Class 12: Functions – Exercise 2.2 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Find $gof \text{ and } fog \text{ when } f:R \rightarrow R \text{ and } g:R \rightarrow R$ are defined by:

$\text{(i) } f(x) = 2x+3 \text{ and } g(x) = x^2+5$

$\text{(ii) }f(x) = 2x+x^2 \text{ and } g(x) = x^3$

$\text{(iii) }f(x) = x^2+8 \text{ and } g(x) = 3x^3+1$

$\text{(iv) }f(x) = x \text{ and } g(x) = |x|$

$\text{(v) }f(x) = x^2+2x-3 \text{ and } g(x) = 3x-4$

$\text{(vi) } f(x) = 8x^3 \text{ and } g(x) = x^{1/3}$

Answer:

$\text{(i) } \text{Now, } f(x) = 2x+3 \text{ and } g(x) = x^2+5$

$gof(x) = g(2x+3) = (2x+3)^2 + 5 = 4x^2 + 12x + 14$

$fog(x) = f(g(x)) = f(x^2+5) = 2(x^2+5) + 3 = 2x^2 + 13$

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$\text{(ii) } \text{Now, }f(x) = 2x+x^2 \text{ and } g(x) = x^3$

$gof(x) = g(f(x)) = g(2x+x^2) = (2x+x^2)^3$

$fog(x) = f(g(x)) = f(x^3) = 2x^3 + x^6$

$\\$

$\text{(iii) } \text{Now, }f(x) = x^2+8 \text{ and } g(x) = 3x^3+1$

$gof(x) = g(f(x)) = g(x^2 + 8) = 3(x^2+8)^3 + 1$

$fog(x) = f(g(x)) = f( 3x^3 + 1) = ( 3x^3+1)^2 + 8 = 9x^6 + 6x^3 + 9$

$\\$

$\text{(iv) } \text{Now, }f(x) = x \text{ and } g(x) = |x|$

$gof(x) = g(f(x)) = g(x) = |x|$

$fog(x) = = f(g(x)) = f(|x|) = |x|$

$\\$

$\text{(v) } \text{Now, }f(x) = x^2+2x-3 \text{ and } g(x) = 3x-4$

$gof(x) = g(f(x)) = g(x^2+2x-3) = 3(x^2+2x-3) - 4 = 3x^2+ 6x-13$

$fog(x) = f(g(x)) = f(3x-4) = (3x-4)^2 + 2(3x-4) - 3 = 9x^2 - 18x + 5$

$\\$

$\text{(vi) } \text{Now, }f(x) = 8x^3 \text{ and } g(x) = x^{1/3}$

$\displaystyle gof(x) = g(f(x)) = g(8x^3) = ( 8x^3)^{\frac{1}{3}} = [(2x)^3]^{\frac{1}{3}} = 2x$

$\displaystyle fog(x) = f(g(x)) = f\Big(x^{\frac{1}{3}}\Big) = 8\Big(x^{\frac{1}{3}}\Big)^3 = 8x$

$\\$

Question 2: Let $f =\{ (3,1),(9,3),(12,4) \} \text{ and } g =\{ (1, 3),(3, 3), (4,9), (5,9) \}.$ Show that $gof \text{ and } fog$ are both defined. Also, find $fog \text{ and } gof.$

Answer:

$\text{ Given function } f = \{(3, 1), (9, 3), (12, 4)\} \text{ and } g = \{(1, 3), (3, 3),(4, 9),(5, 9)\}$

$f: \{3, 9, 12\} \rightarrow \{1, 3, 4\} \text{ and } g: \{1, 3, 4, 5\} \rightarrow \{3, 9 \}$

$\text{ Co-domain of } f \text{ is a subset of the domain of } g$

$\text{ Therefore, } gof \text{ exists and } gof: \{3, 9, 12\} \rightarrow \{3, 9 \}$

$(gof) (3) = g(f(3)) = g(1) = 3$

$(gof) (9) = g(f(9)) = g(3) = 3$

$(gof) (12) = g(f(12)) = g(4) = 9$

$\Rightarrow gof = \{(3, 3), (9, 3), (12, 9) \}$

$\text{ Co-domain of } g \text{ is a subset of the domain of } f.$

$\text{ Therefore, } fog \text{ exists and } fog: \{1, 3, 4, 5\} \rightarrow \{3, 9, 12\}$

$(fog) (1) = f(g(1)) = f(3) = 1$

$(fog) (3) = f(g(3)) = f(3) = 1$

$(fog) (4) = f(g(4)) = f(9) = 3$

$(fog) (5) = f(g(5)) = f(9) = 3$

$\Rightarrow fog = \{(1, 1), (3, 1), (4, 3), (5, 3)\}$

$\\$

Question 3: Let $f =\{ (1,-1), (4,-2),(9,-3),(1,6,4)\} \text{ and } g= \{ (-1, -2),(-2,-4), (-3,-6),(4,8) \}.$ Show that $gof$ is defined while $fog$ is not defined. Also, find $gof.$

Answer:

$\text{ Given as } \\ \\ f = \{(1, -1), (4, -2), (9, -3), (16, 4)\} \text{ and } g = \{(-1, -2), (-2, -4), (-3, -6), (4, 8)\}$

$f: \{1, 4, 9, 16\} \rightarrow \{-1, -2, -3, 4\} \text{ and } g: \{-1, -2, -3, 4\} \rightarrow \{-2, -4, -6, 8\}$

$\text{ Co-domain of } f = \text{ domain of } g$

$\text{ Therefore, } gof \text{ exists and } gof: \{1, 4, 9, 16\} \rightarrow \{-2, -4, -6, 8\}$

$(gof)(1) = g(f(1)) = g(-1) = -2$

$(gof)(4) = g(f(4)) = g(-2) = -4$

$(gof)(9) = g(f(9)) = g(-3) = -6$

$(gof)(16) = g(f(16)) = g(4) = 8$

$\text{ Therefore, } gof = \{(1, -2), (4, -4), (9, -6), (16, 8)\}$

$\text{ Since, the co-domain of } g \text{ is not same as the domain of } f.$

$\text{ Therefore, } fog \text{ does not exist. }$

$\\$

Question 4: Let $A = \{a, b, c\}, B = \{u , v, w \}$ and let $f$ and $g$ be two functions from $A$ to $B$ and from $B$ to $A$ respectively defined as: $f = \{(a, v), (b, u), (c, w)\}, g = \{(u, b), (v, a), (w, c)\}.$ Show that $f$ and $g$ both are bijections and find $fog$ and $gof.$

Answer:

$\text{ Given as } f = \{(a, v), (b, u), (c, w)\}, g = \{(u, b), (v, a), (w, c)\}$

$\text{ Also given that } A = \{a, b, c\}, B = \{uv, w\}$

$\text{ Let us show } f \text{ and } g \text{ both are bijective }$

$\text{ Consider that } f = \{(a, v), (b, u), (c, w)\} \text{ and } f: A \rightarrow B$

$\text{ Injectivity of } f$

$\text{ No two elements of } A \text{ have the same image in } B$

$\text{ Therefore, } f \text{ is one-one }$

$\text{ Surjectivity of } f:$

$\text{ Co-domain of } f = \{uv, w\}$

$\text{ Range of } f = \{uv, w\}$

$\text{ Both are same }$

$\text{ Therefore, } f \text{ is onto }$

$\text{ Hence, } f \text{ is a bijection. }$

$\text{ Now consider that } g = \{(u, b), (v, a), (w, c)\} \text{ and } g: B \rightarrow$

$\text{ Injectivity of } g$

$\text{ No two elements of } B \text{ have the same image in } A.$

$\text{ Therefore, } g \text{ is one-one. }$

$\text{ Surjectivity of } g$

$\text{ Co-domain of } g = \{a, b, c\}$

$\text{ Range of } g = \{a, b, c\}$

$\text{ Both are the same }$

$\text{ Therefore, } g \text{ is onto. }$

$\text{ Hence, } g \text{ is a bijection. }$

$\text{ Let us find } fog,$

$\text{ We known, Co-domain of } g \text{ is same as the domain of } f.$

$\text{ Therefore, } fog \text{ exists and } fog: \{uv, w\} \rightarrow \{uv, w\}$

$(fog)(u) = f(g(u)) = f(b) = u$

$(fog)(v) = f(g(v)) = f(a) = v$

$(fog)(w) = f(g(w)) = f(c) = w$

$\text{ Therefore, } fog = \{(u, u), (v, v), (w, w)\}$

$\text{ Let us find }gof,$

$\text{ Co-domain of } f \text{ is same as the domain of } g.$

$\text{ Therefore, } fog \text{ exists and } gof: \{a, b, c\} \rightarrow \{a, b, c\}$

$(gof)(a) = g(f(a)) = g(v) = a$

$(gof)(b) = g(f(b)) = g(u) = b$

$(gof)(c) = g(f(c)) = g(w) = c$

$\text{ Therefore, } gof = \{(a, a), (b, b), (c, c)\}$

$\\$

Question 5: Find $fog(2) \text{ and } gof(1)$ when: $f:R \rightarrow R; f (x)=x^2 +8 \text{ and } g:R \rightarrow R; g(x)=3x^3 +1$

Answer:

$\text{Given as } f: R \rightarrow R; f(x) = x^2 + 8 \text{ and } g: R \rightarrow R; g(x) = 3x^3 + 1.$

$(fog)(2) = f(g(2)) = f(3 \times 2^3 + 1) = f(3 \times 8 + 1) = f(25) = 25^2 + 8 = 633$

$(gof)(1) = g(f(1)) = g(1^2 + 8) = g(9) = 3 \times 9^3 + 1 = 2188$

$\\$

Question 6: Let $R^+$ be the set of non-negative real numbers. If $f:R^+ \rightarrow R^+ \text{ and } g: R^+ \rightarrow R^+ \text{ are defined as } f(x) = x^2 \text{ and } g(x) = + \sqrt{x}$ . Find $fog \text{ and } gof.$ Are they equal functions.

Answer:

$\text{ Given that } f: R^+ \rightarrow R^+ \text{ and } g: R^+ \rightarrow R^+$

$\text{ Therefore, } fog: R^+ \rightarrow R^+ \text{ and } gof: R^+ \rightarrow R^+$

$\text{ Domains of } fog \text{ and } gof \text{ are the same. }$

$\text{ Let us find } fog \text{ and } gof \text{ also check whether they are equal or not, }$

$(fog)(x) = f(g(x)) = f(\sqrt{x}) = \sqrt{x^2} = x$

$(gof)(x) = g(f(x)) = g(x^2) = \sqrt{x^2} = x$

$\text{ Therefore, } (fog)(x) = (gof)(x), \forall \ x \in R^+$

$\text{ Hence, } fog = gof$

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$\displaystyle \text{Question 7: Let } f:R \rightarrow R \text{ and } g: R \rightarrow R \text{ be defined by } f(x) = x^2 \\ \\ \text{ and } g(x) = x+ 1 . \text{ Show the } fog \neq gof .$

Answer:

$\text{ Given as } f: R \rightarrow R \text{ and } g: R \rightarrow R.$

Therefore, the domains of $f \text{ and } g$ are the same.

$(fog) (x) = f(g(x)) = f(x + 1) = (x + 1)^2 = x^2 + 1 + 2x$

$(gof)(x) = g(f(x)) = g(x^2) = x^2 + 1$

$\text{ Therefore, } fog \neq gof$

$\\$

$\displaystyle \text{Question 8: Let } f:R \rightarrow R \text{ and } g: R \rightarrow R \text{ be defined by } \\ \\ f(x) = x+1 \text{ and } g(x) = x- 1 . \text{ Show the } fog = gof = I_R .$

Answer:

$\text{ Given as } f: R \rightarrow R \text{ and } g: R \rightarrow R.$

$\Rightarrow fog : R \rightarrow \text{ and } gof : R \rightarrow R \text{ ( Also, we know that } I_R : R \rightarrow R \text{ ) }$

$\text{ So, the domain of all } fog , gof \text{ and } I_R \text{ are the same }$

$(fog)(x) = f(g(x)) = f( x-1 ) = x-1 + 1 = x = I_R(x)$

$(gof) (x) = g(f(x)) = g(x+1) = x+1 -1=x=I_R(x)$

$\therefore (fog) (x)= ( gof) (x) = I_R (x) , \forall \ x \in R$

$\text{ Hence, } fog = gof = I_R$

$\\$

$\displaystyle \text{Question 9: Verify associativity for the following three mappings: } \\ \\ f : N \rightarrow Z_0 \text{ (the set of non-zero integers), } g : Z_0 \rightarrow Q \text{ and } h : Q \rightarrow R \text{ given by } \\ \\ f(x) = 2x , g(x) = \frac{1}{x} \text{ and } h (x) = e^x.$

Answer:

$\text{ Given } f : N \rightarrow Z_0, \ \ g : Z_0 \rightarrow Q \text{ and } h : Q \rightarrow R$

$gof : N \rightarrow Q \text{ and } hog : Z_0 \rightarrow R$

$\Rightarrow ho(gof) : N \rightarrow R \text{ and } (hog)of : N \rightarrow R$

So, both have the same domains.

$\displaystyle (gof)(x) = g(f(x)) = g(2x) = \frac{1}{2x}$

$\displaystyle (hog)(x) = h( g(x)) = h \Big( \frac{1}{x} \Big) = e^{\frac{1}{x}}$

Now

$\displaystyle (ho(gof))(x) = h((gof)(x)) = h \Big( \frac{1}{2x} \Big) = e^{\frac{1}{2x}}$

$\displaystyle ((hog)of)(x) = (hog)(f(x)) = (hog)(2x) = e^{\frac{1}{2x}}$

$\Rightarrow ho(gof) =(hog)of$

Hence, the associative property has been verified.

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$\displaystyle \text{Question 10: Consider } f: N \rightarrow N, g: N \rightarrow N \text{ and } h : N \rightarrow R \\ \\ \text{ defined as } f(x) = 2x , g(y) = 3y + 4 \text{ and } h(z) = \sin z \text{ for all } x, y, z \in N. \\ \\ \text{Show that } ho(gof) = ( hog)of.$

Answer:

$\text{ Given } f: N \rightarrow N, g: N \rightarrow N \text{ and } h : N \rightarrow R$

$\Rightarrow gof : N \rightarrow N \text{ and } hog : N \rightarrow R$

$\Rightarrow ho(gof) : N \rightarrow R \text{ and } (hog) of : N \rightarrow R$

So, both have the same domains.

$(gof)(x) = g(f(x)) = g(2x) = 3(2x) + 4 = 6x + 4$

$(hog)(x) = h(g(x)) = h ( 3x+4) = \sin (3x+4)$

Now,

$\displaystyle (ho(gof))(x) = h((gof)(x)) = h (6x+4) = \sin (6x+4)$

$\displaystyle ((hog)of)(x) = (hog)(f(x)) = (hog)(2x) = \sin (6x+4)$

$\Rightarrow (ho(gof))(x) =((hog)of)(x) \ \forall \ x \in N$

$\text{ Hence, } ho(gof) =(hog)of$

$\\$

Question 11: Give examples of two functions $f: N \rightarrow N \text{ and } g : N \rightarrow N$ such that $gof$ is onto but $f$ is not onto.

Answer:

Let us consider $g: N \rightarrow N$ given by

$\displaystyle g(x) = \Bigg\{ \begin{array}{rr} x-1, & \text{ if } x > 1 \\ \\ 1, & \text{ if } x=1 \end{array}$

$\text{ Case 1: } x > 1$

$(gof)(x) = g(f(x)) = g(x+1) = x+ 1 - 1 = x$

$\text{ Case 2: } x = 1$

$(gof)(x) = g(f(x)) = g(x+1) = 1$

Therefore $(gof)(x) = x \ \forall \ x \in N$ which is an identity function and hence it is onto.

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Question 12: Give examples of two functions $f: N \rightarrow N \text{ and } g : Z \rightarrow Z$ such that $gof$ is injective but $g$ is not injective.