Ncert Solutions Maths class 12th

Required area

$={\displaystyle \underset{-2}{\overset{4}{\int }}\left(\frac{3x}{2}+6-\frac{3x^2}{4}\right)dx=27}$

$\frac{3x^2}{4}=\frac{3x}{2}+6\Rightarrow x=-2,4$

$2sin\frac{\pi }{8}sin\frac{2\pi }{8}tan\frac{3\pi }{8}sin\left(\pi -\frac{5\pi }{8}\right)sin\left(\pi -\frac{6\pi }{8}\right)sin\left(\pi -\frac{7\pi }{8}\right)$

$={\left(\frac{1}{\sqrt[]{2}}\right)}^2.2si{n}^2\frac{\pi }{8}si{n}^2\frac{3\pi }{8}$

^{$=\frac{1}{4}.{\left(2sin\frac{\pi }{8}sin\frac{3\pi }{8}\right)}^2$}

$=\frac{1}{4}.{\left(2sin\frac{\pi }{8}sin\frac{3\pi }{8}\right)}^2$

${\displaystyle \underset{0}{\overset{5}{\int }}\frac{x+\left[x\right]}{e^{x-\left[x\right]}}}dx$

$={\displaystyle \sum _{r=1}^5{\displaystyle \underset{r-1}{\overset{r}{\int }}\frac{\left(x+1-r\right)}{e^{x+1-r}}}dx}$

Put x + 1 – r = t ->dx = dt

$=5{\left[-t{e}^{-t}-e^{-t}\right]}_0^1$              

$=5\left[-1e^{-1}-e^{-1}+e^0\right]=5\left(1-\frac{2}{e}\right)$

$\Rightarrow \alpha =-10\beta =5\Rightarrow {\left(\alpha +\beta \right)}^2=25$

$f\left(x\right)=\left\{x\right\},g\left(x\right)=1?\left\{x\right\}$

$f\left(x\right)=e^{x^{2}ln\left(\frac{2}{x}\right)}=e^{x^2\left(ln2-lnx\right)}$

$f\text{'}\left(x\right)=e^{x^2\left(ln2-lnx\right)}\left[2x\left(ln2-lnx\right)-x\right]$            

$={\left(\frac{2}{x}\right)}^{x^2}x\left[2\left(ln2-lnx\right)-1\right]$          

$f\text{'}\left(x\right)=0\Rightarrow 2\left(ln2-lnx\right)-1=0$   

2ln x = $ln\frac{4}{e}\Rightarrow x^2=\frac{4}{e}$  

System of equations can be written as

  $\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 1\hfill & \hfill 3\hfill & \hfill \lambda \hfill \end{array}\right)\left(\begin{array}{l}x\\ y\\ z\end{array}\right)=\left(\begin{array}{l}5\\ \mu \\ 1\end{array}\right)$             

R₃ – R₂, R₂ – R₁

For unique solution $\lambda \ne 5,\mu \in R.P=\frac{5}{6}$  

For no solution $\lambda =5,\mu \ne 3q=\frac{1}{6}*\frac{5}{6}=\frac{5}{36}$  

$P\left(\frac{x\ge 5}{x>2}\right)=\frac{P\left(x\ge 5\cap x>2\right)}{P\left(x>2\right)}=\frac{P\left(x\ge 5\right)}{P\left(x>2\right)}$

$=\frac{1-P\left(x\le 4\right)}{1-P\left(x\le 2\right)}$

$=\frac{1-\left[{\left(\frac{5}{6}\right)}^3*\frac{1}{6}+{\left(\frac{5}{6}\right)}^2*\frac{1}{6}+\left(\frac{5}{6}\right)\frac{1}{6}+\frac{1}{6}\right]}{1-\left[\frac{5}{6}*\frac{1}{6}+\frac{1}{6}\right]}$

$={\left(\frac{5}{6}\right)}^4*{\left(\frac{6}{5}\right)}^2={\left(\frac{5}{6}\right)}^2=\frac{25}{36}$

${\overrightarrow{n}}_1=6\widehat{i}+7\widehat{j}+8\widehat{k}{\overrightarrow{n}}_2=3\widehat{i}+5\widehat{j}+7\widehat{k}$

${\overrightarrow{n}}_1*{\overrightarrow{n}}_2=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 6\hfill & \hfill 7\hfill & \hfill 8\hfill \\ \hfill 3\hfill & \hfill 5\hfill & \hfill 7\hfill \end{array}\right|=9\widehat{i}-18\widehat{j}+9k$              

$\therefore$ the normal to required plane is $\widehat{i}-2\widehat{j}+\widehat{k}$

Equation of plane 1(x + 1) – 2(y – 1) + (z – 3) = 0

x – 2y + z = 0

P (7, -2, 13)

$\therefore PQ=\left|\frac{7+4+13}{\sqrt[]{1+4+1}}\right|=\frac{24}{\sqrt[]{6}}$

${\left(PQ\right)}^2=\frac{24*24}{6}=96$

$\left|2A\right|=2^3\left|A\right|$

replace A by adj 2A

  $\left|2adj2A\right|=2^3\left|adj2A\right|$           

$=2^3{\left|2A\right|}^2=2^3{\left(2^3\left|A\right|\right)}^2$

$=2^9{\left|A\right|}^2$              

Again replace A by (adj A)

|2adj 2 (adj (adj 2A)| = 2⁹ |adj 2A|⁴

= 2⁹ (|2A|²)⁴

->|A²| = 4

$f\left(x\right)=2x^3-3x^2-12x$  

  $f\text{'}\left(x\right)=6x^2-6x-12$             

= 6 (x – 2) (x + 1)

a = -1, b = 2

$A={\displaystyle \underset{-1}{\overset{0}{\int }}\left(2x^3-3x^2-12x\right)dx-{\displaystyle \underset{0}{\overset{2}{\int }}\left(2x^3-3x^2-12x\right)dx=\frac{57}{2}}}$

->4A = 114