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NCERT Exemplar · Class 12 Mathematics Relations and Functions

62 questions · 62 still being checked

EXERCISE 1.3 41–50 (part 5 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 41

    Let f:A→B\displaystyle f: \mathrm{A} \rightarrow \mathrm{B} and g:B→C\displaystyle g: \mathrm{B} \rightarrow \mathrm{C} be the bijective functions. Then (gof)−1\displaystyle (g o f)^{-1} is
    (A)
    f−1og−1\displaystyle f^{-1} o g^{-1}
    (B)
    fog\displaystyle f o g
    (C)
    g−1of−1\displaystyle g^{-1} o f^{-1}
    (D)
    gof\displaystyle g o f

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    NCERT’s answer
    A
    (A) \(\displaystyle f^{-1}\circ g^{-1}\): composing it with \(\displaystyle g\circ f\) on either side gives the identity.\[(g\circ f)\circ(f^{-1}\circ g^{-1}) = g\circ(f\circ f^{-1})\circ g^{-1} = g\circ g^{-1} = I_C \]\[(f^{-1}\circ g^{-1})\circ(g\circ f) = f^{-1}\circ(g^{-1}\circ g)\circ f = f^{-1}\circ f = I_A \]\[(g\circ f)^{-1} = f^{-1}\circ g^{-1} \]

  2. Exercise 42

    Let f:R−{35}→R\displaystyle f: \mathbf{R}-\left\{\frac{3}{5}\right\} \rightarrow \mathbf{R} be defined by f(x)=3x+25x−3\displaystyle f(x)=\frac{3 x+2}{5 x-3}. Then
    (A)
    f−1(x)=f(x)\displaystyle f^{-1}(x)=f(x)
    (B)
    f−1(x)=−f(x)\displaystyle f^{-1}(x)=-f(x)
    (C)
    (fof)x=−x\displaystyle (f o f) x=-x
    (D)
    f−1(x)=119f(x)\displaystyle f^{-1}(x)=\frac{1}{19} f(x)

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    NCERT’s answer
    A
    (A) \(\displaystyle f^{-1}(x)=f(x)\), because \(\displaystyle f\circ f\) is the identity.\[f(f(x)) = \frac{3\cdot\dfrac{3x+2}{5x-3}+2}{5\cdot\dfrac{3x+2}{5x-3}-3} \]\[= \frac{3(3x+2)+2(5x-3)}{5(3x+2)-3(5x-3)} = \frac{19x}{19} = x \]\[f\circ f = I \;\Rightarrow\; f^{-1}=f \]

  3. Exercise 43

    Let f:[0,1]→[0,1]\displaystyle f:[0,1] \rightarrow[0,1] be defined by f(x)={x, if x is rational 1−x, if x is irrational \displaystyle f(x)=\left\{\begin{array}{c}x, \text { if } x \text { is rational } \\ 1-x, \text { if } x \text { is irrational }\end{array}\right. Then (fof)x\displaystyle (f o f) x is
    (A)
    constant
    (B)
    1+x\displaystyle 1+x
    (C)
    x\displaystyle x (D) none of these

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    NCERT’s answer
    C
    (C) \(\displaystyle x\). Treat rational and irrational \(\displaystyle x\) separately.\[x\in\mathbf{Q}:\quad f(x)=x\in\mathbf{Q} \;\Rightarrow\; f(f(x)) = x \]\[x\notin\mathbf{Q}:\quad f(x)=1-x\notin\mathbf{Q} \;\Rightarrow\; f(f(x)) = 1-(1-x) = x \]\[(f\circ f)(x) = x \quad \text{for all } x\in[0,1] \]

  4. Exercise 44

    Let f:[2,∞)→R\displaystyle f:[2, \infty) \rightarrow \mathbf{R} be the function defined by f(x)=x2−4x+5\displaystyle f(x)=x^2-4 x+5, then the range of f\displaystyle f is
    (A)
    R\displaystyle \mathbf{R}
    (B)
    [1,∞)\displaystyle [1, \infty)
    (C)
    [4,∞)\displaystyle [4, \infty)
    (B)
    [5,∞)\displaystyle [5, \infty)

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    NCERT’s answer
    B
    (B) \(\displaystyle [1,\infty)\). Complete the square, then solve for \(\displaystyle x\) in terms of \(\displaystyle y\).\[y = x^2-4x+5 = (x-2)^2+1 \]\[x\ge 2 \;\Rightarrow\; (x-2)^2\ge 0 \;\Rightarrow\; y\ge 1 \]\[x = 2+\sqrt{y-1}\ \ge 2 \quad \text{exists for every } y\ge 1 \]\[\text{Range of } f = [1,\infty) \]

  5. Exercise 45

    Let f:N→R\displaystyle f: \mathbf{N} \rightarrow \mathbf{R} be the function defined by f(x)=2x−12\displaystyle f(x)=\frac{2 x-1}{2} and g:Q→R\displaystyle g: \mathbf{Q} \rightarrow \mathbf{R} be another function defined by g(x)=x+2\displaystyle g(x)=x+2. Then (gof)32\displaystyle (g o f) \frac{3}{2} is
    (A)
    1\displaystyle 1 (B) 1\displaystyle 1 (C) 72\displaystyle \frac{7}{2}
    (B)
    none of these

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    NCERT’s answer
    D
    (D) none of these.\[\frac{3}{2}\notin\mathbf{N} \quad \text{(domain of } f\text{)} \]So \(\displaystyle (g\circ f)\left(\frac{3}{2}\right)\) is undefined. Even the bare formula gives a value that is not listed:\[\frac{2\cdot\dfrac{3}{2}-1}{2}+2 = 1+2 = 3 \]

  6. Exercise 46

    Let f:R→R\displaystyle f: \mathbf{R} \rightarrow \mathbf{R} be defined by f(x)={2x:x>3x2:1<x≤33x:x≤1f(x)=\left\{\begin{array}{c} 2 x: x>3 \\ x^2: 1<x \leq 3 \\ 3 x: x \leq 1 \end{array}\right. Then f(−1)+f(2)+f(4)\displaystyle f(-1)+f(2)+f(4) is
    (A)
    9\displaystyle 9 (B) 14\displaystyle 14
    (C)
    5\displaystyle 5 (D) none of these

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    NCERT’s answer
    A
    (A) \(\displaystyle 9\). Use the branch that matches each argument.\[f(-1) = 3(-1) = -3 \qquad (x\le 1) \]\[f(2) = 2^2 = 4 \qquad (1<x\le 3) \]\[f(4) = 2(4) = 8 \qquad (x>3) \]\[f(-1)+f(2)+f(4) = -3+4+8 = 9 \]

  7. Exercise 47

    Let f:R→R\displaystyle f: \mathbf{R} \rightarrow \mathbf{R} be given by f(x)=tan⁡x\displaystyle f(x)=\tan x. Then f−1(1)\displaystyle f^{-1}(1) is
    (A)
    π4\displaystyle \frac{\pi}{4}
    (B)
    {nπ+π4:n∈Z}\displaystyle \left\{n \pi+\frac{\pi}{4}: n \in Z\right\}
    (C)
    does not exist
    (D)
    none of these

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    NCERT’s answer
    B
    (B) \(\displaystyle f\) is not one-one, so \(\displaystyle f^{-1}(1)\) is the set of all \(\displaystyle x\) with \(\displaystyle f(x)=1\).\[\tan x = 1 = \tan\frac{\pi}{4} \]\[x = n\pi+\frac{\pi}{4}, \quad n\in\mathbf{Z} \]\[f^{-1}(1) = \left\{ n\pi+\frac{\pi}{4} : n\in\mathbf{Z} \right\} \]
  8. Fill in the blanks in each of the Exercises $\displaystyle 48$ to 52.

    Exercise 48

    Let the relation R be defined in N\displaystyle \mathbf{N} by aRb\displaystyle a \mathrm{R} b if 2a+3b=30\displaystyle 2 a+3 b=30. Then R=\displaystyle \mathrm{R}= ____.

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    NCERT’s answer
    \(\displaystyle \mathrm{R}=\{(3,8),(6,6),(9,4),(12,2)\}\)
    {$\displaystyle (3, 8)$, $\displaystyle (6, 6)$, $\displaystyle (9, 4)$, $\displaystyle (12, 2)$}\[2a+3b=30 \;\Rightarrow\; b=\frac{30-2a}{3} \]For \(\displaystyle b\) to be a natural number, \(\displaystyle 3\mid a\), and \(\displaystyle b>0\) needs \(\displaystyle a<15\):\[a=3,\,6,\,9,\,12 \;\Rightarrow\; b=8,\,6,\,4,\,2 \]\[a=15 \;\Rightarrow\; b=0\notin\mathbf{N} \]\[\mathrm{R}=\{(3,8),\,(6,6),\,(9,4),\,(12,2)\} \]

  9. Exercise 49

    Let the relation R be defined on the set A={1,2,3,4,5}\displaystyle \mathrm{A}=\{1,2,3,4,5\} by R={(a,b):∣a2−b2∣<8\displaystyle \mathrm{R}=\left\{(a, b):\left|a^2-b^2\right|<8\right.. Then R is given by ____.

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    NCERT’s answer
    \(\displaystyle \mathrm{R}=\{(1,1),(1,2),(2,1),(2,2),(2,3),(3,2),(3,3),(3,4),(4,3),(4,4),(5,5)\}\)
    \(\displaystyle \mathrm{R}=\{(1,1),(2,2),(3,3),(4,4),(5,5),(1,2),(2,1),(2,3),(3,2),(3,4),(4,3)\}\)\[a=b \;\Rightarrow\; |a^2-b^2|=0<8 \] \[|1^2-2^2|=3,\quad |2^2-3^2|=5,\quad |3^2-4^2|=7 \quad (<8) \] \[|1^2-3^2|=8,\quad |4^2-5^2|=9,\quad |2^2-4^2|=12 \quad (\ge 8) \]Every other pair gives a difference of at least $\displaystyle 8$, so it is left out.

  10. Exercise 50

    Let f={(1,2),(3,5),(4,1)\displaystyle f=\{(1,2),(3,5),(4,1) and g={(2,3),(5,1),(1,3)}\displaystyle g=\{(2,3),(5,1),(1,3)\}. Then gof=\displaystyle g o f= ____ and fog=\displaystyle f o g= ____.

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    NCERT’s answer
    \(\displaystyle g o f=\{(1,3),(3,1),(4,3)\}\) and \(\displaystyle f o g=\{(2,5),(5,2),(1,5)\}\)
    \(\displaystyle g\circ f=\{(1,3),(3,1),(4,3)\}\) and \(\displaystyle f\circ g=\{(1,5),(2,5),(5,2)\}\)\[(g\circ f)(x)=g(f(x)) \] \[f(1)=2,\; g(2)=3 \;\Rightarrow\; (1,3) \] \[f(3)=5,\; g(5)=1 \;\Rightarrow\; (3,1) \] \[f(4)=1,\; g(1)=3 \;\Rightarrow\; (4,3) \] \[(f\circ g)(x)=f(g(x)) \] \[g(1)=3,\; f(3)=5 \;\Rightarrow\; (1,5) \] \[g(2)=3,\; f(3)=5 \;\Rightarrow\; (2,5) \] \[g(5)=1,\; f(1)=2 \;\Rightarrow\; (5,2) \]