Class 11 Chapter 2 Exercise 2.3 Solution

Since, \( 2,5,8,11,14 \) and 17 are the elements of domain R having their unique images. Hence, this relation R is a function.
As Domain of \( \mathrm{R}= \) set of all first elements of the order pairs in the relation.
\( \Rightarrow \) Domain of \( \mathrm{R}=\{2,5,8,11,14,17\} \)
Range of \( \mathrm{R}= \) set of all second elements of the order pairs in the relation.
\( \Rightarrow \) range of \( \mathrm{R}=\{1\} \).

Since, \( 2,5,8,11,14 \) and 17 are the elements of domain R having their unique images. Hence, this relation R is a function.
As Domain of \( \mathrm{R}= \) set of all first elements of the order pairs in the relation.
\( \Rightarrow \) Domain of \( \mathrm{R}=\{2,4,6,8,10,12,14\} \)
Range of \( \mathrm{R}= \) set of all second elements of the order pairs in the relation.
\( \Rightarrow \) range of \( \mathrm{R}=\{1,2,3,4,5,6,7\} \).

Since, the same first element 1 corresponds to two different images 3 and 5 . Hence, this relation is not a function.

Given: \( \mathrm{f}(x)=-|x| \)
As we know,
\(|x|=\left\{\begin{array}{l}
x, \text { if } x \geq 0 \\
-x \text { if } x < 0
\end{array}\right.\)
For a function \( \mathrm{f}(x),-|x|=\left\{\begin{array}{cl}-x & \text { if } x \geq 0 \\ x & \text { if } x 0\end{array}\right. \)
Domain: The values that can be put in the function to obtain a real value. For example: \( f(x)=x \), now we can put any value in place of \( x \) and we will get a real value. Hence, domain of this function will be Real Numbers.
Range: The values that we obtain of the function after putting the value from domain. For Example: \( f(x)=x+1 \), now if we put \( x=0, f(x)=1 \). This 1 is a value of Range that we obtained.
Since, \( f(x) \) is defined for \( x \in \mathrm{R} \), the domain of f is R .
It can be observed that the range of \( f(x)=-|x| \) is all real numbers except positive real numbers. Because will always get a negative number when we put a value from domain.

Given: \( f(x)=\sqrt{9-x^{2}} \)
Domain: These are the values of \( x \) for which \( f(x) \) is defined. from the given \( f(x) \) we can say that, \( f(x) \) should be real and for that, \(9 - x^{2} \geq 0 \) [Since a value less than 0 will give an imaginary value]
\( (3+x)(3-x) \geq 0 \) Now there are two critical points, \( x=+3 \) and \( x=-3 \) Taking a value less than \(-3\) and putting in the expression we get, \( (3-5) (3+5)=-\mathrm{ve} \) value and thus Plotting these on number line we get,

Since, \( f(x) \) is defined for all real numbers that are greater than or equal to -3 and less than or equal to 3 , the domain of \( f(x) \) is \( [-3,3] \).
Range: The values of \( \mathrm{f}(x) \) obtained by putting possible values of.
From the \( f(x) \) we can see that, the values obtained will only be positive and can be any positive number. Hence, Range of \( f(x)=[0, \infty) \)

Given: \( \mathrm{f}(x)=2 x-5 \)
(i) \( \mathrm{f}(0)=2 \times 0-5 \)
\(=0-5=-5\)
(i) \( \mathrm{f}(7)=2 \times 7-5 \)
\(=14-5=9\)
(i) \( f(-3)=2 \times(-3)-5 \)
\(=-6-5=-11\)

Given: \( \mathrm{t}(\mathrm{c})=\frac{9 c}{5}+32 \)
(i) \( \mathrm{t}(0) \)
\(\mathrm{t}(\mathrm{c})=\frac{9 \times 0}{5}+32\)
\(\Rightarrow \mathrm{t}(0)=32\)
(ii) \(t (28)\)
\(\mathrm{t}(28)=\frac{9 \times 25}{5}+32\)
\(\mathrm{t}(28)=\frac{252}{5}+32\)
\(\mathrm{t}(28)=\frac{252+160}{5}\)
\(\mathrm{t}(28)=\frac{412}{5}\)
(iii) \( t(-10) \)
\(\mathrm{t}(-10)=\frac{9 \times(-10)}{5}\)
\(\Rightarrow \mathrm{t}(-10)=-18+32=14\)
(iv) \( \mathrm{t}(\mathrm{C})=212 \)
\(\mathrm{t}(\mathrm{C})=\frac{9 \times C}{5}+32\)
\(212=\frac{9 C}{5}+32\)
\(\frac{9 C}{5}=212-32\)
\(\frac{9 C}{5}=180\)
\(C=\frac{180 \times 5}{9}=100\)
\( \Rightarrow \) Value of C , when \( \mathrm{t}(\mathrm{C}) \) is \( 212=100 \).

Given: \( f(x)=2-3 x, x \in R, x > 0 \)
As \( x > 0 \)
\(\Rightarrow 3 x > 0\)
\(\Rightarrow-3 x < 0\)
\(\Rightarrow-3 x+2 < 0+2\)
\(\Rightarrow 2-3 x < 2\)
\(\Rightarrow \mathrm{f}(x) < 2\)
Hence, range of \( f(x)=(-\infty, 2) \)

Given: \( f(x)=x^2+2, x \) is a real number.
As \( x^2 > 0 \)
\(\Rightarrow x^2+2 > 0+2\)
\(\Rightarrow x^2+2 > 2\)
\(\Rightarrow \mathrm{f}(x) > 2\)
Hence, range of \( \mathrm{f}(x)=[2, \infty) \).

Given: \( f(x)=x, x \) is a real number.
It is clear that range of \( f(x) \) is set of all real numbers.
Hence, range of \( \mathrm{f}(x)=\mathrm{R} \).