NCERT Solutions

30. 

30. For inequality y+8 ≤ 2x, the equation of the line is y+8=2x. We consider the table below to plot of y+8=2x.

$\begin{array}{l}x\\ y\end{array}|\begin{array}{l}0\\ -8\end{array}|\begin{array}{l}4\\ 0\end{array}|$

The line devides the xy-plane into half planer I and II. We select a point (0,0) and check the correctness of the inequality.

i.e., 0+8 ≤ 2 * 0

0 ≤ 0 which is true.

So, the solution region is I which includes the rigin (0,0). The continuous line also indicates that any points on the line also satisfy the given inequality.

29. For inequality 3x+4y≥ 12 the equation of the line is 3x+4y=12

We consider the table below to plot 3x+4y=12.

$\begin{array}{l}x\\ y\end{array}|\begin{array}{l}0\\ 3\end{array}|\begin{array}{l}4\\ 0\end{array}|$

This line devides the xy-plane into half planer I and II.

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

i.e., 3 * 0+4 * 0 ≤ 12.

0+0 ≤ 12

0 ≤12 which is true.

So, the solution region is I which includes the origin (0,0). The continuous line also indicates that any points in the line also satisfy the given inequality.

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 $\frac{5}{6}$ .