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NCERT Exemplar · Class 12 Mathematics Relations and Functions

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EXERCISE 1.3 1–10 (part 1 of 6)

  1. Exercise 1

    Let A={a,b,c}\displaystyle \mathrm{A}=\{a, b, c\} and the relation R be defined on A as follows: R={(a,a),(b,c),(a,b)}.\mathrm{R}=\{(a, a),(b, c),(a, b)\} . Then, write minimum number of ordered pairs to be added in R\displaystyle \mathbf{R} to make R\displaystyle \mathbf{R} reflexive and transitive.

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    NCERT’s answer
    \(\displaystyle (b,b), (c,c), (a,c)\)
    Reflexive on \(\displaystyle A\) needs \(\displaystyle (a,a),(b,b),(c,c)\); only the first is present. \[\text{missing: } (b,b),\ (c,c) \] Transitive: the chain below forces one more pair. \[(a,b)\in R,\ (b,c)\in R \Rightarrow (a,c) \] After adding the three pairs: \[R'=\{(a,a),(b,b),(c,c),(a,b),(b,c),(a,c)\} \] \[(x,y),(y,z)\in R' \Rightarrow (x,z)\in R' \quad \text{for every chain} \] Each added pair is forced, so none can be dropped. Answer: $\displaystyle 3$ ordered pairs, \(\displaystyle (b,b),\ (c,c),\ (a,c)\).

  2. Exercise 2

    Let D be the domain of the real valued function f\displaystyle f defined by f(x)=25−x2\displaystyle f(x)=\sqrt{25-x^2}. Then, write D.

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    NCERT’s answer
    [-$\displaystyle 5{,}5$]
    The square root is real only when the radicand is non-negative. \[25-x^2\ge 0 \] \[x^2\le 25 \] \[|x|\le 5 \] \[-5\le x\le 5 \] Answer: \(\displaystyle D=[-5,\,5]\)

  3. Exercise 3

    Let f,g:R→R\displaystyle f, g: \mathbf{R} \rightarrow \mathbf{R} be defined by f(x)=2x+1\displaystyle f(x)=2 x+1 and g(x)=x2−2,∀x∈R\displaystyle g(x)=x^2-2, \forall x \in \mathbf{R}, respectively. Then, find gof\displaystyle g o f.

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    NCERT’s answer
    \(\displaystyle 4 x^2+4 x-1\)
    \[(g\circ f)(x)=g(f(x)) \] \[=g(2x+1) \] \[=(2x+1)^2-2 \] \[=4x^2+4x+1-2 \] \[=4x^2+4x-1 \] Answer: \(\displaystyle (g\circ f)(x)=4x^2+4x-1\), \(\displaystyle x\in\mathbb{R}\)

  4. Exercise 4

    Let f:R→R\displaystyle f: \mathbf{R} \rightarrow \mathbf{R} be the function defined by f(x)=2x−3∀x∈R\displaystyle f(x)=2 x-3 \forall x \in \mathrm{R}. write f−1\displaystyle f^{-1}.

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    NCERT’s answer
    \(\displaystyle f^{-1}(x)=\frac{x+3}{2}\)
    \(\displaystyle f^{-1}\) exists only if \(\displaystyle f\) is bijective. \[f(x_1)=f(x_2) \Rightarrow 2x_1-3=2x_2-3 \Rightarrow x_1=x_2 \] \[y=2x-3 \Rightarrow x=\frac{y+3}{2}\in\mathbb{R} \] \[f\!\left(\frac{y+3}{2}\right)=y \] So \(\displaystyle f\) is one-one and onto, and the inverse is the \(\displaystyle x\) found above. \[f^{-1}(y)=\frac{y+3}{2} \] Answer: \(\displaystyle f^{-1}(x)=\dfrac{x+3}{2}\)

  5. Exercise 5

    If A={a,b,c,d}\displaystyle \mathrm{A}=\{a, b, c, d\} and the function f={(a,b),(b,d),(c,a),(d,c)}\displaystyle f=\{(a, b),(b, d),(c, a),(d, c)\}, write f−1\displaystyle f^{-1}.

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    NCERT’s answer
    \(\displaystyle f^{-1}\{(b, a),(d, b),(a, c),(c, d)\}\)
    \[f(a)=b,\quad f(b)=d,\quad f(c)=a,\quad f(d)=c \] The images \(\displaystyle b,d,a,c\) are all different and fill \(\displaystyle A\), so \(\displaystyle f\) is bijective. Reverse each pair: \[f^{-1}=\{(b,a),(d,b),(a,c),(c,d)\} \] Answer: \(\displaystyle f^{-1}=\{(b,a),(d,b),(a,c),(c,d)\}\)

  6. Exercise 6

    If f:R→R\displaystyle f: \mathbf{R} \rightarrow \mathbf{R} is defined by f(x)=x2−3x+2\displaystyle f(x)=x^2-3 x+2, write f(f(x))\displaystyle f(f(x)).

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    NCERT’s answer
    \(\displaystyle f(f(x))=x^4-6 x^3+10 x^2-3 x\)
    \[f(f(x))=(x^2-3x+2)^2-3(x^2-3x+2)+2 \] \[(x^2-3x+2)^2=x^4-6x^3+13x^2-12x+4 \] \[f(f(x))=x^4-6x^3+13x^2-12x+4-3x^2+9x-6+2 \] \[=x^4-6x^3+10x^2-3x \] Answer: \(\displaystyle f(f(x))=x^4-6x^3+10x^2-3x\)

  7. Exercise 7

    Is g={(1,1),(2,3),(3,5),(4,7)}\displaystyle g=\{(1,1),(2,3),(3,5),(4,7)\} a function? If g\displaystyle g is described by g(x)=αx+β\displaystyle g(x)=\alpha x+\beta, then what value should be assigned to α\displaystyle \alpha and β\displaystyle \beta.

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    NCERT’s answer
    \(\displaystyle \alpha=2, \beta=-1\)
    Yes, \(\displaystyle g\) is a function: each of \(\displaystyle 1,2,3,4\) appears once as a first element, so each has exactly one image. Use two pairs to fix \(\displaystyle \alpha,\beta\). \[g(1)=\alpha+\beta=1 \] \[g(2)=2\alpha+\beta=3 \] Subtract: \[\alpha=2,\qquad \beta=1-2=-1 \] Check on the other pairs: \[g(3)=2(3)-1=5,\qquad g(4)=2(4)-1=7 \] Answer: \(\displaystyle g\) is a function; \(\displaystyle \alpha=2,\ \beta=-1\), so \(\displaystyle g(x)=2x-1\)

  8. Exercise 8

    Are the following set of ordered pairs functions? If so, examine whether the mapping is injective or surjective.
    (i)
    {(x,y):x\displaystyle \{(x, y): x is a person, y\displaystyle y is the mother of x}\displaystyle x\}.
    (ii)
    {(a,b):a\displaystyle \{(a, b): a is a person, b\displaystyle b is an ancestor of a}\displaystyle a\}.

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    NCERT’s answer
    (i)
    represents function which is surjective but not injective
    (ii)
    does not represent function.
    (i) A function: every person \(\displaystyle x\) has exactly one mother. The codomain is the set of mothers. Not injective: siblings share a mother. \[x_1\ne x_2,\quad f(x_1)=f(x_2) \] Surjective: every mother \(\displaystyle y\) is the mother of her own child. \[y\ \text{a mother},\ x\ \text{her child} \Rightarrow f(x)=y \] (ii) Not a function: a person has several ancestors, so one \(\displaystyle a\) is paired with many \(\displaystyle b\). \[(a,b_1),\ (a,b_2)\ \text{in the set, with}\ b_1\ne b_2 \] Answer: (i) a function, surjective but not injective; (ii) not a function

  9. Exercise 9

    If the mappings f\displaystyle f and g\displaystyle g are given by 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)\}, write fog\displaystyle f o g.

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    NCERT’s answer
    \(\displaystyle f o g=\{(2,5),(5,2),(1,5)\}\)
    Compose by applying \(\displaystyle g\) first, then \(\displaystyle f\), on the domain of \(\displaystyle g\). \[(f\circ g)(x) = f(g(x)) \] \[(f\circ g)(2) = f(3) = 5 \] \[(f\circ g)(5) = f(1) = 2 \] \[(f\circ g)(1) = f(3) = 5 \] Answer: \(\displaystyle f\circ g=\{(2,5),(5,2),(1,5)\}\)

  10. Exercise 10

    Let C\displaystyle \mathbf{C} be the set of complex numbers. Prove that the mapping f:C→R\displaystyle f: \mathbf{C} \rightarrow \mathbf{R} given by f(z)=∣z∣,∀z∈C\displaystyle f(z)=|z|, \forall z \in \mathbf{C}, is neither one-one nor onto.

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    Not one-one: \[f(1) = |1| = 1, \qquad f(-1) = |-1| = 1 \] \[f(1) = f(-1) \quad\text{but}\quad 1 \ne -1 \] Not onto: \[f(z) = |z| \ge 0 \quad \forall z \in \mathbf{C} \] \[-1 \in \mathbf{R}, \qquad f(z) = -1 \Rightarrow |z| = -1 < 0 \quad \text{(impossible)} \] So \(\displaystyle -1\) has no pre-image. Answer: \(\displaystyle f\) is neither one-one nor onto.