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NCERT Exemplar · Class 11 Mathematics Sets

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EXERCISE 1.3 1–10 (part 1 of 6)

  1. Exercise 1

    Write the following sets in the roaster from
    (i)
    A={x:x∈R,2x+11=15}\displaystyle \mathrm{A}=\{x: x \in \mathbf{R}, 2 x+11=15\}
    (ii)
    B={x∣x2=x,x∈R}\displaystyle \mathrm{B}=\left\{x \mid x^2=x, x \in \mathbf{R}\right\}
    (iii)
    C={x∣x\displaystyle \mathrm{C}=\{x \mid x is a positive factor of a prime number p}\displaystyle p\}

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    (i)
    \(\displaystyle \{2\}\) (ii) \(\displaystyle \{0,1\}\)
    (iii)
    \(\displaystyle \{1, p\}\)
    (i)
    \[2x + 11 = 15 \]
    \[2x = 4 \Rightarrow x = 2 \]
    \[A = \{2\} \] (ii) \[x^2 = x \Rightarrow x(x-1) = 0 \]
    \[x = 0 \ \text{or}\ x = 1 \]
    \[B = \{0, 1\} \]
    (iii)
    A prime \(\displaystyle p\) has exactly two positive factors.
    \[C = \{1, p\} \]
    Answer: \(\displaystyle A=\{2\}\), \(\displaystyle B=\{0,1\}\), \(\displaystyle C=\{1,p\}\)

  2. Exercise 2

    Write the following sets in the roaster form :
    (i)
    D={t∣t3=t,t∈R}\displaystyle \mathrm{D}=\left\{t \mid t^3=t, t \in \mathrm{R}\right\}
    (ii)
    E={w∣ w−2w+3=3,w∈R}\displaystyle \mathrm{E}=\left\{w \left\lvert\, \frac{w-2}{w+3}=3\right., w \in \mathbf{R}\right\}
    (iii)
    F={x∣x4−5x2+6=0,x∈R}\displaystyle \mathrm{F}=\left\{x \mid x^4-5 x^2+6=0, x \in \mathbf{R}\right\}

    Check this one against your book

    NCERT’s printed answer for this exercise does not match its own question. This working follows the question as printed.

    (i)
    \[t^3 = t \Rightarrow t(t-1)(t+1) = 0 \]
    \[D = \{-1, 0, 1\} \]
    (ii)
    \[\frac{w-2}{w+3} = 3, \quad w \neq -3 \]
    \[w - 2 = 3w + 9 \]
    \[-2w = 11 \Rightarrow w = -\frac{11}{2} \]
    This is not \(\displaystyle -3\), so it is admissible.
    \[E = \left\{-\frac{11}{2}\right\} \]
    (iii)
    \[x^4 - 5x^2 + 6 = (x^2 - 2)(x^2 - 3) = 0 \]
    \[x^2 = 2 \ \text{or}\ x^2 = 3 \]
    \[F = \{-\sqrt{3},\, -\sqrt{2},\, \sqrt{2},\, \sqrt{3}\} \]
    Answer: \(\displaystyle D=\{-1,0,1\}\), \(\displaystyle E=\left\{-\tfrac{11}{2}\right\}\), \(\displaystyle F=\{-\sqrt3,-\sqrt2,\sqrt2,\sqrt3\}\)

  3. Exercise 3

    If Y={x∣x\displaystyle \mathrm{Y}=\left\{x \mid x\right. is a positive factor of the number 2p−1(2p−1)\displaystyle 2^{p-1}\left(2^p-1\right), where 2p−1\displaystyle 2^p-1 is a prime number }\displaystyle \}. Write Y in the roaster form.

    Check this one against your book

    NCERT’s printed answer for this exercise does not match its own question. This working follows the question as printed.

    Let \(\displaystyle q = 2^p - 1\), an odd prime, so \(\displaystyle q \neq 2\). Then \[N = 2^{p-1} q \] Its positive factors are \(\displaystyle 2^a\) and \(\displaystyle 2^a q\) for \(\displaystyle a = 0, 1, \ldots, p-1\). \[Y = \{1,\, 2,\, 2^2,\, \ldots,\, 2^{p-1},\ q,\, 2q,\, 2^2 q,\, \ldots,\, 2^{p-1} q\} \] \[\text{number of factors} = p \cdot 2 = 2p \]Answer: \(\displaystyle Y=\{1,2,2^2,\ldots,2^{p-1},\,(2^p-1),\,2(2^p-1),\ldots,2^{p-1}(2^p-1)\}\)

  4. Exercise 4

    State which of the following statements are true and which are false. Justify your answer.
    (i)
    35∈{x∣x\displaystyle 35 \in\{x \mid x has exactly four positive factors }\displaystyle \}.
    (ii)
    128∈{y∣\displaystyle 128 \in\{y \mid the sum of all the positive factors of y\displaystyle y is 2y}\displaystyle 2 y\}
    (iii)
    3∉{x∣x4−5x3+2x2−112x+6=0}\displaystyle 3 \notin\left\{x \mid x^4-5 x^3+2 x^2-112 x+6=0\right\}
    (iv)
    496∉{y∣\displaystyle 496 \notin\{y \mid the sum of all the positive factors of y\displaystyle y is 2y}\displaystyle 2 y\}.

    Check this one against your book

    NCERT’s printed answer for this exercise does not match its own question. This working follows the question as printed.

    (i)
    True
    \[35 = 5 \times 7 \]
    \[\text{factors: } 1,\ 5,\ 7,\ 35 \]
    Exactly four.
    (ii)
    False
    \[128 = 2^7 \]
    \[1 + 2 + 2^2 + \cdots + 2^7 = 2^8 - 1 = 255 \]
    \[255 \neq 2 \times 128 = 256 \]
    (iii)
    True
    \[3^4 - 5\cdot 3^3 + 2\cdot 3^2 - 112\cdot 3 + 6 = 81 - 135 + 18 - 336 + 6 \]
    \[= -366 \neq 0 \]
    So \(\displaystyle 3\) is not in the set.
    (iv)
    False
    \[496 = 2^4 \times 31 \]
    \[(1 + 2 + 4 + 8 + 16)(1 + 31) = 31 \times 32 = 992 = 2 \times 496 \]
    So \(\displaystyle 496\) is in the set.
    Answer: (i) True, (ii) False, (iii) True, (iv) False

  5. Exercise 5

    Given L={1,2,3,4},M={3,4,5,6}\displaystyle \mathrm{L}=\{1,2,3,4\}, \mathrm{M}=\{3,4,5,6\} and N={1,3,5}\displaystyle \mathrm{N}=\{1,3,5\} Verify that L−(M∪N)=(L−M)∩(L−N)\displaystyle \mathrm{L}-(\mathrm{M} \cup \mathrm{N})=(\mathrm{L}-\mathrm{M}) \cap(\mathrm{L}-\mathrm{N})

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    \[M \cup N = \{1, 3, 4, 5, 6\} \] \[L - (M \cup N) = \{2\} \] \[L - M = \{1, 2\} \] \[L - N = \{2, 4\} \] \[(L - M) \cap (L - N) = \{2\} \] \[L - (M \cup N) = (L - M) \cap (L - N) = \{2\} \]Answer: Both sides equal \(\displaystyle \{2\}\), so the result is verified.

  6. Exercise 6

    If A and B are subsets of the universal set U, then show that
    (i)
    A⊂A∪B\displaystyle \mathrm{A} \subset \mathrm{A} \cup \mathrm{B}
    (ii)
    A⊂B⇔A∪B=B\displaystyle \mathrm{A} \subset \mathrm{B} \Leftrightarrow \mathrm{A} \cup \mathrm{B}=\mathrm{B}
    (iii)
    (A∩B)⊂A\displaystyle (\mathrm{A} \cap \mathrm{B}) \subset \mathrm{A}

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    (i)
    \[x \in A \Rightarrow x \in A \ \text{or}\ x \in B \]
    \[\Rightarrow x \in A \cup B \]
    \[A \subset A \cup B \]
    (ii)
    (\(\displaystyle \Rightarrow\)) Let \(\displaystyle A \subset B\).
    \[x \in A \cup B \Rightarrow x \in A \ \text{or}\ x \in B \]
    \[x \in A \Rightarrow x \in B \quad \text{(} A \subset B \text{)} \]
    \[A \cup B \subset B \]
    \[B \subset A \cup B \quad \text{(as in (i))} \]
    \[A \cup B = B \]
    (\(\displaystyle \Leftarrow\)) Let \(\displaystyle A \cup B = B\).
    \[A \subset A \cup B \quad \text{(by (i))} \]
    \[A \subset B \quad \text{(since } A \cup B = B \text{)} \]
    (iii)
    \[x \in A \cap B \Rightarrow x \in A \ \text{and}\ x \in B \]
    \[\Rightarrow x \in A \]
    \[A \cap B \subset A \]
    Answer: (i), (ii) and (iii) are proved.

  7. Exercise 7

    Given that N={1,2,3,…,100}\displaystyle \mathrm{N}=\{1,2,3, \ldots, 100\}. Then write
    (i)
    the subset of N whose elements are even numbers.
    (ii)
    the subset of N whose element are perfect square numbers.

    Check this one against your book

    NCERT’s printed answer for this exercise does not match its own question. This working follows the question as printed.

    (i)
    Even numbers from $\displaystyle 1$ to $\displaystyle 100$:
    \[\{2, 4, 6, \ldots, 100\} \]
    (ii)
    Perfect squares up to $\displaystyle 100$:
    \[1^2, 2^2, \ldots, 10^2 \quad (11^2 = 121 > 100) \]
    \[\{1, 4, 9, 16, 25, 36, 49, 64, 81, 100\} \]
    Answer: (i) \(\displaystyle \{2,4,6,\ldots,100\}\); (ii) \(\displaystyle \{1,4,9,16,25,36,49,64,81,100\}\)

  8. Exercise 8

    If X={1,2,3}\displaystyle \mathrm{X}=\{1,2,3\}, if n\displaystyle n represents any member of X, write the following sets containing all numbers represented by
    (i)
    4n\displaystyle 4 n
    (ii)
    n+6\displaystyle n+6
    (iii)
    n2\displaystyle \frac{n}{2}
    (iv)
    n−1\displaystyle n-1

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    (i)
    \(\displaystyle \{4,8,12\}\)
    (ii)
    \(\displaystyle \{7,8,9\}\)
    (iii)
    \(\displaystyle \left\{\frac{1}{2}, 1, \frac{3}{2}\right\}\)
    (iv)
    \(\displaystyle \{0, 1, 2\}\)
    Put \(\displaystyle n = 1, 2, 3\).
    (i)
    \[\{4\cdot 1,\ 4\cdot 2,\ 4\cdot 3\} = \{4, 8, 12\} \]
    (ii)
    \[\{1+6,\ 2+6,\ 3+6\} = \{7, 8, 9\} \]
    (iii)
    \[\left\{\frac12,\ \frac22,\ \frac32\right\} = \left\{\frac12,\ 1,\ \frac32\right\} \]
    (iv)
    \[\{1-1,\ 2-1,\ 3-1\} = \{0, 1, 2\} \]
    Answer: (i) \(\displaystyle \{4,8,12\}\); (ii) \(\displaystyle \{7,8,9\}\); (iii) \(\displaystyle \left\{\tfrac12,1,\tfrac32\right\}\); (iv) \(\displaystyle \{0,1,2\}\)

  9. Exercise 9

    If Y={1,2,3,…10}\displaystyle \mathrm{Y}=\{1,2,3, \ldots 10\}, and a\displaystyle a represents any element of Y, write the following sets, containing all the elements satisfying the given conditions.
    (i)
    a∈Y\displaystyle a \in \mathrm{Y} but a2∉Y\displaystyle a^2 \notin \mathrm{Y}
    (ii)
    a+1=6,a∈Y\displaystyle a+1=6, a \in \mathrm{Y}
    (iii)
    a\displaystyle a is less than 6\displaystyle 6 and a∈Y\displaystyle a \in \mathrm{Y}

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    (i)
    \(\displaystyle \{4,5,6, \ldots .10\}\)
    (ii)
    \(\displaystyle \{5\}\)
    (iii)
    \(\displaystyle \{1,2,3,4,5\}\)
    (i) \(\displaystyle a^2\notin Y\) means \(\displaystyle a^2>10\). \[3^2 = 9 \in Y, \qquad 4^2 = 16 \notin Y \] \[\{a : a\in Y,\ a^2\notin Y\} = \{4,5,6,7,8,9,10\} \] (ii) \[a+1 = 6 \Rightarrow a = 5 \] \[\{a : a+1=6,\ a\in Y\} = \{5\} \] (iii) \[a<6,\ a\in Y \Rightarrow a = 1,2,3,4,5 \] \[\{a : a<6,\ a\in Y\} = \{1,2,3,4,5\} \] Answer: (i) \(\displaystyle \{4,5,6,7,8,9,10\}\); (ii) \(\displaystyle \{5\}\); (iii) \(\displaystyle \{1,2,3,4,5\}\).

  10. Exercise 10

    A, B and C are subsets of Universal Set U. If A={2,4,6,8,12,20}\displaystyle \mathrm{A}=\{2,4,6,8,12,20\} B={3,6,9,12,15},C={5,10,15,20}\displaystyle \mathrm{B}=\{3,6,9,12,15\}, \mathrm{C}=\{5,10,15,20\} and U is the set of all whole numbers, draw a Venn diagram showing the relation of U, A, B and C.

    Check this one against your book

    NCERT’s printed answer for this exercise does not match its own question. This working follows the question as printed.

    \[A\cap B = \{6,12\}, \quad B\cap C = \{15\}, \quad A\cap C = \{20\} \] \[A\cap B\cap C = \varnothing \] \[A\setminus(B\cup C) = \{2,4,8\}, \quad B\setminus(A\cup C) = \{3,9\}, \quad C\setminus(A\cup B) = \{5,10\} \] \[A\cup B\cup C = \{2,3,4,5,6,8,9,10,12,15,20\} \] \[U\setminus(A\cup B\cup C) = \{0,1,7,11,13,14,16,17,18,19,21,22,\dots\} \] Regions of the Venn diagram:
    RegionElements
    \(\displaystyle A\) only\(\displaystyle \{2,4,8\}\)
    \(\displaystyle B\) only\(\displaystyle \{3,9\}\)
    \(\displaystyle C\) only\(\displaystyle \{5,10\}\)
    \(\displaystyle A\cap B\) only\(\displaystyle \{6,12\}\)
    \(\displaystyle B\cap C\) only\(\displaystyle \{15\}\)
    \(\displaystyle A\cap C\) only\(\displaystyle \{20\}\)
    \(\displaystyle A\cap B\cap C\)\(\displaystyle \varnothing\)
    Outside \(\displaystyle A\cup B\cup C\) (in \(\displaystyle U\))\(\displaystyle \{0,1,7,11,13,14,16,17,18,19,21,22,\dots\}\)
    Answer: Inside \(\displaystyle U\), the circles \(\displaystyle A\), \(\displaystyle B\), \(\displaystyle C\) overlap pairwise and \(\displaystyle A\cap B\cap C=\varnothing\); each region holds the elements in the table. NCERT prints: a diagram with \(\displaystyle A\) and \(\displaystyle C\) disjoint and $\displaystyle 20$ written in both -- but \(\displaystyle 20\in A\cap C\), so \(\displaystyle A\) and \(\displaystyle C\) must overlap.