Class 11 Maths Ncert Solutions Chapter 2 Ex 2.3 Relations & Functions PDF

# Class 11 Maths Ncert Solutions Ex 2.3

## Class 11 Maths Ncert Solutions Chapter 2 Ex 2.3

Q.1: Which of the given relations are functions? Answer with reason. If it is a function, find its range and domain.

(i) {(12, 11), (15, 11), (18, 11), (1, 11), (4, 11), (7, 11)}

(ii) {(12, 11), (14, 12), (16, 13), (18, 14), (0, 15), (2, 16), (4, 17)}

(iii) {(11, 13), (11, 15), (12, 15)}

Sol:

(i).   {(12, 11), (15, 11), (18, 11), (1, 11), (4, 11), (7, 11)}

Since, 12, 15, 18, 1, 4 and 7 are the elements of the domain of the given relation which have unique images.

Therefore, the given relation is a function.

Domain = {12, 15, 18, 1, 4, 7}

Range = {11}

(ii).   {(12, 11), (14, 12), (16, 13), (18, 14), (0, 15), (2, 16), (4, 17)}

Since, 12, 14, 16, 18, 0, 2 and 4 are the elements of the domain of the given relation which have unique images.

Therefore, the given relation is a function.

Domain = {12, 14, 16, 18, 0, 2, 4}

Range = {11, 12, 13, 14, 15, 16, 17}

(iii).  {(11, 13), (11, 15), (12, 15)}

Since, the same first element that is 11 corresponds to two images that is 13 and 15.

Therefore, the given relation is not a function.

Q.2: Find the range and domain of the given real function:

(i) f(y) = -|y|

(ii) f(y) = 9y2$\sqrt{9 – y^{2}}$

Sol:

(i) f(y) = -|y|, y $\in$ R

As we know:

|y| = {y,ify0y,ify<0$\left\{\begin{matrix} y,\; i\! f \; \; y\geq 0\\ -y,\; i\! f \; \; y< 0 \end{matrix}\right.\\$

Therefore,

f(y) = -|y| = {y,ify0y,ify<0$\left\{\begin{matrix} -y,\; i\! f \; \; y\geq 0\\ y,\; i\! f \; \; y< 0 \end{matrix}\right.\\$

Since, f(y) is defined for y $\in$ R, the domain of ‘f’ is R.

The range of f(y) = -|y|is all the real number except positive real number.

Therefore, range of ‘f’ is (,0]$(-\infty , 0]$.

(ii) f(y) = 9y2$\sqrt{9 – y^{2}}\\$

9y2$\sqrt{9 – y^{2}}$ is defined for the real number which are greater than or equal to -3 and less than or equal to 3.

Therefore, Domain of f(y) = {y: -3 $\leq$ y $\leq$ 3} or [-3, 3].

For any value of ‘y’ such that 3y3$-3 \; \leq \; y \; \leq 3$, the value of f(y) will lie between 0 and 3.

Therefore, the Range of f(y) = {y: 0 $\leq$ y $\leq$ 3} or [0, 3]

Q.3: A function f is f(y) = 3y – 6. Find the values of the following:

(i) f(1)

(ii) f(8)

(iii) f(-2)

Sol:

The function of ‘f’ is:

f(y) = 3y – 6

(i) f(1) = (3 × 1) – 6

= 3 – 6 = -3

(ii) f(8) = (3 × 8) – 6

= 24 – 6= 18

(iii) f(-2)  = (3 × -2) – 6

= -6 – 6= -12

Q.4: The function ‘f’ which shows temperature in degree Celsius into temperature in degree Fahrenheit is expressed as: 9C5+32$\frac{9C}{5}\; + \; 32$.

Find for the following values:

(i) f(0)

(ii) f(28)

(iii) f(-10)

(iv) The value of C, when f(C) = 212

Sol:

F = 9C5+32$\frac{9C}{5}\; + \; 32$

(i) f(0):

= 9×05+32$\frac{9 \times 0}{5}\; + \; 32$

= 0 + 32= 32

(ii) f(28):

= 9×285+32$\frac{9 \times 28}{5}\; + \; 32$

= 252+1605$\frac{252 + 160}{5}$

= 4125$\frac{412}{5}$

(iii) f(-10):

= 9×105+32$\frac{9 \times -10}{5}\; + \; 32$

= 9×(2)+32$9 \times (-2) \;+ \; 32$

= -18 + 32= 14

(iv) The value of C, when f(C) = 212:

$\Rightarrow$ 212=9×C5+32$212=\frac{9 \times C}{5}+32\\$

$\Rightarrow$ 9×C5=21232$\\\frac{9 \times C}{5}=212-32\\$

$\Rightarrow$ 9×C5=180$\\\frac{9 \times C}{5} = 180\\$

$\Rightarrow$ 9×C=180×5$\\9 \times C = 180 \times 5\\$

$\Rightarrow$ C=180×59$\\C=\frac{180\times5}{9}\\$

i.e.  C=100

Therefore, The value of f, when f(C) = 212 is 100.

Q.5: Calculate range of the given functions:

(i) f(y) = 2 – 3y, y $\in$ R, y > 0.

(ii) f(y) = y2+2$y^{2}\; + \;2$, is a real no.

(iii) f(y)  = y, y is a real no.

Sol:

(i)  f(y)=23y,yR,y>0$f\left ( y \right ) = 2 – 3y, \; y \in R, \; y > 0$

We can write the value of f(y) for different real numbers x > 0 in tabular form as:

 y 0.01 0.1 0.9 1 2 2.5 4 5 … f(y) 1.97 1.7 – 0.7 – 1 – 4 – 5.5 – 10 – 13 …

Thus, we can clearly observe that the range of ‘f’ forms the set for all real numbers which are less than 2.

i.e. Range of f = (,2)$\left ( -\infty , 2 \right )$

Alternative:

Let, y > 0

3y > 0

2 – 3y < 2

i.e f(y) < 2

Therefore, Range of f = (,2)$\left ( -\infty , 2 \right )$

(ii) f(y)=y2+2$f(y) = y^{2} + 2$, y, is a real number.

We can write the value of f(y) for different real numbers x, in tabular form as:

 y 0 ±$\pm$0.3 ±$\pm$0.8 ±$\pm$1 ±$\pm$2 ±$\pm$3 . . . f(y) 2 2.09 2.64 3 6 11 . . .

Thus, we can clearly observe that the range of f forms the set for all real numbers which are less than 2.

i.e. Range of f = (,2)$\left ( -\infty , 2 \right )$

Alternative:

Let ‘y’ be any real number. Then,

y20=>y2+20+2=>y2+22f(y)2$\\ y^{2} \geq 0 \\ => y^{2} + 2 \geq 0 + 2 \\ => y^{2} + 2 \geq 2 \\ f(y) \geq 2$

Therefore, Range of f = [2,)$\left [ 2, \; \infty \right )$

(iii) f(y) = y, where y is a real number

Here, we can see that the range of f is the set of all the real numbers.

Therefore, Range of f = R