Class 12 ncert solution math exercise 1.2

              Exercise 1.2

Question 1: Show that the function f: R_{*} \rightarrow R_{*} defined by (x)=\frac{1}{x} is one-one and onto, where R_{\text {o is }} the set of all non-zero real numbers. Is the result true, if the domain R_{\cdot} is replaced by \mathrm{N} with codomain being same as R \cdot ?(Class 12 ncert solution math exercise 1.2 )

Solution: f: R_{\bullet} \rightarrow R_{\bullet} \text { is by } f(x)=\frac{1}{x}

For one-one:

x, y \in R _* \text { such that } f(x)=f(y)
\Rightarrow \frac{1}{x}=\frac{1}{y}
\Rightarrow x=y

\therefore f is one-one.

For onto:

For y \in R, there exists x=\frac{1}{y} \in R \cdot[ as y \neq 0] such that

f(x)=\frac{1}{\left(\frac{1}{y}\right)}=y

\therefore f is onto.

Given function f is one-one and onto. Consider function g: N \rightarrow R • defined by g(x)=\frac{1}{x}

We have, g\left(x_{1}\right)=g\left(x_{2}\right) \Rightarrow \frac{1}{x_{1}}=\frac{1}{x_{2}} \Rightarrow x_{1}=x_{2}

\therefore g is one-one.

g is not onto as for 1.2 \in R. there exist any x in N such that g(x)=\frac{1}{1.2}

Function g is one-one but not onto.

Question 2:Check the injectivity and surjectivity of the following functions:

i. f: N \rightarrow N given by f(x)=x^{2}

ii. f: Z \rightarrow Z given by f(x)=x^{2}

iii.  f: R \rightarrow R given by f(x)=x^{2}

 iv  f: N \rightarrow N given by f(x)=x^{3}

v. f: Z \rightarrow Z given by f(x)=x^{3}

Solution:i. For f: N \rightarrow N given by f(x)=x^{2}

x, y \in N

f(x)=f(y) \Rightarrow x^{2}=y^{2} \Rightarrow x=y

\therefore f is injective. 2 \in N. But, there does not exist any x in N such that f(x)=x^{2}=2\Rightarrow x=\sqrt2\notin N

\therefore f is not surjective

Function f is injective but not surjective.

ii. f: Z \rightarrow Z given by f(x)=x^{2}

f(-1)=f(1)=1 but -1 \neq 1

\therefore f is not injective.

-2 \in Z But, there does not exist any x \in Z such that f(x)=-2 \Rightarrow x^{2}=-2

\therefore f is not surjective.

Function f is neither injective nor surjective.

iii. f: R \rightarrow R given by f(x)=x^{2}

f(-1)=f(1)=1 but -1 \neq 1

\therefore f is not injective.

-2 \in Z But, there does not exist any x \in Z such that f(x)=-2 \Rightarrow x^{2}=-2

\therefore f is not surjective.

Function f is neither injective nor surjective.

iv. f: N \rightarrow N given by f(x)=x^{3}

x, y \in N

f(x)=f(y) \Rightarrow x^{3}=y^{3} \Rightarrow x=y

\therefore f is injective.

2 \in N. But, there does not exist any x in N such that f(x)=x^{3}=2

\therefore f is not surjective

Function f is injective but not surjective.

v. f: Z \rightarrow Z given by f(x)=x^{3}

x, y \in Z

f(x)=f(y) \Rightarrow x^{3}=y^{3} \Rightarrow x=y

\therefore f is injective.

2 \in Z. But, there does not exist any x in Z such that f(x)=x^{3}=2

\therefore f is not surjective.

Function f is injective but not surjective.

Question 3:Prove that the greatest integer function f: R \rightarrow R given by f(x)=[x]_{\text {is neither one-one nor }} onto, where [x] denotes the greatest integer less than or equal to x.

Solution:f: R \rightarrow R given by f(x)=[x]

f(1.2)=[1.2]=1, f(1.9)=[1.9]=1

\therefore f(1.2)=f(1.9), but 1.2 \neq 1.9

\therefore f is not one-one.

Consider 0.7 \in R

f(x)=[x]_{\text {is an integer. There does not exist any element }} x \in R such that f(x)=0.7

\therefore f is not onto.

The greatest integer function is neither one-one nor onto.

Question 4:Show that the modulus function f: R \rightarrow R given by f(x)=|x| is neither one-one nor onto, where |x| is x, if x is positive or 0 and |x| is -x, if x is negative.

Solution:f: R \rightarrow R is f(x)=|x|=\left\{\begin{array}{l}\mathrm{x}, \text { if } \mathrm{x} \geq 0 \\ -x, \text { if } \mathrm{x}<0\end{array}\right\}

f(-1)=|-1|=1 and f(1)=|1|=1

\therefore f(-1)=f(1) but -1 \neq 1

\therefore f is not one-one.

Consider -1 \in R

f(x)=|x| is non-negative. There exist any element x in domain R such that f(x)=|x|=-1

\therefore f is not onto.

The modulus function is neither one-one nor onto.

Question 5:Show that the signum function f: R \rightarrow R given by onto.

f(x)=\left\{\begin{array}{c} 1, \text { if } x>0 \\ 0, \text { if } x=0 \\ -1, \text { if } x<0 \end{array}\right\}\text{ is neither one-one nor }

Solution: f(x)=\left\{\begin{array}{l} 1, \text { if } \mathrm{x}>0 \\ 0, \text { if } \mathrm{x}=0 \\ -1, \text { if } \mathrm{x}<0 \end{array}\right\}

f(1)=f(2)=1, but 1 \neq 2

\therefore f is not one-one.

f(x) takes only 3 values (1,0,-1) for the element -2 in co-domain

\mathrm{R}, there does not exist any x in domain \mathrm{R} such that f(x)=-2.

\therefore f is not onto.

The signum function is neither one-one nor onto.

Question 6: Let A=\{1,2,3\}, B=\{4,5,6,7\} and let f=\{(1,4),(2,5),(3,6)\} be a function from A to B. Show that \mathrm{f} is one-one.

Solution: A=\{1,2,3\}, B=\{4,5,6,7\}

f: A \rightarrow B is defined as f=\{(1,4),(2,5),(3,6)\}

\therefore f(1)=4, f(2)=5, f(3)=6

It is seen that the images of distinct elements of A under f are distinct.

\therefore f is one-one.

Question 7: In each of the following cases, state whether the function is one-one, onto or bijective.

Justify your answer.

i). \quad f: R \rightarrow R defined by f(x)=3-4 x

ii). f: R \rightarrow R defined by f(x)=1+x^{2}

Solution: i. f: R \rightarrow R defined by f(x)=3-4 x

x_{1}, x_{2} \in R such that f\left(x_{1}\right)=f\left(x_{2}\right)

\Rightarrow 3-4 x_{1}=3-4 x_{2}

\Rightarrow-4 x=-4 x_{2}

\Rightarrow x_{1}=x_{2}

\therefore f is one-one.

For any real number (y) in R, there exists \frac{3-y}{4} in R such that f\left(\frac{3-y}{4}\right)=3-4\left(\frac{3-y}{4}\right)=y

\therefore f is onto.

Hence, f is bijective.

ii. f: R \rightarrow R defined by f(x)=1+x^{2}

x_{1}, x_{2} \in R such that f\left(x_{1}\right)=f\left(x_{2}\right)

\Rightarrow 1+x_{1}^{2}=1+x_{2}^{2}

\Rightarrow x_{1}^{2}=x_{2}^{2}

\Rightarrow x_{1}=\pm x_{2}

\therefore f\left(x_{1}\right)=f\left(x_{2}\right) does not imply that x_{1}=x_{2}

Consider f(1)=f(-1)=2

\therefore f is not one-one.

Consider an element -2 in co domain R.

It is seen that f(x)=1+x^{2} is positive for all x \in R.

\therefore f is not onto.

Hence, f is neither one-one nor onto.

Question 8: Let A and B be sets. Show that f: A \times B \rightarrow B \times A such that (a, b)=(b, a) is a bijective function.

Solution: f: A \times B \rightarrow B \times A is defined as (a, b)=(b, a).

\left(a_{1}, b_{1}\right),\left(a_{2}, b_{2}\right) \in A \times B_{\text {such that }} f\left(a_{1}, b_{1}\right)=f\left(a_{2}, b_{2}\right) \Rightarrow\left(b_{1}, a_{1}\right)=\left(b_{2}, a_{2}\right)

\Rightarrow b_{1}=b_{2} and a_{1}=a_{2}

\Rightarrow\left(a_{1}, b_{1}\right)=\left(a_{2}, b_{2}\right)

\therefore f is one-one.

(b, a) \in B \times A there exist (a, b) \in A \times B such that f(a, b)=(b, a)

\therefore f is onto.

f is bijective.

Question 9: Let f: N \rightarrow N be defined as

f(n)=\left\{\begin{array}{l} \frac{n+1}{2}, \text { if } n \text { is odd } \\ \frac{n}{2}, \text { if } n \text { is even } \end{array}\right\}
for all n \in N.

State whether the function f is bijective. Justify your answer.

Solution: f: N \rightarrow N be defined as

f(n)=\left\{\begin{array}{l} \frac{n+1}{2}, \text { if } n \text { is odd } \\ \frac{n}{2}, \text { if } n \text { is even } \end{array}\right\} \text { for all } n \in N .

f(1)=\frac{1+1}{2}=1 and f(2)=\frac{2}{2}=1

f(1)=f(2), where 1 \neq 2

\therefore f is not one-one.

Consider a natural number n in co domain N.

Case I: n is odd

\therefore n=2 r+1 for some r \in N there exists 4 r+1 \in N such that

f(4 r+1)=\frac{4 r+1+1}{2}=2 r+1

Case II: n is even

\therefore n=2 r for some r \in N there exists 4 r \in N such that

f(4 r)=\frac{4 r}{2}=2 r

\therefore f is onto.

f is not a bijective function.

Question 10: Let A=R-\{3\}, B=R-\{1\} and f: A \rightarrow B defined by f(x)=\left(\frac{x-2}{x-3}\right). Is f one-one and onto? Justify your answer.

Solution: A=R-\{3\}, B=R-\{1\} \text { and } f: A \rightarrow B \text { defined by } f(x)=\left(\frac{x-2}{x-3}\right)
x, y \in A \text { such that } f(x)=f(y)
\Rightarrow \frac{x-2}{x-3}=\frac{y-2}{y-3}
\Rightarrow(x-2)(y-3)=(y-2)(x-3)
\Rightarrow x y-3 x-2 y+6=x y-3 y-2 x+6
\Rightarrow-3 x-2 y=-3 y-2 x
\Rightarrow 3 x-2 x=3 y-2 y
\Rightarrow x=y

\therefore f is one-one.

Let y \in B=R-\{1\}, then y \neq 1

The function f is onto if there exists x \in A such that f(x)=y.

Now,

\begin{array}{ll} & f(x)=y \\ \Rightarrow & \frac{x-2}{x-3}=y \\ \Rightarrow & x-2=x y-3 y \\ \Rightarrow & x(1-y)=-3 y+2 \\ \Rightarrow & x=\frac{2-3 y}{1-y} \in A \quad[y \neq 1] \end{array}

Thus, for any y \in B, there exists \frac{2-3 y}{1-y} \in A such that

f\left(\frac{2-3 y}{1-y}\right)=\frac{\left(\frac{2-3 y}{1-y}\right)-2}{\left(\frac{2-3 y}{1--y}\right)-3}=\frac{2-3 y-2+2 y}{2-3 y-3+3 y}=\frac{-y}{-1}=y

\therefore f is onto.

Hence, the function is one-one and onto.

Question 11: Let f: R \rightarrow R defined as f(x)=x^{4}.Choose the correct answer.
A. f is one-one onto
B. f is many-one onto
C. f is one-one but not onto
D. f is neither one-one nor onto

Solution: f: R \rightarrow R defined as f(x)=x^{4}

x, y \in R\text { such that } f(x)=f(y)

\Rightarrow x^{4}=y^{4}

\Rightarrow x=\pm y

\therefore f(x)=f(y) does not imply that x=y.

For example f(1)=f(-1)=1

\therefore f is not one-one.

Consider an element 2 in co domain R there does not exist any x in domain R such that f(x)=2.

\therefore f is not onto.

Function f is neither one-one nor onto.

The correct answer is D.

Question 12: Let f: R \rightarrow R defined as f(x)=3 x.Choose the correct answer.
A. f is one-one onto
B. f is many-one onto
C. f is one-one but not onto
D. f is neither one-one nor onto

Solution: f: R \rightarrow R defined as f(x)=3 x

x, y \in R_{\text {such that }} f(x)=f(y)

\Rightarrow 3 x=3 y

\Rightarrow x=y \therefore f is one-one.

For any real number y in co domain \mathrm{R}, there exist \frac{y}{3} in \mathrm{R} such that f\left(\frac{y}{3}\right)=3\left(\frac{y}{3}\right)=y

\therefore f is onto.

Hence, function f is one-one and onto.

The correct answer is A.

Class 12 ncert solution math exercise 1.1

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