Maths

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

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

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}Givenequationoflinesarey=x+1,x=2andx=3\\ \therefore Requiredarea={\displaystyle \underset{2}{\overset{3}{\int }}\left(x+1\right)}dx\\ ={\left[\frac{x^2}{2}+x\right]}_2^3=\left(\frac{9}{2}+3\right)-\left(\frac{4}{2}+2\right)\\ =\frac{15}{2}-4=\frac{7}{4}sq.units\\ Hence,thecorrectoptionis\left(a\right).\end{array}$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}Givenequationofcircleis{x}^2+y^2=1\\ \Rightarrow y=\sqrt[]{1-x^2}\\ \therefore Sincethecircleissymmetricalabouttheaxes.\\ \therefore Requiredarea=4*{\displaystyle \underset{0}{\overset{1}{\int }}\sqrt[]{1-x^2}}dx\\ =4{\left[\frac{x}{2}\sqrt[]{1-x^2}+\frac{1}{2}si{n}^{-1}x\right]}_0^1\\ =4\left[0+\frac{1}{2}si{n}^{-1}\left(1\right)-0-0\right]=4*\frac{1}{2}*\frac{\pi }{2}=\pi sq.units\\ Hence,thecorrectoptionis\left(b\right).\end{array}$