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NCERT Exemplar · Class 11 Mathematics Relations and Functions

42 questions · 42 still being checked

EXERCISE 2.3 31–42 (part 4 of 4)

  1. Choose the correct answers in Exercises from $\displaystyle 24$ to $\displaystyle 35$ (M.C.Q.)

    Exercise 31

    The domain and range of the real function f\displaystyle f defined by f(x)=4−xx−4\displaystyle f(x)=\frac{4-x}{x-4} is given by
    (A)
    Domain =R\displaystyle =\mathbf{R}, Range ={−1,1}\displaystyle =\{-1,1\}
    (B)
    Domain =R−{1}\displaystyle =\mathbf{R}-\{1\}, Range =R\displaystyle =\mathbf{R}
    (C)
    Domain =R−{4}\displaystyle =\mathbf{R}-\{4\}, Range ={−1}\displaystyle =\{-1\}
    (D)
    Domain =R−{−4}\displaystyle =\mathbf{R}-\{-4\}, Range ={−1,1}\displaystyle =\{-1,1\}

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    NCERT’s answer
    C
    (C) Domain \(\displaystyle =\mathbf{R}-\{4\}\), Range \(\displaystyle =\{-1\}\). The denominator must not vanish; elsewhere the quotient is constant.\[x-4\ne 0 \Rightarrow x\ne 4 \] \[f(x)=\frac{4-x}{x-4}=\frac{-(x-4)}{x-4}=-1 \quad (x\ne 4) \]

  2. Exercise 32

    The domain and range of real function f\displaystyle f defined by f(x)=x−1\displaystyle f(x)=\sqrt{x-1} is given by
    (A)
    Domain =(1,∞)\displaystyle =(1, \infty), Range =(0,∞)\displaystyle =(0, \infty)
    (B)
    Domain =[1,∞)\displaystyle =[1, \infty), Range =(0,∞)\displaystyle =(0, \infty)
    (C)
    Domain =[1,∞)\displaystyle =[1, \infty), Range =[0,∞)\displaystyle =[0, \infty)
    (D)
    Domain =[1,∞)\displaystyle =[1, \infty), Range =[0,∞)\displaystyle =[0, \infty)

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    NCERT’s answer
    C
    (C) Domain \(\displaystyle =[1,\infty)\), Range \(\displaystyle =[0,\infty)\) (option (D) is printed identically). The radicand must be non-negative, and every \(\displaystyle y\ge 0\) is attained.\[x-1\ge 0 \Rightarrow x\ge 1 \] \[f(1)=0,\qquad \sqrt{x-1}\ge 0 \] \[y=\sqrt{x-1}\Rightarrow x=y^2+1\ge 1 \quad (y\ge 0) \]

  3. Exercise 33

    The domain of the function f\displaystyle f given by f(x)=x2+2x+1x2−x−6\displaystyle f(x)=\frac{x^2+2 x+1}{x^2-x-6}
    (A)
    R−{3,−2}\displaystyle \mathbf{R}-\{3,-2\}
    (B)
    R−{−3,2}\displaystyle \mathbf{R}-\{-3,2\}
    (C)
    R−[3,−2]\displaystyle \mathbf{R}-[3,-2]
    (D)
    R−(3,−2)\displaystyle \mathbf{R}-(3,-2)

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    NCERT’s answer
    A
    (A) \(\displaystyle \mathbf{R}-\{3,-2\}\). Exclude the zeros of the denominator; the numerator \(\displaystyle (x+1)^2\) shares no factor with it, so nothing cancels.\[x^2-x-6=(x-3)(x+2) \] \[(x-3)(x+2)=0 \Rightarrow x=3,\ -2 \]

  4. Exercise 34

    The domain and range of the function f\displaystyle f given by f(x)=2−∣x−5∣\displaystyle f(x)=2-|x-5| is
    (A)
    Domain =R+\displaystyle =\mathbf{R}^{+}, Range =(−∞,1]\displaystyle =(-\infty, 1]
    (B)
    Domain =R\displaystyle =\mathbf{R}, Range =(−∞,2]\displaystyle =(-\infty, 2]
    (C)
    Domain =R\displaystyle =\mathbf{R}, Range =(−∞,2)\displaystyle =(-\infty, 2)
    (D)
    Domain =R+\displaystyle =\mathbf{R}^{+}, Range =(−∞,2]\displaystyle =(-\infty, 2]

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    NCERT’s answer
    B
    (B) Domain \(\displaystyle =\mathbf{R}\), Range \(\displaystyle =(-\infty,2]\). \(\displaystyle |x-5|\) is defined for every real \(\displaystyle x\) and is never negative, so \(\displaystyle 2\) is the maximum, reached at \(\displaystyle x=5\).\[|x-5|\ge 0 \Rightarrow f(x)=2-|x-5|\le 2 \] \[f(5)=2 \] \[f\bigl(5+(2-y)\bigr)=2-(2-y)=y \quad \text{for every } y\le 2 \]

  5. Exercise 35

    The domain for which the functions defined by f(x)=3x2−1\displaystyle f(x)=3 x^2-1 and g(x)=3+x\displaystyle g(x)=3+x are equal is
    (A)
    {−1,43}\displaystyle \left\{-1, \frac{4}{3}\right\}
    (B)
    {−1,43}\displaystyle \left\{-1, \frac{4}{3}\right\}
    (C)
    (−1,43)\displaystyle \left(-1, \frac{4}{3}\right)
    (D)
    [−1,43)\displaystyle \left[-1, \frac{4}{3}\right)

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    NCERT’s answer
    A
    (A) \(\displaystyle \left\{-1,\tfrac43\right\}\) (option (B) is printed identically). The functions agree exactly where their values are equal.\[3x^2-1=3+x \] \[3x^2-x-4=0 \] \[(3x-4)(x+1)=0 \] \[x=\tfrac43 \ \text{or}\ x=-1 \]
  6. Fill in the blanks :

    Exercise 36

    Let f\displaystyle f and g\displaystyle g be two real functions given by f={(0,1),(2,0),(3,−4),(4,2),(5,1)}g={(1,0),(2,2),(3,−1),(4,4),(5,3)}\begin{gathered} f=\{(0,1),(2,0),(3,-4),(4,2),(5,1)\} \\ g=\{(1,0),(2,2),(3,-1),(4,4),(5,3)\} \end{gathered} then the domain of f.g\displaystyle f . g is given by ____\displaystyle \_\_\_\_.

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    NCERT’s answer
    \(\displaystyle \{2, 3, 4, 5\}\)
    \(\displaystyle \{2,3,4,5\}\)\[\operatorname{dom}(f\cdot g)=\operatorname{dom}f\cap\operatorname{dom}g \]\[\operatorname{dom}f=\{0,2,3,4,5\},\qquad \operatorname{dom}g=\{1,2,3,4,5\} \]\[\operatorname{dom}(f\cdot g)=\{2,3,4,5\} \]

  7. Exercise 37

    Let f={(2,4),(5,6),(8,−1),(10,−3)}\displaystyle f=\{(2,4),(5,6),(8,-1),(10,-3)\} g={(2,5),(7,1),(8,4),(10,13),(11,5)}g=\{(2,5),(7,1),(8,4),(10,13),(11,5)\} be two real functions. Then, match the following :
    (a) f−g\displaystyle f-g(i) {(2,45),(8,−14),(10,−313)}\displaystyle \left\{\left(2, \frac{4}{5}\right),\left(8, \frac{-1}{4}\right),\left(10, \frac{-3}{13}\right)\right\}
    (b) f+g\displaystyle f+g(ii) {(2,20),(8,−4),(10,−39)}\displaystyle \{(2,20),(8,-4),(10,-39)\}
    (c) f.g\displaystyle f . g(iii) {(2,−1),(8,−5),(10,−16)}\displaystyle \{(2,-1),(8,-5),(10,-16)\}
    (d) fg\displaystyle \frac{f}{g}(iv) {(2,9),(8,3),(10,10)}\displaystyle \{(2,9),(8,3),(10,10)\}

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    NCERT’s answer
    (a)
    \(\displaystyle \leftrightarrow\) (iii)
    (b)
    \(\displaystyle \leftrightarrow\) (iv)
    (c)
    \(\displaystyle \leftrightarrow\) (ii)
    (d)
    \(\displaystyle \leftrightarrow\) (i)
    (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)
    Each operation is defined only on the common domain:
    \[\operatorname{dom}f\cap\operatorname{dom}g=\{2,8,10\} \]
    (a)
    \[f-g:\ (2,\,4-5),\ (8,\,-1-4),\ (10,\,-3-13)=\{(2,-1),(8,-5),(10,-16)\} \]
    (b)
    \[f+g:\ (2,\,4+5),\ (8,\,-1+4),\ (10,\,-3+13)=\{(2,9),(8,3),(10,10)\} \]
    (c)
    \[f\cdot g:\ (2,\,4\cdot 5),\ (8,\,-1\cdot 4),\ (10,\,-3\cdot 13)=\{(2,20),(8,-4),(10,-39)\} \]
    (d)
    \(\displaystyle g\neq 0\) on \(\displaystyle \{2,8,10\}\), so
    \[\frac fg:\ \left(2,\tfrac45\right),\ \left(8,\tfrac{-1}{4}\right),\ \left(10,\tfrac{-3}{13}\right) \]
  8. State True or False for the following statements given in Exercises $\displaystyle 38$ to $\displaystyle 42$ :

    Exercise 38

    The ordered pair (5,2)\displaystyle (5,2) belongs to the relation R={(x,y):y=x−5,x,y∈Z}\displaystyle \mathrm{R}=\{(x, y): y=x-5, x, y \in \mathbf{Z}\}

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    NCERT’s answer
    False
    False. For \(\displaystyle x=5\):\[y=x-5=5-5=0 \]\[0\neq 2 \]so \(\displaystyle (5,2)\notin R\).

  9. Exercise 39

    If P={1,2}\displaystyle \mathrm{P}=\{1,2\}, then P×P×P={(1,1,1),(2,2,2),(1,2,2),(2,1,1)}\displaystyle \mathrm{P} \times \mathrm{P} \times \mathrm{P}=\{(1,1,1),(2,2,2),(1,2,2),(2,1,1)\}

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    NCERT’s answer
    False
    False. The product has \(\displaystyle 2\cdot2\cdot2\) ordered triples, but only $\displaystyle 4$ are listed.\[n(P\times P\times P)=2\cdot 2\cdot 2=8 \]\[P\times P\times P=\{(1,1,1),(1,1,2),(1,2,1),(1,2,2),(2,1,1),(2,1,2),(2,2,1),(2,2,2)\} \]

  10. Exercise 40

    If A={1,2,3},B={3,4}\displaystyle \mathrm{A}=\{1,2,3\}, \mathrm{B}=\{3,4\} and C={4,5,6}\displaystyle \mathrm{C}=\{4,5,6\}, then (A×B)∪(A×C)\displaystyle (\mathrm{A} \times \mathrm{B}) \cup(\mathrm{A} \times \mathrm{C}) ={(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,3),(3,4),(3,5),(3,6)}\displaystyle =\{(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,3),(3,4),(3,5),(3,6)\}.

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    NCERT’s answer
    True
    True. Product distributes over union.\[(A\times B)\cup(A\times C)=A\times(B\cup C) \]\[B\cup C=\{3,4,5,6\} \]\[n\big(A\times(B\cup C)\big)=3\cdot 4=12 \]Pairing each of \(\displaystyle 1,2,3\) with each of \(\displaystyle 3,4,5,6\) gives exactly the $\displaystyle 12$ pairs listed.

  11. Exercise 41

    If (x−2,y+5)=(−2,13)\displaystyle (x-2, y+5)=\left(-2, \frac{1}{3}\right) are two equal ordered pairs, then x=4,y=−143\displaystyle x=4, y=\frac{-14}{3}

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    NCERT’s answer
    False
    False. Equal ordered pairs have equal components.\[x-2=-2 \;\Rightarrow\; x=0 \]\[y+5=\frac13 \;\Rightarrow\; y=\frac13-5=-\frac{14}{3} \]So \(\displaystyle x=0\), not \(\displaystyle 4\).

  12. Exercise 42

    If A×B={(a,x),(a,y),(b,x),(b,y)}\displaystyle \mathrm{A} \times \mathrm{B}=\{(a, x),(a, y),(b, x),(b, y)\}, then A={a,b},B={x,y}\displaystyle \mathrm{A}=\{a, b\}, \mathrm{B}=\{x, y\}

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    NCERT’s answer
    True.
    True. The first coordinates give \(\displaystyle A\), the second coordinates give \(\displaystyle B\).\[A=\{a,b\},\qquad B=\{x,y\} \]\[A\times B=\{(a,x),(a,y),(b,x),(b,y)\} \]