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NCERT Exemplar · Class 12 Mathematics Relations and Functions

62 questions · 62 still being checked

EXERCISE 1.3 51–62 (part 6 of 6)

  1. Fill in the blanks in each of the Exercises $\displaystyle 48$ to 52.

    Exercise 51

    Let f:R→R\displaystyle f: \mathbf{R} \rightarrow \mathbf{R} be defined by f(x)=x1+x2\displaystyle f(x)=\frac{x}{\sqrt{1+x^2}}. Then (fofof)(x)=\displaystyle (f o f o f)(x)= ____

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    NCERT’s answer
    \(\displaystyle (f o f o f)(x)=\frac{x}{\sqrt{3 x^2+1}}\)
    \(\displaystyle \dfrac{x}{\sqrt{1+3x^2}}\)\[1+f(x)^2 = 1+\frac{x^2}{1+x^2}=\frac{1+2x^2}{1+x^2} \] \[f(f(x))=\frac{f(x)}{\sqrt{1+f(x)^2}}=\frac{x}{\sqrt{1+x^2}}\cdot\sqrt{\frac{1+x^2}{1+2x^2}}=\frac{x}{\sqrt{1+2x^2}} \] \[1+f(f(x))^2=1+\frac{x^2}{1+2x^2}=\frac{1+3x^2}{1+2x^2} \] \[(f\circ f\circ f)(x)=\frac{f(f(x))}{\sqrt{1+f(f(x))^2}}=\frac{x}{\sqrt{1+2x^2}}\cdot\sqrt{\frac{1+2x^2}{1+3x^2}}=\frac{x}{\sqrt{1+3x^2}} \]

  2. Exercise 52

    If f(x)=(4−(x−7)3}\displaystyle f(x)=\left(4-(x-7)^3\right\}, then f−1(x)=\displaystyle f^{-1}(x)= ____.

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    NCERT’s answer
    \(\displaystyle f^{-1}(x)=7+(4-x)^{\frac{1}{3}}\)
    \(\displaystyle 7+(4-x)^{1/3}\)\[y=4-(x-7)^3 \] \[(x-7)^3=4-y \] \[x=7+(4-y)^{1/3} \] \[f^{-1}(x)=7+(4-x)^{1/3} \]A cube root exists for every real number, so \(\displaystyle f\) is one-one and onto \(\displaystyle \mathbf{R}\). Check: \[f\big(f^{-1}(x)\big)=4-\big((4-x)^{1/3}\big)^3=4-(4-x)=x \]
  3. State True or False for the statements in each of the Exercises $\displaystyle 53$ to 63.

    Exercise 53

    Let R={(3,1),(1,3),(3,3)}\displaystyle R=\{(3,1),(1,3),(3,3)\} be a relation defined on the set A={1,2,3}\displaystyle A=\{1,2,3\}. Then R\displaystyle R is symmetric, transitive but not reflexive.

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    NCERT’s answer
    False
    False. \(\displaystyle R\) is symmetric but not transitive.\[(1,3)\in R,\; (3,1)\in R \;\Rightarrow\; (1,1)\in R \ \text{would be required} \] \[(1,1)\notin R \]

  4. Exercise 54

    Let f:R→R\displaystyle f: \mathbf{R} \rightarrow \mathbf{R} be the function defined by f(x)=sin⁡(3x+2)∀x∈R\displaystyle f(x)=\sin (3 x+2) \forall x \in \mathrm{R}. Then f\displaystyle f is invertible.

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    NCERT’s answer
    False
    False. \(\displaystyle f\) is periodic, so it is not one-one.\[f\!\left(x+\tfrac{2\pi}{3}\right)=\sin(3x+2+2\pi)=\sin(3x+2)=f(x) \]Different inputs give the same output. Also the range is \(\displaystyle [-1,1]\ne\mathbf{R}\), so \(\displaystyle f\) is not onto either.

  5. Exercise 55

    Every relation which is symmetric and transitive is also reflexive.

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    NCERT’s answer
    False
    False. Take \(\displaystyle A=\{1,2,3\}\) and \[R=\{(1,1),(1,2),(2,1),(2,2)\} \]\(\displaystyle R\) is symmetric and transitive, but \[(3,3)\notin R \] so \(\displaystyle R\) is not reflexive.

  6. Exercise 56

    An integer m\displaystyle m is said to be related to another integer n\displaystyle n if m\displaystyle m is a integral multiple of n\displaystyle n. This relation in Z\displaystyle \mathbf{Z} is reflexive, symmetric and transitive.

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    NCERT’s answer
    False
    False. The relation is not symmetric.\[4=2\cdot 2 \;\Rightarrow\; (4,2)\in R \] \[2=4k \;\Rightarrow\; k=\tfrac12\notin\mathbf{Z} \;\Rightarrow\; (2,4)\notin R \]

  7. Exercise 57

    Let A={0,1}\displaystyle \mathrm{A}=\{0,1\} and N\displaystyle \mathbf{N} be the set of natural numbers. Then the mapping f:N→\displaystyle f: \mathbf{N} \rightarrow A defined by f(2n−1)=0,f(2n)=1,∀n∈N\displaystyle f(2 n-1)=0, f(2 n)=1, \forall n \in \mathbf{N}, is onto.

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    NCERT’s answer
    True
    True — every element of \(\displaystyle A\) has a preimage in \(\displaystyle \mathbf{N}\). \[f(1) = f(2\cdot 1-1) = 0 \] \[f(2) = f(2\cdot 1) = 1 \] \[\text{range of } f = \{0,1\} = A \] So \(\displaystyle f\) is onto.

  8. Exercise 58

    The relation R\displaystyle R on the set A={1,2,3}\displaystyle A=\{1,2,3\} defined as R={{1,1),(1,2),(2,1),(3,3)}\displaystyle R=\{\{1,1),(1,2),(2,1),(3,3)\} is reflexive, symmetric and transitive.

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    NCERT’s answer
    False
    False — \(\displaystyle 2\in A\) but \(\displaystyle (2,2)\notin R\), so \(\displaystyle R\) is not reflexive. \[(2,1)\in R, \quad (1,2)\in R \] \[\text{but } (2,2)\notin R \quad \Rightarrow \quad R \text{ is not transitive} \]

  9. Exercise 59

    The composition of functions is commutative.

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    NCERT’s answer
    False
    False — take \(\displaystyle f(x)=x+1\) and \(\displaystyle g(x)=x^2\) on \(\displaystyle \mathbf{R}\). \[(f\circ g)(x) = f(x^2) = x^2+1 \] \[(g\circ f)(x) = g(x+1) = (x+1)^2 \] \[(f\circ g)(1) = 2 \neq 4 = (g\circ f)(1) \] So \(\displaystyle f\circ g \neq g\circ f\).

  10. Exercise 60

    The composition of functions is associative.

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    NCERT’s answer
    True
    True — evaluate both sides at any \(\displaystyle x\). \[((f\circ g)\circ h)(x) = (f\circ g)(h(x)) = f(g(h(x))) \] \[(f\circ(g\circ h))(x) = f\big((g\circ h)(x)\big) = f(g(h(x))) \] \[(f\circ g)\circ h = f\circ(g\circ h) \]

  11. Exercise 61

    Every function is invertible.

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    NCERT’s answer
    False
    False — a function is invertible only if it is one-one and onto. Take \(\displaystyle f:\mathbf{R}\to\mathbf{R}\), \(\displaystyle f(x)=x^2\). \[f(1) = 1 = f(-1) \] \[1 \neq -1 \quad \Rightarrow \quad f \text{ is not one-one} \] So \(\displaystyle f\) has no inverse.

  12. Exercise 62

    A binary operation on a set has always the identity element.

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    NCERT’s answer
    False
    False — take addition on \(\displaystyle \mathbf{N}\). \[a * b = a + b \] \[a * e = a \;\Rightarrow\; a + e = a \;\Rightarrow\; e = 0 \notin \mathbf{N} \] So this operation has no identity element.