Maths

Point of intersection of

$\frac{x^2}{9}+y^2=1and{x}^2+y^2=3inthe1stquadrantis\left(\frac{3}{2},\frac{\sqrt[]{3}}{2}\right)$                

$m_1=-\frac{1}{3\sqrt[]{3}},m_2=-\sqrt[]{3}$                

$tan\theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|=\frac{2}{\sqrt[]{3}}$                

$S_{20}={\displaystyle \underset{n=1}{\overset{20}{?}}\frac{1}{d}\left(\frac{1}{a_n}?\frac{1}{a_{n+1}}\right)}$

$=\frac{1}{d}\left(\frac{1}{a_1}?\frac{1}{a_{21}}\right)$

$?a\left(a+20d\right)=45.......\left(i\right)$

$?a+10d=9.........\left(ii\right)$

$\left(i\right)\&\left(ii\right)?d^2=\frac{36}{100}$

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$\overrightarrow{a}*\left[\left(\overrightarrow{r}-\overrightarrow{b}\right)*\overrightarrow{a}\right]+\overrightarrow{b}*\left[\left(\overrightarrow{r}-\overrightarrow{c}\right)*\overrightarrow{b}\right]+\overrightarrow{c}*\left[\left(\overrightarrow{r}-\overrightarrow{a}\right)*\overrightarrow{c}\right]=\overrightarrow{0}$

As ${\left|\overrightarrow{a}\right|}^2={\left|\overrightarrow{b}\right|}^2={\left|\overrightarrow{c}\right|}^2$

$={\left|\overrightarrow{a}\right|}^2\left(3\overrightarrow{r}-\left(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right)\right)-\left(\left(\overrightarrow{a}.\overrightarrow{r}\right)\overrightarrow{a}+\left(\overrightarrow{a}.\overrightarrow{r}\right)\overrightarrow{b}+\left(\overrightarrow{c}.\overrightarrow{r}\right)\overrightarrow{c}\right)$

Let $\overrightarrow{r}=\frac{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}}{2}$

$DE:\frac{dy}{dx}+\frac{y}{x^2}=\frac{1}{x^3}$  

IF =    $e^{-\frac{1}{x}}$

Solution : $y{e}^{-\frac{1}{x}}={\displaystyle \int e^{-\frac{1}{x}}.\frac{1}{x^3}dx=\frac{1}{x}{e}^{-\frac{1}{x}}+e^{-\frac{1}{x}}+C}$  

Point (1, 1) -> C = $-\frac{1}{e}$  

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

$\alpha =\underset{x\to \pi /4}{lim}\frac{ta{n}^{3}x-tanx}{cos\left(x+\pi /4\right)}$

a = -4

$\beta =\underset{x\to 0}{lim}{\left(cosx\right)}^{cotx}$      

$\beta =e^0=1$

Equation whose roots are a and b

x² + 3x – 4 = 0

a = 1, b = 3

$32^{ta{n}^{2}x}+32^{se{c}^{2}x}=81$

$32^{ta{n}^{2}x}+32^{1+ta{n}^{2}x}=81$

$33.32^{ta{n}^{2}x}=81$

$32^{ta{n}^{2}x}=\frac{81}{33}$   

$forx\in \left[0,\pi /4\right]ta{n}^{2}x\in \left[0,1\right]$

One solution

Given expression

= 2π - 5 + 6 - 2π - (12 - 4π) = 4π - 11

f (m + n) = f (m) + f (n)

put m = n = 1, f (2) = f (1) + f (1)

again put m = 2, n = 1, f (3) = f (2) + f (1)

and put m = 3, n = 3, f (3 + 3) = f (3) + f (3), 2f (3) = f (6) = 18 Þ f (3) = 9

f (3) = 3f (1)

f (1) = 3, f (2) = 6

f (2).f (3) = 54

^{$y.\frac{dy}{dx}=x\left[\frac{y^2}{x^2}+\frac{?\left(y^2/x^2\right)}{?\text{'}\left(y^2/x^2\right)}\right],x>0,?>0$}

$Let\frac{y}{x}=t$

$\frac{dy}{dx}=t+x.\frac{dt}{dx}$

$ln\left(?\left(\frac{y^2}{4}\right)\right)=ln4+ln\left(?\left(1\right)\right)$

$=ln4\left(?\left(1\right)\right)$

$?\left(\frac{y^2}{4}\right)=4.?\left(1\right)$