Ncert Solutions Maths class 12th

$l_{n,m}={\displaystyle \underset{0}{\overset{\frac{1}{2}}{\int }}\frac{x^n}{x^m-1}dx,m,n\in N,n>m}$

$l_{6+i,3}-l_{3+i,3}$

$A=\frac{1}{2^5}\left[\begin{array}{ccc}\hfill \frac{1}{5}\hfill & \hfill \frac{1}{5}\hfill & \hfill \frac{1}{5}\hfill \\ \hfill 0\hfill & \hfill \frac{1}{12}\hfill & \hfill \frac{1}{12}\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill \frac{1}{28}\hfill \end{array}\right]$ $=\frac{B}{32}$

$\left|A\right|={\left(\frac{1}{32}\right)}^3\left|B\right|=\frac{1}{105.2^{19}}$

f (0) = 0, f (1) = 1, f (2) = 2

h (x) = f (x) – x has three roots

$\Rightarrow h\text{'}\left(x\right)=f\text{'}\left(x\right)-1$ , has at least two roots

$\Rightarrow h\text{'}\text{'}\left(x\right)=f\text{'}\text{'}\left(x\right)$ has at least one root.

$\frac{S_{10}}{S_P}=\frac{100}{P^2}$

$S_P=\frac{S_{10}.P^2}{100}\Rightarrow S_{11}=\frac{S_{10}.121}{100}$

$\frac{a_{11}}{a_{10}}=\frac{S_{11}-S_{10}}{S_{10}-S_9}=\frac{S_{10}.\frac{121}{100}-S_{10}}{S_{10}-\frac{S_{10}.81}{100}}$

= $\frac{21}{19}$

$\alpha +\beta +\gamma =2\pi$

$\Delta =\left|\begin{array}{ccc}\hfill 1\hfill & \hfill cos\gamma \hfill & \hfill cos\beta \hfill \\ \hfill cos\gamma \hfill & \hfill 1\hfill & \hfill cos\alpha \hfill \\ \hfill cos\beta \hfill & \hfill cos\alpha \hfill & \hfill 1\hfill \end{array}\right|$

$=1-co{s}^2\alpha -co{s}^2\beta -co{s}^2\gamma +2cos\alpha .cos\beta .cos\gamma$

$=-cos\gamma cos\left(\alpha -\beta \right)+cos\gamma cos\left(\alpha -\beta \right)=0$

  $\underset{x\to \frac{\pi }{4}}{lim}\frac{\frac{\pi }{4}.f\left(se{c}^{2}x\right)2se{c}^{2}xtanx}{2x}$ = 2f (2)

Required area

$A=2{\displaystyle {\int }_{}^{}\left(\left(sinx+cosx\right)-\left(cosx-sinx\right)\right)dx}$

$=2{\displaystyle {\int }_0^{\pi /4}sinxdx}$

$=2\sqrt[]{2}\left(\sqrt[]{2}-1\right)$    

Total ways = 6!

Ways satisfying g (3) = 2g (1) is 3

Number of onto function 3 * 4!

Probability = $\frac{1}{10}$

$\Delta =0\Rightarrow a=3,4$  

 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 $\overrightarrow{PQ}$

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

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

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

Point of intersection of

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

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

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