Class 11 Maths Ch 2 Miscellaneous Exercise Solutions

Given: \( \mathrm{f}(x)=\left\{\begin{array}{c}x^{2}, 0 \leq x \leq 3 \\ 3 x, 3 \leq x \leq 10\end{array}\right. \) and \( \mathrm{g}(x)=\left\{\begin{array}{c}x^{2}, 0 \leq x \leq 2 \\ 3 x, 2 \leq x \leq 10\end{array}\right. \)
As \( \mathrm{f}(x)=\left\{\begin{array}{l}x^{2}, 0 \leq x \leq 3 \\ 3 x, 3 \leq x \leq 10\end{array}\right. \)
\( \Rightarrow \mathrm{f}(x)=x^2 \) for \( 0 \leq x < 3 \)
And \( f(x)=3 x \) for \( 3 \leq x < 10 \)
At \( x=3, \mathrm{f}(x)=x^2=3^2=9 \)
Also, at \( x=3, \mathrm{f}(x)=3 x=3 \times 3=9 \)
Hence, we see for \( 0 \leq x \leq 10, f(x) \) has unique images. Thus, by definition of a function the given relation is function.
Now,
As \( \mathrm{g}(x)=\left\{\begin{array}{l}x^{2}, 0 \leq x \leq 2 \\ 3 x, 2 \leq x \leq 10\end{array}\right. \)
\( \Rightarrow \mathrm{g}(x)=x^2 \) for \( 0 \leq x < 2 \)
And \( \mathrm{g}(x)=3 \) xfor \( 2 \leq x < 10 \)
At \( x=2, \mathrm{~g}(x)=x^2=2^2=4 \)
Also, at \( x=2, \mathrm{~g}(x)=3 x=3 \times 2=6 \)
Hence, element 2 of the domain of relation \( g(x) \) corresponds to two different images i.e. 4 and 6.
because for \( 0 \leq x \leq 10, \mathrm{f}(x) \) does not have unique images. Thus, by definition of a function the given relation is not a function.

Given: \( \mathrm{f}(x)=x^2 \)
Then
\(\frac{f(1.1)-f(1)}{(1.1-1)}=\frac{(1.1)^{2}-(1)^{2}}{0.1}\)
\(=\frac{1.21-1}{0.1}=\frac{0.21}{0.1}=2.1\)

Given: \( \mathrm{f}(x)=\frac{x^{2}+2 x+1}{x^{2}-8 x+12} \)
\(f(x)=\frac{x^{2}+2 x+1}{x^{2}-8 x+12}\)
\(=\frac{x^{2}+2 x+1}{x^{2}-6 x-2 x+12}=\frac{x^{2}+2 x+1}{x(x-6)-2(x-6)}\)
\(\quad \mathrm{f}(x)=\frac{x^{2}+2 x+1}{(x-2)(x-6)}\)
Hence, we see \( f(x) \) can be defined at all real numbers except at \( x=2 \) and \( x=6 \). Hence the domain of \( f(x) \) is \( (R-\{2,6\}) \).

Given: \( \mathrm{f}(x)=\sqrt{(x-1)} \)
To Find: Domain and Range of \( f(x) \) Now for real values of \( f(x), x \geq 1 \).
Thus,
\( \sqrt{ }(x-1) \) is defined for \( x \geq 1 \).
Hence, the domain of f is the set of all real numbers greater than or equal to 1 i.e., the domain of \( \mathrm{f}=[1, \infty) \).
Now for Range,
As, \( x \geq 1, x-1 \geq 0 \) And thus \( \sqrt{ } x-1 \geq 0 \)
Hence, the Range of \( f \) is the set of all real numbers greater than or equal to 0 i.e., the Range of \( f=[0, \infty) \).

Given: \( \mathrm{f}(x)=|x-1| \)
As we see \( |x-1| \) is defined for all real numbers (R).
Hence the domain of \( \mathrm{f}=\mathrm{R} \).
Also, for \( x \in \mathrm{R},|x-1| \) assumes all real numbers.
Hence, the range of \( f \) is set of all non-negative real numbers.
Range of \( \mathrm{f}=[0, \infty) \)

Given: \( \mathrm{f}=\left\{\left(x, \frac{x^{2}}{1+x^{2}}\right): x \in R\right\} \)
To Find: \( \mathrm{f}=\left\{\left(x, \frac{x^{2}}{1+x^{2}}\right): x \in R\right\} \)
Range of any function is the set of values obtained after mapping is done in domain of the function So every value of the co domain that are being mapped is Range of the function for finding the range of any function Let \( y=f(x) \) and then find the values of \( y \) for
Let, \( y=\frac{x^{2}}{1+x^{2}} \)
Cross multiplying we get,
\(y\left(1+x^{2}\right)=x^{2}\)
which \(x\) exists. \( y+yx^{2}=x^{2} \)
\(x^{2}-yx^{2}=y\)
\(x^{2}=\frac{y}{1-y}\)
\(x= \pm \sqrt{\frac{y}{1-y}}\)
Now, \(1 - y\) should be greater than zero and \(y\) should be greater than and equal to zero for \(x\) to exist because other than those values the \(x\) will be imaginary
Thus, \( 1-y > 0 y < 1 \) and, \( y \geq 0 \) Hence, \( 0 \leq y < 1 \) will be the range of \( f(x) \)

Given: \( \mathrm{f}(x)=x+1, \mathrm{~g}(x)=2 x-3 \).
To find: \( f+g, f-g \),
\((\mathrm{f}+\mathrm{g}) x=\mathrm{f}(x)+\mathrm{g}(x)\)
\((\mathrm{f}+\mathrm{g}) x=x+1+2 x-3\)
\(\Rightarrow(\mathrm{f}+\mathrm{g}) x=3 x-2\)
And,
\((\mathrm{f}-\mathrm{g}) x=\mathrm{f}(x)-\mathrm{g}(x)\)
\((f-g) x=(x+1)-(2 x-3)\)
\(\Rightarrow(\mathrm{f}-\mathrm{g}) x=x+1-2 x+3\)
\(\Rightarrow(\mathrm{f}-\mathrm{g}) x=4-x\)
\(\text { And, }\left(\frac{f}{g}\right) x=\frac{f(x)}{g(x)}, g(x) \neq 0, x \in R\)
\(\left(\frac{f}{g}\right) x=\frac{x+1}{2 x-3}, 2 x-3 \neq 0 \text {, or } 2 x \neq 3 \text { or } x \neq \frac{3}{2}\)

Given: \( \mathrm{f}=\{(1,1),(2,3),(0,-1),(-1,-3)\} \)
\(\mathrm{f}(x)=\mathrm{a}x+\mathrm{b}\)
\((1,1) \in \mathrm{f} \Rightarrow \text { for } x=1, \mathrm{f}(x)=1\)
\(\Rightarrow 1=\mathrm{a}(1)+\mathrm{b}\)
\(\Rightarrow \mathrm{a}+\mathrm{b}=1 \ldots (1)\)
Similarly, \( (0,-1) \in \mathrm{f} \Rightarrow \) for \( x=0, \mathrm{f}(x)=-1 \)
\(\Rightarrow-1=\mathrm{a}(0)+\mathrm{b}\)
\(\Rightarrow \mathrm{b}=-1 \Rightarrow \mathrm{a}-1=1 \text { (from 1) }\)
\(\Rightarrow \mathrm{a}=2\)
Hence \( \mathrm{a}=2 \) and \( \mathrm{b}=-1 \).

Given: \( \mathrm{R}=\{(\mathrm{a}, \mathrm{b}): \mathrm{a}, \mathrm{b} \in \mathrm{N} \) and \( \mathrm{a}=\mathrm{b}^2\} \)
(i) \((a, a) € R\), for all a \( \epsilon \mathrm{N} \)
As we can see that \( 3 \in \mathrm{N} \) but \( 3 \neq 3^2=9 \).
Hence, the statement is not true.
(ii) (a, b) \( \in \) R, implies (b, a) \( \in R \)
As we can see that \( (4,2) \in \mathrm{N} \) and \( 4=2^2=4 \) but \( 2 \neq 4^2=16 \) hence, \( (2,4) \) does not belong to N .
Hence, the statement is not true.
(iii) \((a, b) \in R , (b, c) R \) implies (a, c) \( \in \) R.
As we see, \( (9,3) \in R,(16,4) \in R \) because \( 3,4,9,16 \in \mathrm{N} \) and \( 9=3^2 \) and \( 16=4^2 \).
Now, \( 9 \neq 4^2=16 \); therefore, \( (9,4) \) does not belong to N .
Hence, the statement is not true.

Given: \( \mathrm{A}=\{1,2,3,4\}, \mathrm{B}=\{1,5,9,11,15,16\} \) and \( \mathrm{f}=\{(1,5),(2,9),(3,1) \), \( (4,5),(2,11)\} \)
Now, \( \mathrm{A} \times \mathrm{B}=\{(1,1),(1,5),(1,9),(1,11),(1,15),(1,16),(2,1),(2,5) ,\) \( (2,9),(2,11),(2,15),(2,16),(3,1),(3,5),(3,9),(3,11),(3,15),(3,16) ,\) \( (4,1),(4,5),(4,9),(4,11),(4,15),(4,16) \).
(i) f is a relation from A to \( B \).
\(f=\{(1,5),(2,9),(3,1),(4,5),(2,11)\}\)
A relation from a non -empty set A to a non \( = \) empty set B is a subset of the Cartesian product. \( \mathrm{A} \times \mathrm{B} \).
And we can see \( f \) is a subset of \( \mathrm{A} \times \mathrm{B} \).
Hence \( f \) is a relation from \( A \) to \( B \) statement is true.
(ii) f is a function from A to \( B \).
\(f=\{(1,5),(2,9),(3,1),(4,5),(2,11)\}\)
As we observe that same first element i.e. 2 corresponds to two different images that is 9 and 11 . Thus f is not a function from A to B .

Given: \( \mathrm{f}=\{(\mathrm{ab}, \mathrm{a}+\mathrm{b}): \mathrm{a}, \mathrm{b} \in Z\} \)
For value \( 2,6,-2,-6 \in Z \),
\(f=\{(2 \times 6,2+6),(-2 \times-6,-2-6),(2 \times-6,2-6),(-2 \times 6,-2+6) .\)
\(\Rightarrow \mathrm{f}=\{(12,8),(12,-8),(-12,-4),(-12,4)\)
Here we observe that same first element i.e. 12 corresponds to two different images that is 8 and -8 . Thus f is not a function.

Given: \( A=\{9,10,11,12,13\} \)
\( f: A \rightarrow N \) be defined by \( f(n)= \) the highest prime factor of \( n .\)
Prime factor of \( 9=3 \)
Prime factor of \( 10=2,5 \)
Prime factor of \( 11=1 \)
Prime factor of \( 12=2,3 \)
Prime factor of \( 13=13 \)
\( \mathrm{f}(\mathrm{n})= \) the highest prime factor of n .
Hence,
\( \mathrm{f}(9)= \) the highest prime factor of \( 9=3 \)
\( \mathrm{f}(10)= \) the highest prime factor of \( 10=5 \)
\( \mathrm{f}(11)= \) the highest prime factor of \( 11=1 \)
\( \mathrm{f}(12)= \) the highest prime factor of \( 12=3 \)
\( \mathrm{f}(13)= \) the highest prime factor of \( 13=13 \)
As the range of f is the set of all \( \mathrm{f}(\mathrm{n}) \), where \( \mathrm{n} \in \mathrm{A} \).
Thus, the range of f is: \( \{3,5,11,13\} \).