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NCERT Exemplar · Class 12 Mathematics Relations and Functions

62 questions · 62 still being checked

EXERCISE 1.3 31–40 (part 4 of 6)

  1. Choose the correct answer out of the given four options in each of the Exercises from $\displaystyle 28$ to $\displaystyle 47$ (M.C.Q.).

    Exercise 31

    If a relation R on the set {1,2,3}\displaystyle \{1,2,3\} be defined by R={(1,2)}\displaystyle \mathrm{R}=\{(1,2)\}, then R is
    (A)
    reflexive
    (B)
    transitive
    (C)
    symmetric
    (D)
    none of these

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    NCERT’s answer
    B
    (B) transitive\[(1,1)\notin R \quad \text{(not reflexive)} \] \[(1,2)\in R,\ (2,1)\notin R \quad \text{(not symmetric)} \] \[(1,2)\in R,\ (2,c)\notin R \text{ for every } c \quad \text{(no pairs to chain, so transitive)} \]

  2. Exercise 32

    Let us define a relation R in R\displaystyle \mathbf{R} as aRb\displaystyle a \mathrm{R} b if a≥b\displaystyle a \geq b. Then R is
    (A)
    an equivalence relation
    (B)
    reflexive, transitive but not symmetric
    (C)
    symmetric, transitive but not reflexive
    (D)
    neither transitive nor reflexive but symmetric.

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    NCERT’s answer
    B
    (B) reflexive, transitive but not symmetric\[a \ge a \quad \text{(reflexive)} \] \[a \ge b,\ b \ge c \Rightarrow a \ge c \quad \text{(transitive)} \] \[2 \ge 1 \text{ but } 1 \not\ge 2 \quad \text{(not symmetric)} \]

  3. Exercise 33

    Let A={1,2,3}\displaystyle \mathrm{A}=\{1,2,3\} and consider the relation
    R={1,1),(2,2),(3,3),(1,2),(2,3),(1,3)}.\mathrm{R}=\{1,1),(2,2),(3,3),(1,2),(2,3),(1,3)\} .
    Then R is
    (A)
    reflexive but not symmetric
    (B)
    reflexive but not transitive
    (C)
    symmetric and transitive
    (D)
    neither symmetric, nor transitive

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    NCERT’s answer
    A
    (A) reflexive but not symmetric\[(1,1),(2,2),(3,3)\in R \quad \text{(reflexive)} \] \[(1,2)\in R,\ (2,1)\notin R \quad \text{(not symmetric)} \] \[(1,2),(2,3)\in R \Rightarrow (1,3)\in R \quad \text{(transitive)} \]Every other chain \(\displaystyle (a,b),(b,c)\) also gives a pair already in \(\displaystyle R\), so \(\displaystyle R\) is transitive and (B), (C), (D) fail.

  4. Exercise 34

    The identity element for the binary operation ∗\displaystyle * defined on Q∼{0}\displaystyle \mathrm{Q} \sim\{0\} as a∗b=ab2∀a,b∈Q∼{0}\displaystyle a * b=\frac{a b}{2} \quad \forall a, b \in \mathrm{Q} \sim\{0\} is
    (A)
    1\displaystyle 1 (B) 0\displaystyle 0 (C) 2\displaystyle 2 (D) none of these

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    NCERT’s answer
    C
    (C) \(\displaystyle 2\)\[a*e=a \Rightarrow \frac{ae}{2}=a \] \[e=2 \quad (a\neq 0) \] \[e*a=\frac{2a}{2}=a \]

  5. Exercise 35

    If the set A contains 5\displaystyle 5 elements and the set B contains 6\displaystyle 6 elements, then the number of one-one and onto mappings from A to B is
    (A)
    720\displaystyle 720
    (B)
    120\displaystyle 120
    (C)
    0\displaystyle 0 (D) none of these

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    NCERT’s answer
    C
    (C) \(\displaystyle 0\)\[n(A)=5,\quad n(B)=6 \] \[n\big(f(A)\big)\le n(A)=5<6=n(B) \]The image can never cover \(\displaystyle B\), so no map from \(\displaystyle A\) to \(\displaystyle B\) is onto.

  6. Exercise 36

    Let A={1,2,3,…n}\displaystyle \mathrm{A}=\{1,2,3, \ldots n\} and B={a,b}\displaystyle \mathrm{B}=\{a, b\}. Then the number of surjections from A into B is
    (A)
    nP2\displaystyle { }^n \mathrm{P}_2
    (B)
    2n−2\displaystyle 2^n-2
    (C)
    2n−1\displaystyle 2^n-1
    (D)
    None of these

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    NCERT’s answer
    B
    (B) \(\displaystyle 2^n-2\)\[\text{maps } A\to B = 2^n \] \[\text{not onto: all to } a \text{ or all to } b = 2 \] \[\text{surjections}=2^n-2 \]

  7. Exercise 37

    Let f:R→R\displaystyle f: \mathbf{R} \rightarrow \mathbf{R} be defined by f(x)=1x∀x∈R\displaystyle f(x)=\frac{1}{x} \forall x \in \mathbf{R}. Then f\displaystyle f is
    (A)
    one-one
    (B)
    onto
    (C)
    bijective
    (D)
    f\displaystyle f is not defined

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    NCERT’s answer
    D
    (D) \(\displaystyle f\) is not defined\[f(0)=\frac{1}{0} \ \text{is undefined, but } 0\in\mathbf{R} \]So \(\displaystyle f\) is not a function on \(\displaystyle \mathbf{R}\); one-one, onto and bijective do not apply.

  8. Exercise 38

    Let f:R→R\displaystyle f: \mathbf{R} \rightarrow \mathbf{R} be defined by f(x)=3x2−5\displaystyle f(x)=3 x^2-5 and g:R→R\displaystyle g: \mathbf{R} \rightarrow \mathbf{R} by g(x)=xx2+1\displaystyle g(x)=\frac{x}{x^2+1}. Then gof\displaystyle g o f is
    (A)
    3x2−59x4−30x2+26\displaystyle \frac{3 x^2-5}{9 x^4-30 x^2+26}
    (B)
    3x2−59x4−6x2+26\displaystyle \frac{3 x^2-5}{9 x^4-6 x^2+26}
    (C)
    3x2x4+2x2−4\displaystyle \frac{3 x^2}{x^4+2 x^2-4}
    (D)
    3x29x4+30x2−2\displaystyle \frac{3 x^2}{9 x^4+30 x^2-2}

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    NCERT’s answer
    A
    (A) \(\displaystyle \dfrac{3x^2-5}{9x^4-30x^2+26}\)\[(g\circ f)(x)=g\big(f(x)\big)=\frac{f(x)}{\big(f(x)\big)^2+1} \] \[=\frac{3x^2-5}{(3x^2-5)^2+1} \] \[(3x^2-5)^2=9x^4-30x^2+25 \] \[(g\circ f)(x)=\frac{3x^2-5}{9x^4-30x^2+26} \]

  9. Exercise 39

    Which of the following functions from Z\displaystyle \mathbf{Z} into Z\displaystyle \mathbf{Z} are bijections?
    (A)
    f(x)=x3\displaystyle f(x)=x^3
    (B)
    f(x)=x+2\displaystyle f(x)=x+2
    (C)
    f(x)=2x+1\displaystyle f(x)=2 x+1
    (D)
    f(x)=x2+1\displaystyle f(x)=x^2+1

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    NCERT’s answer
    B
    (B) \(\displaystyle f(x)=x+2\)\[x_1+2=x_2+2 \Rightarrow x_1=x_2 \quad \text{(one-one)} \] \[x+2=y \Rightarrow x=y-2\in\mathbf{Z} \quad \text{(onto)} \]The others fail: \[x^3=2,\quad 2x+1=2 \ \text{have no integer solution (not onto)} \] \[f(1)=f(-1)=2 \ \text{for } x^2+1 \quad \text{(not one-one)} \]

  10. Exercise 40

    Let f:R→R\displaystyle f: \mathbf{R} \rightarrow \mathbf{R} be the functions defined by f(x)=x3+5\displaystyle f(x)=x^3+5. Then f−1(x)\displaystyle f^{-1}(x) is
    (A)
    (x+5)13\displaystyle (x+5)^{\frac{1}{3}}
    (B)
    (x−5)13\displaystyle (x-5)^{\frac{1}{3}}
    (C)
    (5−x)13\displaystyle (5-x)^{\frac{1}{3}}
    (D)
    5−x\displaystyle 5-x

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    NCERT’s answer
    B
    (B) \(\displaystyle (x-5)^{\frac{1}{3}}\)\[y=x^3+5 \] \[x^3=y-5 \] \[x=(y-5)^{\frac{1}{3}} \] \[f^{-1}(x)=(x-5)^{\frac{1}{3}} \]Check: \[f\big((x-5)^{\frac{1}{3}}\big)=(x-5)+5=x \]