Question 13 Marks
Show that f : R → R, given by f(x) = x - [x], is neither one-one nor onto.
Answerf : R → R, given by f(x) = x - [x]Injectivity: f(x) = 0 for all $\text{x}\in\text{Z}$
Therefore, f is not one-one.Surjectivity: Range of f = (0, 1) ≠ R.
Co-domain of f = R Both are not same. Therefore, f is not onto.
View full question & answer→Question 23 Marks
State with reasons whether the following functions have inverse:
g : {5, 6, 7, 8} → {1, 2, 3, 4} with g = {(5, 4), (6, 3), (7, 4), (8, 2)}
Answerg : {5, 6, 7, 8} → {1, 2, 3, 4} with g = {(5, 4), (6, 3), (7, 4), (8, 2)}
g(5) = g(7) = 4
⇒ f is not one-one.
⇒ f is not a bijection.
So, f does not have an inverse.
View full question & answer→Question 33 Marks
Let A = [-1, 1]. Then, discuss whether the following functions from A to itself are one-one, onto or bijective:
g(x) = |x|
Answerg(x) = |x|Injection test: Let x and y be any two elements in the domain (A), such that f(x) = f(y).
f(x) = f(y) |x| = |y| $\text{x}=\pm\text{y}$ So, f is not one-one. Surjection test: For y = -1, there is no value of x in A. So, f is not onto. So, f is not bijective.
View full question & answer→Question 43 Marks
Classify the following functions as injection, surjection or bijection:
$f : R \rightarrow R$, defined by $f(x) = x^3 + 1$
Answer$f : R \rightarrow R,$ defined by $f(x) = x^3 + 1$
Injection test: Let x and y be any two elements in the domain (R), such that f(x) = f(y).
$f(x) = f(y)$
$x^3 + 1 = y^3 + 1$
$x^3 = y^3$
$x = y$
So, f is an injection.
Surjection test: Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).
$f(x) = y$
$x^3 + 1 = y$
$\text{x}=\sqrt[3]{\text{y}-1}\in\text{R}$
So, f is a surjection.
So, f is a bijection.
View full question & answer→Question 53 Marks
Classify the following functions as injection, surjection or bijection:
f : R → R, defined by f(x) = |x|
Answerf : R → R, given by f(x) = |x|
Injectivity: Let $\text{x, y}\in\text{R}$ such that
x = y but if y = -x
⇒ |x| = |y| ⇒ |y| = |-x| = x
$\therefore$ f is not one-one.
Surjective: Since f attains only positive values, for negative real numbers in R, there is no pre-image in domain R.
$\therefore$ f is not onto.
View full question & answer→Question 63 Marks
Find fog and gof if:$\text{f}(\text{x})=\text{c},\text{c}\in \text{R},\text{g(x)}=\sin \text{x}^2$
Answer$\text{f} \ \text{x}=\text{c} = \sin \text{x} \ 2\ \text{f}:\text{R}\ \rightarrow{\ } \ \text{c};\text{g}:\text{R}\ \rightarrow{\ } \ 0,1$
Computing fog: Clearly, the range of g is a subset of the domain of f.
$.\text{fog}:\text{R}\ \rightarrow{\ }\ \text{x}=\text{f}\ \text{g}\text{ x }=\text{f} \ \sin \text{x}^2=\text{c}$
Computing gof: Clearly, the range of f is a subset of the domain of g.
$\Rightarrow \text{fog}: \text{R}\ \rightarrow{\ }\text{x}=\text{g}\ \text{f}\ \text{x}=\text{g}\ \text{c}=\sin \text{c}^2$
View full question & answer→Question 73 Marks
Find fog and gof if:f(x) = x + 1, g(x) = 2x + 3
Answerf(x) = x + 1, g(x) = 2x + 3f : R → R; g : R → R
Computing fog: Clearly, the range of g is a subset of the domain of f.
⇒ fog : R → R
(fog)(x) = f(g(x))
= f(2x + 3)
= 2x + 3 + 1
= 2x + 4
Computing gof: Clearly, the range of f is a subset of the domain of g.
⇒ fog : R → R
(gof)(x) = g(f(x))
= g(x + 1)
= 2(x + 1) + 3
= 2x + 5
View full question & answer→Question 83 Marks
If $f : R → (0, 2)$ defined by $\text{f(x)}=\frac{\text{e}^{\text{x}}-\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}+1$ is invertible, find $f^{-1}$.
Answer$\text{A}=\{\text{x}\in\text{R}:-1\leq\text{x}\leq1\}$ and f : A → A, g : A → A are two functions defined by $f(x) = x^2$ and $\text{g(x)}=\sin\Big(\frac{\pi\text{x}}{2}\Big)$Here, f : A → A is defined by
$f(x) = x^2$
Clearly f in not injective,
$\because\ \text{f}(1)=\text{f}(-1)=1$
So, f is not bijective and hence not invertible.
Hence, $f^{-1}$ does not exist.
View full question & answer→Question 93 Marks
Classify the following functions as injection, surjection or bijection:
$f : R \rightarrow R$, defined by $f(x) = sinx$
Answer$f: R \rightarrow R$, given by $f(x)=\sin x$ Injective: Let $x, y \in R$ such that $f(x)=f(y) \Rightarrow \sin x=\sin y \Rightarrow x=n \pi+(-1)^n y$ $\Rightarrow x \neq y$
$\therefore f$ is not one-one.
Surjective: Let $y \in R$ be arbitrary such that
$f(x)=y$
$\Rightarrow \sin x=y \Rightarrow x=\sin ^{-1} y$ Now, for $y>1 \times \notin R$ (domain). $\therefore$ is not onto.
View full question & answer→Question 103 Marks
State with reasons whether the following functions have inverse:
f : {1, 2, 3, 4} → {10} with f = {(1, 10), (2, 10), (3, 10), (4, 10)}
Answerf : {1, 2, 3, 4} → {10} with f = {(1, 10), (2, 10), (3, 10), (4, 10)}
We have,
f(1) = f(2) = f(3) = f(4) = 10
⇒ f is not one-one.
⇒ f is not a bijection.
So, f does not have an inverse.
View full question & answer→Question 113 Marks
Consider f : N → N, g : N → N and h : N → R defined as f(x) = 2x, g(y) = 3y + 4 and $\text{h(z)}=\sin\text{z}$ for all $\text{x, y, z}\in\text{N.}$ Show that ho(gof) = (hog)of.
AnswerGiven, f : N → N, g : N → N and h : N → R
⇒ gof : N → N and hog : N → R
⇒ ho(gof) : N → R and (hog) of : N → R
So, both have the same domains.
(gof)(x) = g(f(x)) = g(2x) = 3(2x) + 4 = 6x + 4 ....(1)
(hog)(x) = h(g(x)) = h(3x + 4) = sin(3x + 4) ......(2)
Now,
(ho(gof))(x) = h((gof)(x)) = h(6x + 4) = sin(6x + 4) [from (1)]
((hog)of)(x) = (hog)(f(x)) = (hog)(2x) = sin(6x + 4) [from (2)]
So, (ho(gof))(x) = ((hog)of)(x), $\forall\text{ x}\in\text{N}$
Hence, ho(gof) = (hog)of
View full question & answer→Question 123 Marks
Find $f^{-1}$ if it exists: $f: A \rightarrow B$, where, $A=\{1,3,5,7,9\} ; B=\{0,1,9,25,49,81\}$ and $f(x)=x^2$.
Answer$A=\{1,3,5,7,9\} ; B=\{0,1,9,25,49,81\}$
$f: A \rightarrow B$ be a function defined by $f(x)=x^2$
Since different elements of A have different images in B .
$\therefore f$ is one-one.
Again, $0 \in B$ does not have a preim-age in A .
$\therefore f$ is not onto.
Hence, $f ^{-1}$ does not exist.
View full question & answer→Question 133 Marks
Are the following set of ordered pairs functions? If so, examine whether the mapping is injective or surjective:
{(x, y): x is a person, y is the mother of x}
Answerf = {(x, y): x is a person, y is the mother of x}
As, for each element x in domain set, there is a unique related element y in co-domain set.
Therefore, f is the function.
Injection test: As, y can be mother of two or more persons.
Therefore,
f is not injective.
Surjection test: For every mother y defined by (x, y), there exists a person x for whom y is mother. Therefore, f is surjective.
View full question & answer→Question 143 Marks
If a function g = {(1, 1), (2, 3), (3, 5), (4, 7)} is described by $\text{g(x)}=\alpha\text{x}+\beta\alpha\text{x}+\beta,$ then find the values of $\alpha$ and $\beta.$
AnswerWe have,
A function g = {(1, 1), (2, 3), (3, 5), (4, 7)} is described by $\text{g(x)}=\alpha\text{x}+\beta$
As, g(1) = 1 and g(2) = 3
Therefore, $\alpha(1)+\beta=1$
$=\alpha+\beta=1\ ....(\text{i})$
and $\alpha(2)-\beta=3$
$2\alpha-\beta=3\ ....(\text{ii})$
(ii) - (i) we get
$2\alpha-\alpha=2$
$\alpha=2$
Substituting $\alpha=2$ in (i), we get
$2+\beta=1$
$\beta=1$
View full question & answer→Question 153 Marks
Let A = {1, 2, 3}. Write all one-one from A to itself.
AnswerWe have,
ho(gof)(x) = h(gof(x)) = h(g(f(x)))
= h(g(2x)) = h(3(2x) + 4)
= h(6x + 4) = sin(6x + 4) $\forall\ \text{x}\in\text{N}$
((hog)of)(x) = (hog)(f(x)) = (hog)(2x)
= h(g(2x)) = h(3(2x) + 4)
= h(6x + 4) = sin(6x + 4) $\forall\ \text{x}\in\text{N}$
This shows, ho(gof) = (hog)of
View full question & answer→Question 163 Marks
Let f, g, h be real functions given by f(x) = sinx, g(x) = 2x and h(x) = cosx. Prove that fog = go(fh).
Answerf, g and h are real fuctions given by f(x) = sinx, g(x) = 2x and h(x) = cosx
To prove: fog = go(fh)
L.H.S. fog(x) = f(g(x))
= f(2x) = sin2x
⇒ fog(x) = 2sinx.cosx .....(A)
R.H.S. go(fh)(x) = go(f(x).h(x))
= g(sinx.cosx)
⇒ go(fh)(x) = 2sinx.cosx ......(B)
from (A) & (B)
fog(x) = go(fh)(x)
View full question & answer→Question 173 Marks
Find fog and gof if:$f(x) = x^2 + 2,$ $\text{g(x)}=1-\frac{1}{1-\text{x}}$
Answer$f(x) = x^2 + 2$ and $\text{g(x)}=1-\frac{1}{1-\text{x}}$
Range of $\text{f}=(2,\infty)\ \subset$ Domain of g = R ⇒ gof exist
Range of g = R - [1] $\subset$ Domain of f = R
⇒ fog exist
Now,
fog(x) = f(g(x))
$=\text{f}\Big(\frac{-\text{x}}{1-\text{x}}\Big)=\frac{\text{x}^2}{(1-\text{x})^2}+2$
And,
gof(x) = g(f(x))
$=\text{g}(\text{x}^2+2)=\frac{-(\text{x}^2+2)}{1-(\text{x}^2+2)}$
$\Rightarrow\ \text{gof(x)}=\frac{\text{x}^2+2}{\text{x}^2+1}$
View full question & answer→Question 183 Marks
If A = {1, 2, 3}, show that a one-one function f : A → A must be onto.
AnswerA = {1, 2, 3}
Number of elements in A = 3
Number of one-one functions = number of ways of arranging 3 elements = 3! = 6
So, the possible one-one functions can be the following:
- {(1, 1), (2, 2), (3, 3)}
- {(1, 1), (2, 3), (3, 2)}
- {(1, 2 ), (2, 2), (3, 3)}
- {(1, 2), (2, 1), (3, 3)}
- {(1, 3), (2, 2), (3, 1)}
- {(1, 3), (2, 1), (3, 2)}
Here, in each function, range = {1, 2, 3}, which is same as the co-domain.
So, all the functions are onto.
View full question & answer→Question 193 Marks
If A = {1, 2, 3}, show that a onto function f : A → A must be one-one.
AnswerA = {1, 2, 3}
Possible onto function from A to A can be the following:
- {(1, 1), (2, 2), (3, 3)}
- {(1, 1), (2, 3), (3, 2)}
- {(1, 2), (2, 2), (3, 3)}
- {(1, 2), (2, 1), (3, 3)}
- {(1, 3), (2, 2), (3, 1)}
- {(1, 3), (2, 1), (3, 2)}
Here, in each function, different elements of the domain have different images.
Therefore, all the function are one-one.
View full question & answer→Question 203 Marks
Let f be a function from R to R, such that f(x) = cos(x + 2). Is f invertible? Justify your answer.
AnswerGiven: A and B are two sets with finite elements.
f : A → B and g : B → A are injective map.
To prove: f is bijective.
Proof: Since, f : A → B is injective we need to show f in surjective only.
Now,
g : B → A is injective.
⇒ Each element of B has image in A.
View full question & answer→Question 213 Marks
Let $f : R \rightarrow R$ and $g : R \rightarrow R$ be defined by $f(x) = x + 1$ and $g(x) = x - 1.$ Show that fog $= gof = I_R.$
AnswerGiven, $f: R \rightarrow R$ and $g: R \rightarrow R$ fog $: R \rightarrow R$ and $g \circ f: R \rightarrow R$
(Also, we know that $I_R: R \rightarrow R$ )
Therefore, the domains of all fog, and $I _{ R }$ are the same.
$(fog)(x) = f(g(x)) = f(x - 1) = x - 1 + 1 = x = I_R(x) ..... (1)$
$(gof)(x) = g(f(x)) = g(x + 1) = x + 1 -1 = x = I_R(x) ..... (2)$
From $(1)$ and $(2), (fog)(x) = (gof)(x) = I_R(x), $$\forall\text{ x}\in\text{R}$
Hence, $fog = gof = I_R$
View full question & answer→Question 223 Marks
Find fog and gof if:f(x) = |x|, g(x) = sinx
Answerf(x) = |x|, g(x) = sinx$\text{f}:\text{R}\rightarrow(0,\infty);\ \text{g}:\text{R}\rightarrow[-1,1]$
Computing fog: Clearly, the range of g is a subset of the domain of f.
⇒ fog : R → R
(fog)(x) = f(g(x))
= f(sinx)
= |sinx|
Computing gof: Clearly, the range of f is a subset of the domain of g.
⇒ fog : R → R
(gof)(x) = g(f(x))
= g(|x|)
= sin|x|
View full question & answer→Question 233 Marks
Let $A=[-1,1]$. Then, discuss whether the following functions from $A$ to itself are one-one, onto or bijective: $h(x)=x^2$
Answer$h(x)=x^2$
Injection test: Let x and y be any two elements in the domain (A),
such that $f(x)=f(y) . f(x)=f(y) x^2$
$=y^2 x= \pm y$
So, f is not one-one.
Surjection test: For $y =-1$, there is no value of x in A .
So, $f$ is not onto. So, $f$ is not bijective.
View full question & answer→Question 243 Marks
Let f be an invertible real function. Write $(f^{-1}of)(1) + (f^{-1}of)(2) + ..... + (f^{-1}of)(100).$
AnswerGiven that f is an invertible real function. $f^{-1}$ of $= I,$
where I is an identity function. Therefore, $(f^{-1}of)(1) + (f^{-1}of)(2) + ....... + (f^{-1}of)(100) = I(1) + I(2) + .... + I(100)$
$= 1 + 2 + .... + 100$ $(\text{As Ix}=\text{x},\ \forall\ \text{x}\in\text{R})$
$=\frac{1001(1001+1)}{2}$
$\Big[$ Sum of first n natural numbers $=\frac{\text{n}(\text{n}+1)}{2}\Big]$= 5050
View full question & answer→Question 253 Marks
Give examples of two functions f : N → N and g : N → N, such that gof is onto but f is not onto.
AnswerLet us consider a function f : N → N given by f(x) = x + 1, which is not onto.
[This not onto because if we take 0 in N (co-domain), then, 0 = x + 1, x = -1 $\notin\text{N}$]
Let us consider,
$\text{g(x)}:\begin{Bmatrix}\text{x}=\text{x}-1&\text{if x}>1\\1,&\text{if x}=1\end{Bmatrix}$
Now, let us find (gof)(x)
Case 1: x > 1
(gof)(x) = g(f(x)) = g(x + 1) = x + 1 - 1 = x
Case 2: x = 1
(gof)(x) = g(f(x)) = g(x + 1) = 1
From case 1 and case 2, x = x, $\forall\ \text{x}\in\text{N,}$
which is an identity function and, hence, it is onto.
View full question & answer→Question 263 Marks
If f : A → B and g : B → C are one-one functions, show that gof is a one-one function.
AnswerGiven, f : A → B and g : B → C are one - one.
Then, gof : A → B
Let us take two elements x and y from A, such that
(gof)(x) = (gof)(y)
⇒ g(f(x)) = g(f(y))
⇒ f(x) = f(y) (As, g is one-one)
⇒ x = y (As, f is one-one)
Hence, gof is one-one.
View full question & answer→Question 273 Marks
Let f be any real function and let g be a function given by g(x) = 2x. Prove that gof = f + f.
Answer Given, f : R → R
Since g(x) = 2x is a polynomial, g : R → R
Clearly, gof : R → R and f + f : R → R
So, domain of gof and f + f are the same.
(gof)(x) = g(f(x)) = 2f(x)
(f + f)(x) = f(x) + f(x) = 2f(x)
⇒ (gof)(x) = (f + f)(x), $\forall\ \text{x}\in\text{R}$
Hence, gof = f + f
View full question & answer→Question 283 Marks
Find$ f^{-1} $if it exists: $f : A \rightarrow B,$ where,$ A = {0, -1, -3, 2}; B = {-9, -3, 0, 6}$ and$ f(x) = 3x.$
Answer$A = {0, -1, -3, 2}; B = {-9, -3, 0, 6}$
$f : A \rightarrow B$ is defined by $f(x) = 3x$
Since different elements of $A$ have different images in $B$.
$\therefore f$ is one-one.
Again, each element in B has a preim-age in $A$.
$\because f$ in one-one bijective.
$\Rightarrow f^{-1}: B \rightarrow A$ exists and is given by
$f^{-1}(x)=\frac{x}{3}$
View full question & answer→Question 293 Marks
Let $f : R → R$ be defined as $\text{f(x)}=\frac{2\text{x}-3}{4}.$ Write $fof^{-1}(1)$.
AnswerLet f : R → R, defined by $\text{f(x)}=\frac{2\text{x}-3}{4}$
$\Rightarrow\ \text{f}^{-1}\frac{(2\text{x}-3)}{4}=\text{x}$
$\Rightarrow\ \text{f}^{-1}(2\text{x})=4\text{x}+3$
$\Rightarrow\ \text{f}^{-1}(\text{x})=\frac{4\text{x}+3}{2}$
Now, $fof^{-1}(x) = f(f^{-1}(x))$
$=\text{f}\Big(\frac{4\text{x}+3}{2}\Big)$
$=\frac{2\big(\frac{4\text{x}+3}{2}\big)-3}{4}$
$\Rightarrow fof^{-1}(x) = x$
$\therefore fof^{-1}(1) = 1$
View full question & answer→Question 303 Marks
Find fog and gof if:$f(x) = e^x, g(x) = log_ex$
Answer$f(x) = e^x, g(x) = log_ex$
$\text{f}:\text{R}\rightarrow0,\infty;\ \text{g}:0,\infty\rightarrow\text{R}$
Computing fog: Clearly, the range of g is a subset of the domain of f.
$\text{fog}(0,\infty)\rightarrow\text{R}$
$fog(x) = f(g(x))$
$= f(log_ex) = log_ee^x$
$= x$
Computing gof: Clearly, the range of f is a subset of the domain of g.
$\Rightarrow fog : R \rightarrow R$
$(gof)(x) = g(f(x))$
$= ge^x$
$= log_ee^x$
$= x$
View full question & answer→Question 313 Marks
If f : {5, 6} → {2, 3} and g : {2, 3} → {5, 6} are given by f = {(5, 2), (6, 3)} and g = {(2, 5), (3, 6)}, then find fog.
AnswerWe have, f : {5, 6} → {2, 3} and g : {2, 3} → {5, 6} are given by f = {(5, 2), (6, 3)} and g = {(2, 5), (3, 6)}
As,
fog(2) = f(g(2)) = f(5) = 2,
fog(3) = f(g(3)) = f(6) = 3,
Therefore, fog : {2, 3} → {2, 3} is defined as fog = {(2, 2), (3, 3)}
View full question & answer→Question 323 Marks
Give examples of two one-one functions $f_1$ and $f_2$ from $R$ to $R$, such that $f_1 + f_2 : R \rightarrow R.$ defined by $(f_1 + f_2)(x) = f_1(x) + f_2(x)$ is not one-one.
AnswerWe know that $f_1: R \rightarrow R$, given by $f_1(x)=x$, and $f_2(x)=-x$ are one-one.
Proving $f _1$ is one-one: Let $f_1(x)=f_1(y)$
$\Rightarrow x=y$
So, $f_1$ is one-one.
Proving $f _2$ is one-one: Let $f _2(x)= f _2(y)$
$\Rightarrow -x = -y$
$\Rightarrow x = y$
So, $f _2$ is one-one.
Proving $\left(f_1+f_2\right)$ is not one-one:
Given: $\left(f_1+f_2\right)(x)=f_1(x)+f_2(x)=x+(-x)=0$
So, for every real number $x,\left(f_1+f_2\right)(x)=0$
So, the image of ever number in the domain is same as 0 .
Thus, $\left(f_1+f_2\right)$ is not one-one.
View full question & answer→Question 333 Marks
Show that the logarithmic function $\text{f}:\text{R}0^+\rightarrow \text{R}$ given by $f(x) = log_a x, a > 0 $ is a bijection.
AnswerWe have, $f : A \rightarrow B$ and $g : B \rightarrow C $are one-one functions.
Now we have to prove: gof :$ A \rightarrow C$ in one-one.
Let $\text{x, y}\in\text{A}$ such that
$gof(x) = gof(y)$
$\Rightarrow g(f(x)) = g(f(y))$
$\Rightarrow f(x) = f(y)$ [$\because$ g in one-one]
$\Rightarrow x = y$ [$\because$ f in one-one]
$\therefore$ gof is one-one function.
View full question & answer→Question 343 Marks
Find fog and gof if:$f(x) = x + 1, g(x) = e^x$
Answer$f(x) = x + 1, g(x) = e^x$ Range of f = R $\subset$ Domain of g = R ⇒ gof exist
Range of $\text{g}=(0,\infty)\ \subset$ Domain of f = R ⇒ fog exist
Now,
$gof(x) = g(f(x)) = g(x + 1) = e^{x+1}$
And,
$fog(x) = f(g(x)) = f(e^x) = e^x + 1$
View full question & answer→Question 353 Marks
If f(x) = sin x and g(x) = 2x be two real functions, then describe gof and fog. Are these equal functions?
Answer We know that f : R → -1, 1 and g : R → R gof clearly, the range of f is a subset of the domain of g. gof : R → R, gof(x) = g(f(x)) = g(sinx) = 2sinx fog clearly, the range of g is a subset of the domain of f. fog : R → RSo,
f : R → R, fog(x) = f(g(x)) = f(2x) = sin2x Clearly, $\text{fog}\neq\text{gof}$
View full question & answer→Question 363 Marks
Classify the following functions as injection, surjection or bijection:f: $R \rightarrow R$, defined by $f(x)=1+x^2$
Answer$f : R \rightarrow R$, defined by $f(x) = 1 + x^2$
Injection test: Let $\text{x, y}\in\text{R,}$ such that,
$f(x) = f(y)$
$\Rightarrow 1 + x^2 = 1 + y^2$
$\Rightarrow x^2 - y^2 = 0$
$\Rightarrow (x - y)(x + y) = 0$
either x = y or x = -y or $\text{x}\neq\text{y}$
Therefore, f is not one-one.
Surjection: Let $\text{y}\in\text{R}$ be arbitrary, then
$f(x) = y$
$\Rightarrow 1 + x^2 = y$
$\Rightarrow x^2 + 1 - y = 0$
$\therefore\ \text{x}\pm\sqrt{\text{y}-1}\notin\text{R}$ or $y < 1$
$\therefore$ f is not onto.
View full question & answer→Question 373 Marks
Find gof and fog when f : R → R and g : R → R are defined by:
f(x) = x and g(x) = |x|
Answer Given, f : R → R and g : R → R Therefore, gof : R → R and fog : R → R f(x) = x and g(x) = |x| (gof)(x) = g(f(x))= g(x)
(fog)(x) = f(g(x)) = f|x| = |x|
View full question & answer→Question 383 Marks
If f : A → B and g : B → C are onto functions, show that gof is a onto function.
AnswerGiven, f : A → B and g : B → C are onto.
Then, gof : A → C
Let us take an element z in the co-domain (C).
Now, z is in C and g : B → C is onto.
So, there exists some element y in B, such that g(y) = z .... (1)
Now, y is in B and f : A → B is onto.
So, there exists some x in A, such that f(x) = y .... (2)
From (1) and (2),
z = g(y) = g(f(x)) = (gof)(x)
So, z = (gof)(x), where x is in A.
Hence, gof is onto.
View full question & answer→Question 393 Marks
Let $f: R \rightarrow R$ be the function defined by $f(x)=4 x-3$ for all $x \in R$. Then write $f^{-1}$.
AnswerWe have,
$f : R \rightarrow R$ is the function defined by f(x) = 4x - 3 for all $\text{x}\in\text{R}$
Let $f(x) = y$. Then,
$y = 4x - 3$
$4x = y + 3$
$\text{x}=\frac{\text{y}+3}{4}$
Therefore, $\text{f}^{-1}(\text{y})=\frac{\text{y}+3}{4}$
or, $\text{f}^{-1}(\text{x})=\frac{\text{x}+3}{4}$
View full question & answer→Question 403 Marks
Classify the following functions as injection, surjection or bijection:
f : Z → Z, defined by f(x) = x - 5
Answerf : Z → Z, defined by f(x) = x - 5Injection test: Let x and y be any two elements in the domain (Z), such that f(x) = f(y).
f(x) = f(y) x - 5 = y - 5 x = y Therefore, f is an injection. Surjection test: Let y be any element in the co domain (Z), such that f(x) = y for some element x in Z (domain). f(x) = y x - 5 = y x = y + 5, which is in Z. Therefore, f is a surjection and f is a bijection.
View full question & answer→Question 413 Marks
Find gof and fog when $f: R \rightarrow R$ and $g: R \rightarrow R$ are defined by: $f(x)=x^2+2 x-3$ and $g(x)=3 x-4$
AnswerGiven, $f: R \rightarrow R$ and $g: R \rightarrow R$
Therefore, gof: $R \rightarrow R$ and fog: $R \rightarrow R$
$f(x)=x^2+2 x-3$ and $g(x)=3 x-4$
Now, $g \circ f(x)=g(f(x))=g\left(x^2+2 x-3\right)$
$\therefore \operatorname{gof}(x)=3\left(x^2+2 x-3\right)-4$
$\Rightarrow g \circ f(x)=3 x^2+6 x-13$
and, $f \circ g(x)=f(g(x))=f(3 x-4)$
$\therefore$ $fog(x) = (3x - 4)^2 + 2(3x - 4) - 3$
$= 9x^2 + 16 - 24x + 6x - 8 - 3$
$\therefore$ $fog(x) = 9x^2 - 18x + 5$
View full question & answer→Question 423 Marks
Classify the following functions as injection, surjection or bijection:
$f : R \rightarrow R,$ defined by $f(x) = \sin^2x + \cos^2x$
Answer$f : R \rightarrow R$, defined by $f(x) = \sin^2x + \cos^2x$
$f(x) = \sin^2x + \cos^2x = 1$
So, f(x) = 1 for every x in $R.$
So, for all elements in the domain, the image is 1.
So, f is not an injection.
Range of $f = {1}$
Co-domain of $f = R$
Both are not same.
So, f is not a surjection and f is not a bijection.
View full question & answer→Question 433 Marks
If f(x) = |x|, prove that fof = f.
AnswerWe have, f(x) = |x|
We assume the domain of f = R
Range of $\text{f}=(0,\infty)$
$\therefore$ Range of f $\subset$ domain of f
$\therefore$ fof exists.
Now,
fof(x) = f(f(x)) = f(|x|) = ||x|| = f(x)
$\therefore$ fof = f
View full question & answer→Question 443 Marks
State with reasons whether the following functions have inverse:
h : {2, 3, 4, 5} → {7, 9, 11, 13} with h = {(2, 7), (3, 9), (4, 11), (5, 13)}
Answerh : {2, 3, 4, 5} → {7, 9, 11, 13} with h = {(2, 7), (3, 9), (4, 11), (5, 13)} Here, different elements of the domain have different images in the co-domain. ⇒ h is one-one. Also, each element in the co-domain has a pre-image in the domain.⇒ h is onto.
⇒ h is bijection. ⇒ h has an inverse and it is given by, h - 1 = {(7, 2), (9, 3), (11, 4), (13, 5)}
View full question & answer→Question 453 Marks
If the mapping f : {1, 3, 4} → {1, 2, 5} and g : {1, 2, 5} → {1, 3}, given by f = {(1, 2), (3, 5), (4, 1)} and g= {(2, 3), (5, 1), (1, 3)}, then write fog.
AnswerWe have,
f : {1, 3, 4} → {1, 2, 5} and g : {1, 2, 5} → {1, 3}, are given by f = {(1, 2), (3, 5), (4, 1)} and g= {(2, 3), (5, 1), (1, 3)} respectively.
As,
fog(2) = f(g(2)) = f(3) = 5,
fog(5) = f(g(5)) = f(1) = 2,
fog(1) = f(g(1)) = f(3) = 5,
Therefore, fog : {1, 2, 5} → {1, 2, 5} is given by fog = {(2, 5), (5, 2), (1, 5)}
View full question & answer→Question 463 Marks
Find gof and fog when $f : R \rightarrow R$ and $g : R \rightarrow R$ are defined by:
$f(x) = x^2 + 8$ and $g(x) = 3x^3 + 1$
AnswerGiven,$ f : R \rightarrow R$ and $g : R \rightarrow R$
Therefore, gof $: R \rightarrow R$ and fog : $R \rightarrow R$
$f(x) = x^2 + 8$ and $g(x) = 3x^3 + 1$
$(gof)(x) = g(f(x))$
$= g(x^2 + 8)$
$= 3(x^2 + 8)^3 + 1$
$(fog)(x) = f(g(x))$
$= f(3x^3 + 1)$
$= (3x^3 + 1)^2 + 8$
$= 9x^6 + 6x^3 + 1 + 8$
$= 9x^6 + 6x^3 + 9$
View full question & answer→Question 473 Marks
If $\text{f(x)}=\sqrt{\text{x}+3}$ and $g(x) = x^2 + 1$ be two real functions, then find fog and gof.
Answer$\text{f(x)}=\sqrt{\text{x}+3}$For domain,
$\text{x}+3\geq0$
$\Rightarrow\ \text{x}\geq-3$
Domain of $\text{f}=[-3,\infty)$
Since f is a square root fuction, range of $\text{f}=[0,\infty)$
$\text{f}:[-3,\infty)\rightarrow[0,\infty)$
$g(x) = x^2 + 1$ is a polynomial.
$\Rightarrow g : R \rightarrow R$
Computation of fog: Range of g is not a subset of the doamin of f .
and domain (fog) $=\{x: x \in$ domain of $g$ and $g(x) \in$ domain of $f(x)\}$
$\Rightarrow$ Domain (fog) $=\left\{x: x \in R\right.$ and $\left.x^2+1 \in[-3, \infty)\right\}$
$\Rightarrow$ Domain (fog) $=\left\{x: x \in R\right.$ and $\left.x^2+1 \geq-3\right\}$
$\Rightarrow$ Domain (fog) $=\left\{x: x \in R\right.$ and $\left.x^2+4 \geq 0\right\}$
$\Rightarrow$ Domain (fog) $=\{x: x \in R$ and $x \in R\}$
$\Rightarrow$ Domain (fog) $=R$
fog :$ R \rightarrow R$
$(fog)(x) = f(g(x))$
$= f(x^2 + 1)$
$=\sqrt{\text{x}^2+1+3}$
$=\sqrt{\text{x}^2+4}$
Computation of gof: Range of f is a subset of the doamin of g.
$\text{gof}:[-3,\infty)\rightarrow\text{R}$
$\Rightarrow\ \text{(gof)(x)}=\text{g(f(x)})$
$=\text{g}(\sqrt{\text{x}+3})$
$=(\sqrt{\text{x}+3})^2+1$
$=\text{x}+3+1$
$=\text{x}+4$
View full question & answer→Question 483 Marks
Let A = [-1, 1]. Then, discuss whether the following functions from A to itself are one-one, onto or bijective:
$\text{f(x)}=\frac{\text{x}}{2}$
Answerf : A → A, given by $\text{f(x)}=\frac{\text{x}}{2}$
Injection test: Let x and y be any two elements in the domain (A), such that f(x) = f(y).
f(x) = f(y)
$\frac{\text{x}}{2}=\frac{\text{y}}{2}$
x = y
So, f is one-one.
Surjection test: Let y be any element in the co-domain (A), such that f(x) = y for some element x in A (domain).
f(x) = y
$\frac{\text{x}}{2}=\text{y}$
x = 2y, which may not be in A.
For example, if y = 1, then
x = 2, which is not in A.
So, f is not onto.
So, f is not bijective.
View full question & answer→Question 493 Marks
Find fog and gof if:f(x) = x + 1, g(x) = sinx
Answerf(x) = x + 1, g(x) = sinxRange of f = R $\subset$ Domain of g = R ⇒ gof exists
Range of g = [-1, 1] $\subset$ Domain of f = R ⇒ fog exists
Now,
fog(x) = f(g(x)) = f(sinx) = sinx + 1
And
gof(x) = g(f(x)) = g(x + 1) = sin(x + 1)
View full question & answer→Question 503 Marks
Let $R^{+}$be the set of all non-negative real numbers. If $f: R^{+} \rightarrow R^{+}$and $g: R^{+} \rightarrow R^{+}$are defined as $f(x)=x^2$ and $\mathrm{g}(\mathrm{x})=+\sqrt{\mathrm{x}}$, find fog and gof. Are they equal functions?
AnswerWe have, $f : R^+ \rightarrow R^+$ given by
$f(x) = x^2$
$g : R^+ \rightarrow R^+$ given by
$\text{g}(\text{x})=\sqrt{\text{x}}$
$\therefore$ fog(x) = f(g(x))
$=\text{f}(\sqrt{\text{x}})=(\sqrt{\text{x}})^2=\text{x}$
Also, gof(x) = g(f(x))
$=\text{g}(\text{x}^2)\sqrt{\text{x}^2}=\text{x}$
Thus, fog(x) = gof(x)
View full question & answer→Question 513 Marks
Find gof and fog when $f : R \rightarrow R$ and $g : R \rightarrow R$ are defined by:
$f(x) = 8x^3$ and $\text{g(x)}=\text{x}^\frac{1}{3}$
AnswerGiven, $f : R \rightarrow R$ and $g : R \rightarrow R$
Therefore,$ gof : R \rightarrow R$ and fog : $R \rightarrow R$
$f(x) = 8x^3$ and $\text{g(x)}=\text{x}^\frac{1}{3}$
$(gof)(x) = g(f(x))$
$= g(8x^3)$
$=(8\text{x}^3)^\frac{1}{3}$
$=[(2\text{x})^3]^\frac{1}{3}$
$=2\text{x}$
(fog)(x) = f(g(x))
$=\text{f}\Big(\text{x}^\frac{1}{3}\Big)$
$=8\Big(\text{x}^\frac{1}{3}\Big)^3$
$=8\text{x}$
View full question & answer→Question 523 Marks
Let $f(x) = x^2 + x + 1$ and $g(x) = sinx$. Show that fog ≠ gof.
Answer$(fog)(x) = f(g(x))$
$f(sinx) = \sin^2x + sinx + 1$
and, (gof)(x) = g(f(x))
$= g(x^2 + x + 1)$
$= sinx^2 + x + 1$
Therefore, fog ≠ gof.
View full question & answer→Question 533 Marks
Find fog and gof if$:f(x) = x^2, g(x) = cosx$
Answer$f(x) = x^2, g(x) = cosx$
Domain of f and Domain of g = R
Range of $\text{f}=(0,\infty)$
Range of g = (-1, 1)
$\therefore$ Range of f $\subset$ domain of g ⇒ gof exist
Range of g $\subset$ domain of f ⇒ fog exist
Now,
$gof(x) = g(f(x)) = g(x^2) = cosx^2$
And
$fog(x) = f(f(x)) = f(cosx) = \cos^2x$
View full question & answer→Question 543 Marks
Let $A = R - {3}$ and $B = R - {1}$. Consider the function $f : A \rightarrow B$ defined by $\text{f(x)}=\frac{\text{x}-2}{\text{x}-3}.$ Show that f is one-one and onto and hence find $f^{-1}.$
AnswerWe have,
$A = R - {3}$ and $B = R - {1}$. Consider the function $f : A \rightarrow B$ defined by
$\text{f}(\text{x})=\frac{\text{x}-2}{\text{x}-3},$ Show that f is one-one and onto and hence find $f^{-1}.$
Let $\text{x, y}\in\text{A}$ such that f(x) = f(y). Then,
$\frac{\text{x}-2}{\text{x}-3}=\frac{\text{y}-2}{\text{y}-3}$
$\Rightarrow xy - 3x - 2y + 6 = xy - 2x - 3y + 6$
$\Rightarrow -x = -y$
$\Rightarrow x = y$
$\therefore$ f is one-one.
View full question & answer→Question 553 Marks
Classify the following functions as injection, surjection or bijection:
$f : Z \rightarrow Z$ given by $f(x) = x^2$
Answer$f: Z \rightarrow Z$ given by $f(x)=x^2$
Injection test: Let $x$ and $y$ be any two elements in the domain $(Z)$, such that $f(x)=f(y) . f(x)=f(y) x^2=y^2 x= \pm y$ So, f is not an injection.
Surjection test: Let $y$ be any element in the co-domain $(Z)$, such that $f(x)=y$ for some element $x$ in $Z$ (domain). $f(x)=$ $y x^2=y x= \pm \sqrt{y}$ which may not be in $Z$.
For example, if $y=3$,
$\text{x}=\pm\sqrt{3}$ is not in Z.
So, f is not a bijection.
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