Ncert Solutions Maths class 12th

$\overrightarrow{v}=l\overrightarrow{a}+m\overrightarrow{b}$  

$=\left(2l+m,-l+2m,2l=m\right)$               

  $\overrightarrow{v}.\left(3,2,-1\right)=0\Rightarrow l=-4m.......\left(i\right)$              

  $\left|\overrightarrow{v}.\widehat{a}\right|=19\Rightarrow 9l-2m=57........\left(ii\right)$              

(i) & (ii) Þ l = 6, m =   $\frac{-3}{2}$

$\Rightarrow 2\overrightarrow{v}=\left(21,-18,27\right)$              

${\left|2\overrightarrow{v}\right|}^2=1494$     &nbs

$\underset{x\to 1}{lim}\frac{x^{n}f\left(1\right)-f\left(x\right)}{x-1}$  

$=\underset{x\to 1}{lim}\frac{9x^n-\left(x^6+2x^4+x^3+2x+3\right)}{x-1}$                

= 9n – 19 = 44 -> n = 7

  $\overline{x}=\sum \times p\left(x\right)$  

$\Rightarrow \frac{23}{10}=\frac{-2}{5}-a+1+\frac{4}{5}+6b$            

  $\Rightarrow 6b-a=\frac{9}{10}.........\left(i\right)$             

Also,   $\frac{1}{5}+a+\frac{1}{3}+\frac{1}{5}+b=1$

$\Rightarrow a+b=\frac{4}{15}.......\left(ii\right)$       

(i) & (ii) -> $a=\frac{1}{10},b=\frac{1}{6}$

$\Rightarrow 100{\sigma }^2=781$          

$\frac{1}{2}{\displaystyle \underset{0}{\overset{3/2}{\int }}\left(1+4x-x^2\right)dx}$        

$=\frac{1}{2}{\left[x+2x^2-\frac{x^3}{3}\right]}_0^{3/2}=\frac{39}{8}$

${\displaystyle \underset{0}{\overset{3/2}{\int }}mx=\frac{9m}{8}}$              

(as per question)

$\Rightarrow \frac{39}{8}=\frac{9m}{4}$

$m=\frac{39}{18}$

12m = 26

${\displaystyle \int \frac{sinx}{si{n}^{3}x+co{s}^{3}x}dx}$

$I={\displaystyle \int \frac{tanxse{c}^{2}x}{ta{n}^{3}x+1}dx}$

Let, tan x = t

sec² xdx = dt

${\displaystyle \int \frac{tdt}{t^3+1}={\displaystyle \int \frac{t}{\left(t+1\right)\left(t^2-t+1\right)}dt}}$    

$\frac{t}{\left(t+1\right)\left(t^2-t+1\right)}=\frac{A}{\left(t+1\right)}+\frac{Bt+c}{t^2-t+1}$

$l=\frac{-1}{3}{\displaystyle \int \frac{1}{t+1}dt+\frac{1}{3}{\displaystyle \int \frac{t+1}{t^2-t+1}dt}}$

The value of 18 $\left(\alpha +\beta +{\gamma }^2\right)$ is equal to 3.

$f\left(x\right)=a{x}^3+b{x}^2+cx+d$   

  $f\text{'}\left(x\right)=3a{x}^2+2bx+c$              

$f\text{'}\text{'}\left(x\right)=6ax+2b$

$f\text{'}\text{'}\left(-1\right)=0$       

-6a + 2b = 0

=> b = 3a

$f\text{'}\left(1\right)=0$      

 -5 + d = -10 . (i)

 f (-1) = 6

11a + d = 6 . (ii)

$a=1,d=-5,b=3,c=-9$

$f\left(3\right)=27+27-27-5=22$

f (x) = x + a sin x

$wherea={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}cosyf\left(y\right)dy}$                

$a={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}cosy\left(y+asiny\right)dy}$                

-> a = p - 2

(2, -2) lie in a plane

=>2 + 6 + 4 + β = 0

=> b = -12

Line is perpendicular to normal

α (1) – 5 (3) + 2 (-2) = 0

α = 19

α + β = 7

$S_n={\displaystyle \sum _{r=1}^{n-1}r\left(n-r\right)}$

$=n\sum r-\sum r^2$

$=n\frac{\left(n-1\right)n}{2}-\frac{\left(n-1\right)n\left(2n-1\right)}{6}=\frac{n^3-n}{6}$

$S={\displaystyle \sum _{n=4}^{\infty }\left(\frac{2}{6}.\frac{n^3-n}{n!}-\frac{1}{\left(n-2\right)!}\right)=\frac{1}{3}}{\displaystyle \sum _{n=4}^{\infty }\frac{1}{\left(n-3\right)!}=\frac{e-1}{3}}$

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

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

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

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