\), and every \(\displaystyle a\ge 0\) is attained since\[(a,|a|)=(a,a)\in R_3 \quad (a\ge 0) \]\[\text{Range}=\{y\in\mathbb{R} : y\ge 0\}=[0,\infty) \]NCERT_Solution_Class11_Maths_Exemplar_Ch2_Ex2-3_Q9Answer: Domain \(\displaystyle =\mathbb{R}\); Range \(\displaystyle =[0,\infty)\)

  10. Exercise 10

    Is the given relation a function? Give reasons for your answer.
    (i)
    h={(4,6),(3,9),(−11,6),(3,11)}\displaystyle h=\{(4,6),(3,9),(-11,6),(3,11)\}
    (ii)
    f={(x,x)∣x\displaystyle f=\{(x, x) \mid x is a real number }\displaystyle \}
    (iii)
    g={(n,1n)∣ n\displaystyle g=\left\{\left.\left(n, \frac{1}{n}\right) \right\rvert\, n\right. is a positive integer }\displaystyle \}
    (iv)
    s={(n,n2)∣n\displaystyle s=\left\{\left(n, n^2\right) \mid n\right. is a positive integer }\displaystyle \}
    (v)
    t={(x,3)∣x\displaystyle t=\{(x, 3) \mid x is a real number.

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    NCERT’s answer
    (i)
    \(\displaystyle h\) is not a function
    (ii)
    \(\displaystyle f\) is a function
    (iii)
    \(\displaystyle g\) is a function
    (iv)
    \(\displaystyle s\) is a function
    (v)
    \(\displaystyle t\) is a constant function
    (i) Not a function: \(\displaystyle 3\) has two images. \[h(3)=9 \quad\text{and}\quad h(3)=11 \](ii) Function: each real \(\displaystyle x\) has exactly one image. \[f(x)=x \](iii) Function: each positive integer \(\displaystyle n\) has exactly one image. \[g(n)=\frac{1}{n} \](iv) Function: each positive integer \(\displaystyle n\) has exactly one image. \[s(n)=n^2 \](v) Function (constant): each real \(\displaystyle x\) has exactly one image. \[t(x)=3 \]Answer: (i) not a function; (ii), (iii), (iv), (v) are functions.