Maths

28. For inequality, 2x+y≥ 6, the equation of line is 2x+y=6.

We consider the table below to pot 2x+y=6.

$\begin{array}{l}x\\ y\end{array}|\begin{array}{l}0\\ 6\end{array}|\begin{array}{l}3\\ 0\end{array}|$

Graph of 2x+y=6 is given as a continuous line in fig 2.

This line divides xy-plane in two half planes I and II.

We select 0 (0,0) and check the correctness of the inequality.

is 2 * 0+0 ≥ 6.

0 ≥ 6 which is false.

So, the solution region is II where origin (0,0) is included.

The continuous line indicates that any point on the line. also satisfy the given inequality.

10. Given, $\frac{x}{3}>\frac{x}{2}+1$

$\Rightarrow \frac{x}{3}-\frac{x}{2}>1$

$\Rightarrow \frac{2x-3x}{6}>1$

$\Rightarrow \frac{(2x-3x)*6}{6}>1*6$

x> 6

x< 6.

So, x (–∞, –6)

7. i. Let the points be P (0, 7, –10), Q (1, 6, –6) and R (4, 9, –6)

So,

6. Here, a_n=n $\frac{x^2+5}{4}$

Putting n=1,2,3,4,5 we get,

$a_1=1*\frac{\left(1^2+5\right)}{4}=1*\frac{6}{4}=\frac{3}{2}$

$a_2=2*\left(\frac{2^2+5}{4}\right)=2*\frac{(4+5)}{4}=\frac{9}{2}$

$a_3=3*\frac{\left(3^2+5\right)}{4}=3*\frac{(9+5)}{4}=3*\frac{14}{4}=\frac{21}{2}$

$a_4=4*\frac{\left(4^2+5\right)}{4}=4*\frac{(16+5)}{4}=4*\frac{21}{4}=21$

$a_5=5*\frac{\left(5^2+5\right)}{4}=5*\frac{(25+5)}{4}=5*\frac{30}{4}=\frac{75}{2}$

Hence, the first five terms are $\frac{3}{2},\frac{9}{2},\frac{21}{2},21,\frac{75}{2}$ .

3. Here a_n=2ⁿ

Substituting n=1,2,3,4,5 we get,

a₁=2¹=2

a₂=2²=4

a₃=2³=8

a₄=a⁴=16

a₅=2⁵=32.

Hence the first five terns are 2,4,16,32 and 64.

2. Here, a₁= $\frac{n}{n+1}$

Substituting n=1,2,3,4,5 we get,

$a_1=\frac{1}{1+1}=\frac{1}{2}$

$a_2=\frac{2}{2+1}=\frac{2}{3}$

$a_3=\frac{3}{3+1}=\frac{3}{4}$

$a_4=\frac{4}{4+1}=\frac{4}{5}$

$a_5=\frac{5}{5+1}=\frac{5}{6}$ .

Hence the first five terns are $\frac{1}{2},\frac{2}{3},\frac{3}{4},\frac{4}{5}$ and $\frac{5}{6}$ .

1. Since the ordered pairs are equal, the corresponding elements are equal.

$\therefore \frac{x}{3}+1=\frac{5}{3}.$ and $y-\frac{2}{3}=\frac{1}{3}$

$\frac{x}{3}=\frac{5}{3}-1$ $y=\frac{1}{3}+\frac{2}{3}$

$\Rightarrow \frac{x}{3}=\frac{5-3}{3}$ $\Rightarrow y=\frac{1+2}{3}$

$\Rightarrow \frac{x}{3}=\frac{2}{3}$ $\Rightarrow y=\frac{3}{3}$

$\Rightarrow x=2$ y = 1.

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}Givenequationoflinesarex=2y+3,y=1andy=-1\\ \therefore Requiredarea={\displaystyle \underset{-1}{\overset{1}{\int }}\left(2y+3\right)}dy\\ =2.\frac{1}{2}{\left[y^2\right]}_{-1}^1+3{\left[y\right]}_{-1}^1=\left(1-1\right)+3\left(1+1\right)=6sq.units\\ Hence,thecorrectoptionis\left(c\right).\end{array}$