3 Marks

Question 13 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

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Question 23 Marks

Find fog and gof if:

f(x) = x², g(x) = cosx

Answer

f(x) = x², 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(x²) = cosx²
And
fog(x) = f(f(x)) = f(cosx) = cos²x

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Question 33 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.
$\Rightarrow \text{fog}: \text{R}\ \rightarrow{\ }\text{x}=\text{g}\ \text{f}\ \text{x}=\text{g}\ \text{c}=\sin \text{c}^2$

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Question 43 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) = x² and $\text{g(x)}=\sin\Big(\frac{\pi\text{x}}{2}\Big)$

Here, f : A → A is defined by

f(x) = x²

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.

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Question 53 Marks

Find fog and gof if:

f(x) = x² + 2, $\text{g(x)}=1-\frac{1}{1-\text{x}}$

Answer

f(x) = x² + 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))
$=\text{g}(\text{x}^2+2)=\frac{-(\text{x}^2+2)}{1-(\text{x}^2+2)}$
$\Rightarrow\ \text{gof(x)}=\frac{\text{x}^2+2}{\text{x}^2+1}$

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Question 63 Marks

Let f : R → R be defined as $\text{f(x)}=\frac{2\text{x}-3}{4}.$ Write fof^-1(1).

Answer

Let 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))
$=\text{f}\Big(\frac{4\text{x}+3}{2}\Big)$
$=\frac{2\big(\frac{4\text{x}+3}{2}\big)-3}{4}$
⇒ fof^-1(x) = x
$\therefore$ fof^-1(1) = 1

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Question 73 Marks

Find fog and gof if:

f(x) = |x|, g(x) = sinx

Answer

f(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|

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Question 83 Marks

Show that f : R → R, given by f(x) = x - [x], is neither one-one nor onto.

Answer

f : 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.

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Question 93 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)}

Answer

g : {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.

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Question 103 Marks

Let A = [-1, 1]. Then, discuss whether the following functions from A to itself are one-one, onto or bijective:
g(x) = |x|

Answer

g(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.

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Question 113 Marks

Classify the following functions as injection, surjection or bijection:
f : R → R, defined by f(x) = |x|

Answer

f : 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.

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Question 123 Marks

Classify the following functions as injection, surjection or bijection:
f : R → R, defined by f(x) = sinx

Answer

f : R → R, given by f(x) = sinx

Injective: Let $\text{x, y}\in\text{R}$ such that

f(x) = f(y)

⇒ sinx = siny

$\Rightarrow\ \text{x}=\text{n}\pi+(-1)^{\text{n}}\text{y}$

$\Rightarrow\ \text{x}\neq\text{y}$

$\therefore$ f is not one-one.

Surjective: Let $\text{y}\in\text{R}$ be arbitrary such that

f(x) = y

⇒ sinx = y

⇒ x = sin^-1y

Now, for $\text{y}>1\times\notin\text{R}$ (domain)

$\therefore$ f is not onto.

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Question 133 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)}

Answer

f : {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.

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Question 143 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.

Answer

Given, 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

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Question 153 Marks

Find f^-1 if it exists: f : A → B, where, A = {1, 3, 5, 7, 9}; B = {0, 1, 9, 25, 49, 81} and f(x) = x².

Answer

A = {1, 3, 5, 7, 9}; B = {0, 1, 9, 25, 49, 81}
f : A → B be a function defined by f(x) = x²
Since different elements of A have different images in B.
$\therefore$ f is one-one.
Again, $0\in\text{B}$ does not have a preim-age in A.
$\therefore$ f is not onto.
Hence, f^-1 does not exist.

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Question 163 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}

Answer

f = {(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.

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Question 173 Marks

Let A = {1, 2, 3}. Write all one-one from A to itself.

Answer

We 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

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Question 183 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).

Answer

f, 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)

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Question 193 Marks

If A = {1, 2, 3}, show that a one-one function f : A → A must be onto.

Answer

A = {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.

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Question 203 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 $\beta.$

Answer

We 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
$2+\beta=1$
$\beta=1$

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Question 213 Marks

Let f be a function from R to R, such that f(x) = cos(x + 2). Is f invertible? Justify your answer.

Answer

Given: 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.

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Question 223 Marks

Let f : R → R and g : R → R be defined by f(x) = x + 1 and g(x) = x - 1. Show that fog = gof = I_R.

Answer

Given, f : R → R and g : R → R

fog : R → R and gof : R → R (Also, we know that I_R : R → 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

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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) = x²

Answer

h(x) = x²

Injection test: Let x and y be any two elements in the domain (A), such that f(x) = f(y).

f(x) = f(y)

x² = y²

$\text{x}=\pm\text{y}$

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.

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Question 243 Marks

If A = {1, 2, 3}, show that a onto function f : A → A must be one-one.

Answer

A = {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.

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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.

Answer

Let 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.

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Question 263 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

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Question 273 Marks

Find f^-1 if it exists: f : A → 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 → 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.
⇒ f^-1 : B → A exists and is given by
$\text{f}^{-1}(\text{x})=\frac{\text{x}}{3}$

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Question 283 Marks

Let f be an invertible real function. Write (f^-1of)(1) + (f^-1of)(2) + ..... + (f^-1of)(100).

Answer

Given that f is an invertible real function.

f^-1of = I, where I is an identity function.

Therefore, (f^-1of)(1) + (f^-1of)(2) + ....... + (f^-1of)(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

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Question 293 Marks

Classify the following functions as injection, surjection or bijection:
f : R → R, defined by f(x) = x³ + 1

Answer

f : R → R, defined by f(x) = x³ + 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)
x³ + 1 = y³ + 1
x³ = y³
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
x³ + 1 = y
$\text{x}=\sqrt[3]{\text{y}-1}\in\text{R}$
So, f is a surjection.
So, f is a bijection.

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Question 303 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.

Answer

We 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)}

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Question 313 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|

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Question 323 Marks

If f : A → B and g : B → C are onto functions, show that gof is a onto function.

Answer

Given, 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.

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Question 333 Marks

Let f : R → R be the function defined by f(x) = 4x - 3 for all x ∈ R. Then write f^-1.

Answer

We have,
f : R → 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, $\text{f}^{-1}(\text{x})=\frac{\text{x}+3}{4}$

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Question 343 Marks

Classify the following functions as injection, surjection or bijection:
f : Z → Z, defined by f(x) = x - 5

Answer

f : Z → Z, defined by f(x) = x - 5

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 - 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.

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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 → R

So,

f : R → R, fog(x) = f(g(x)) = f(2x) = sin2x

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

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Question 363 Marks

Find gof and fog when f : R → R and g : R → R are defined by:
f(x) = x² + 2x - 3 and g(x) = 3x - 4

Answer

Given, f : R → R and g : R → R
Therefore, gof : R → R and fog : R → R
f(x) = x² + 2x - 3 and g(x) = 3x - 4
Now, gof(x) = g(f(x)) = g(x² + 2x - 3)
$\therefore$ gof(x) = 3(x² + 2x - 3) - 4
⇒ gof(x) = 3x² + 6x - 13
and, fog(x) = f(g(x)) = f(3x - 4)
$\therefore$ fog(x) = (3x - 4)² + 2(3x - 4) - 3
= 9x² + 16 - 24x + 6x - 8 - 3
$\therefore$ fog(x) = 9x² - 18x + 5

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Question 373 Marks

Classify the following functions as injection, surjection or bijection:
f : R → R, defined by f(x) = sin²x + cos²x

Answer

f : R → R, defined by f(x) = sin²x + cos²x
f(x) = sin²x + cos²x = 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.

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Question 383 Marks

If f(x) = |x|, prove that fof = f.

Answer

We 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

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Question 393 Marks

If f : A → B and g : B → C are one-one functions, show that gof is a one-one function.

Answer

Given, 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.

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Question 403 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)}

Answer

h : {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)}

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Question 413 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.

Answer

We 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)}

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Question 423 Marks

Find gof and fog when f : R → R and g : R → R are defined by:
f(x) = x² + 8 and g(x) = 3x³ + 1

Answer

Given, f : R → R and g : R → R
Therefore, gof : R → R and fog : R → R
f(x) = x² + 8 and g(x) = 3x³ + 1
(gof)(x) = g(f(x))
= g(x² + 8)
= 3(x² + 8)³ + 1
(fog)(x) = f(g(x))
= f(3x³ + 1)
= (3x³ + 1)² + 8
= 9x⁶ + 6x³ + 1 + 8
= 9x⁶ + 6x³ + 9

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Question 433 Marks

Classify the following functions as injection, surjection or bijection:
f : Z → Z given by f(x) = x²

Answer

f : Z → Z given by f(x) = x²

Injection test: Let x and y be any two elements in the domain (Z), such that f(x) = f(y).

f(x) = f(y)

x² = y²

$\text{x}=\pm\text{y}$

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

x² = y

$\text{x}=\pm\sqrt{\text{y}}$ 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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Question 443 Marks

Let A = [-1, 1]. Then, discuss whether the following functions from A to itself are one-one, onto or bijective:
$\text{f(x)}=\frac{\text{x}}{2}$

Answer

f : 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.

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Question 453 Marks

Let f(x) = x² + x + 1 and g(x) = sinx. Show that fog ≠ gof.

Answer

(fog)(x) = f(g(x))
f(sinx) = sin²x + sinx + 1
and, (gof)(x) = g(f(x))
= g(x² + x + 1)
= sinx² + x + 1
Therefore, fog ≠ gof.

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Question 463 Marks

Give examples of two one-one functions f₁ and f₂ from R to R, such that f₁ + f₂ : R → R. defined by (f₁ + f₂)(x) = f₁(x) + f₂(x) is not one-one.

Answer

We know that f₁: R → R, given by f₁(x) = x, and f₂(x) = -x are one-one.
Proving f₁is one-one: Let f₁(x) = f₁(y)
⇒ x = y
So, f₁ is one-one.
Proving f₂ is one-one: Let f₂(x) = f₂(y)
⇒ -x = -y
⇒ x = y
So, f₂ is one-one.
Proving (f₁ + f₂) is not one-one:
Given: (f₁ + f₂)(x) = f₁(x) + f₂(x)= x + (-x) =0
So, for every real number x, (f₁ + f₂)(x)=0
So, the image of ever number in the domain is same as 0.
Thus, (f₁ + f₂) is not one-one.

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Question 473 Marks

Let A = R - {3} and B = R - {1}. Consider the function f : A → 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.

Answer

We have,
A = R - {3} and B = R - {1}. Consider the function f : A → 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,
$\frac{\text{x}-2}{\text{x}-3}=\frac{\text{y}-2}{\text{y}-3}$
⇒ xy - 3x - 2y + 6 = xy - 2x - 3y + 6
⇒ -x = -y
⇒ x = y
$\therefore$ f is one-one.

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Question 483 Marks

Let R⁺ be the set of all non-negative real numbers. If f : R⁺ → R⁺ and g : R⁺ → R⁺ are defined as f(x) = x² and $\text{g(x)}=+\sqrt{\text{x}},$ find fog and gof. Are they equal functions?

Answer

We have, f : R⁺ → R⁺ given by
f(x) = x²
g : R⁺ → 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)

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Question 493 Marks

Find gof and fog when f : R → R and g : R → R are defined by:
f(x) = 8x³ and $\text{g(x)}=\text{x}^\frac{1}{3}$

Answer

Given, f : R → R and g : R → R
Therefore, gof : R → R and fog : R → R
f(x) = 8x³ and $\text{g(x)}=\text{x}^\frac{1}{3}$
(gof)(x) = g(f(x))
= g(8x³)
$=(8\text{x}^3)^\frac{1}{3}$
$=[(2\text{x})^3]^\frac{1}{3}$
$=2\text{x}$
(fog)(x) = f(g(x))
$=\text{f}\Big(\text{x}^\frac{1}{3}\Big)$
$=8\Big(\text{x}^\frac{1}{3}\Big)^3$
$=8\text{x}$

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Question 503 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.

Answer

We have, f : A → B and g : B → C are one-one functions.
Now we have to prove: gof : A → C in one-one.
Let $\text{x, y}\in\text{A}$ such that
gof(x) = gof(y)
⇒ g(f(x)) = g(f(y))
⇒ f(x) = f(y) [$\because$ g in one-one]
⇒ x = y [$\because$ f in one-one]
$\therefore$ gof is one-one function.

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