Maths

$\Delta =0\Rightarrow a=3,4$  

 For a = 3, no solution

For a = 4, no solution

n (S₁) = 2, n (S₂) = 0,

Equation of intersection of line

$\frac{x-0}{1}=y=\frac{z-0}{-1}$

Let $\overrightarrow{r}=\widehat{i}-\widehat{k}$

Direction ratio of $\overrightarrow{PQ}$

$\Rightarrow \left(\lambda +1\right)\left(1\right)+\left(-2\right)\left(0\right)+\left(2-\lambda \right)\left(-1\right)=0$

$\lambda =\frac{1}{2}\Rightarrow Q\left(\frac{1}{2},0,\frac{-1}{2}\right)$

$PQ=\frac{\sqrt[]{34}}{2}$

$f\left(x\right)=lo{g}_{\sqrt[]{5}}\left(3+\sqrt[]{2}\left(cosx-sinx\right)\right)$

$-\sqrt[]{2}\le cosx-sinx\le \sqrt[]{2}$

$\Rightarrow 0\le f\left(x\right)\le 2$

$\left|\begin{array}{ccc}\hfill \alpha \hfill & \hfill \beta \hfill & \hfill 1\hfill \\ \hfill 5\hfill & \hfill 6\hfill & \hfill 1\hfill \\ \hfill 3\hfill & \hfill 2\hfill & \hfill 1\hfill \end{array}\right|=24$

$4\alpha -2\beta -8=\pm 24$

$4\alpha -2\beta =24+8,4\alpha -2\beta =-24+8$

$2\left(2\alpha -\beta \right)=322\alpha -\beta =-8$

Distance of origin

$D=\sqrt[]{{\alpha }^2+{\left(2\alpha +8\right)}^2}$

$\alpha =-\frac{16}{5}$

$D=\sqrt[]{{\left(\frac{-16}{5}\right)}^2+{\left(\frac{8}{5}\right)}^2}$

if 2 - = 16

$D=\frac{16}{\sqrt[]{5}}$

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