Ncert Solutions Maths class 12th

take z = x + iy

$z^2+\overline{z}=0$               

$\Rightarrow x^2-y^2+x+i2xy-yi=0$               

$\Rightarrow x^2-y^2+x=0andy\left(2x-1\right)=0$             

if y = 0 Þ x = 0, -1

$ifx=\frac{1}{2}\Rightarrow y=\pm \frac{\sqrt[]{3}}{2}$               

$\Sigma \left(Re\left(z\right)+lm\left(z\right)\right)=\left(0-1+\frac{1}{2}+\frac{1}{2}\right)+\left(0+0+\frac{\sqrt[]{3}}{2}-\frac{\sqrt[]{3}}{2}\right)=0$          

$\left|f\left(x\right)\right|\le 800\Rightarrow 2n^2-n-1\le 800$  

$\Rightarrow 2n^2-n-801\le 0$                

${\displaystyle \sum _{x\in S}^{}f\left(x\right)={\displaystyle \sum _{}^{}\left(2x^2-x-1\right)}}$                

$=2\left(19^2+18^2+........1^2+0^2+1^2+.....+20^2\right)$                

= 10620

y⁵ – 9xy + 2x = 0

differentiate 5y⁴ – 9x $\frac{dy}{dx}$ - 9y + 2 = 0

$\frac{dy}{dx}=\frac{9y-2}{5y^4-9x}$        

 For horizontal tangent $\frac{dy}{dx}=0\Rightarrow y=\frac{2}{9}$  which does not satisfy the equation so no horizontal

For vertical tangent -> $5y^4-9x=0$  

->m = 0, N = 2

Plane through 4ax – y + 5z – 7a = 0 = 2x – 5y – z – 3

is    $x\left(4a+2\lambda \right)+y\left(-1-5\lambda \right)+z\left(5-\lambda \right)=7a+3\lambda$

This plane contains 4, -1, 0

->9a + 1 + 10l = 0          …… (i)

Plane contains the line $\frac{x-4}{1}=\frac{y+1}{-2}=\frac{z}{1}$  

-> $4a+11\lambda +7=0$ ……. (ii)

From (i) & (ii) a = 1,   $\lambda$  =-1

Equation of plane $\pi \equiv x+2y+3z-2=0$  

$\Rightarrow 7P+3-2P+4-12P+9-2=0\Rightarrow P=2$

  $f\left(x\right)\{\begin{array}{l}{\displaystyle \underset{0}{\overset{3}{\int }}\left(2-t\right)dt+{\displaystyle \underset{3}{\overset{x}{\int }}\left(8-t\right)dt;x>4}}\\ x^2+bx;x\le 4\end{array}$

f(x) is continuous at x = 4

$16+4b={\displaystyle \underset{0}{\overset{3}{\int }}\left(2-t\right)dt+{\displaystyle \underset{3}{\overset{4}{\int }}\left(8-t\right)dt}}$           

16 + 4b = 15

$\Rightarrow b=\frac{-1}{4}$               

$forx\le 4,$ $\begin{array}{l}f\left(x\right)=x^2-\frac{x}{4}\\ f\text{'}\left(x\right)=2x-\frac{1}{4}\end{array}$                 

f(x) is increasing in  $\left(\frac{1}{8},\infty \right)$  

 rate change at $x=\frac{1}{8},$  from -ve

${\pi }_1\equiv 2x+ky-5z=1$  

${\pi }_2\equiv 3kx-ky+z=5$                

Given  ${\pi }_1\perp {\pi }_2\Rightarrow 6k-k^2-5=0$  

k = 1, 5

Now required plane is p₁ +   $\lambda {\pi }_2=0$

y intercept = $\frac{1+5\lambda }{1-\lambda }$  

Draw the truthtable

              p            q       $p\wedge \sim q$ $p\wedge q$ $\left(p\wedge q\right)\iff \left(p\wedge \sim q\right)$                     &nb

Find total (5 digit) number of divisors of 7. Find total (5 digit) number of divisors of 35

  $\overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill \alpha \hfill & \hfill 1\hfill & \hfill \beta \hfill \\ \hfill 3\hfill & \hfill -5\hfill & \hfill 4\hfill \end{array}\right|=\left(4+5\beta \right)\widehat{i}+\left(3\beta -4\alpha \right)\widehat{j}+\left(-5\alpha -3\right)\widehat{k}$                

Compare with  $\overrightarrow{a}\times \overrightarrow{b}=-\widehat{i}+9\widehat{j}+12\widehat{k}$  

b = -1, a = -3

Projection of  $\overrightarrow{b}-2\overrightarrow{a}on\overrightarrow{b}+\overrightarrow{a}$  

$=\frac{\left(\overrightarrow{b}-2\overrightarrow{a}\right).\left(\overrightarrow{b}+\overrightarrow{a}\right)}{\left|\overrightarrow{b}+\overrightarrow{a}\right|}$   

${\left(x+2\sqrt[]{3}\right)}^2+y^2=4^2$                

$y^2=8\left(x+\frac{1}{2}\right)$

Point of intersection are (0, 2) and (0, -2)

  $2{\displaystyle \underset{0}{\overset{2}{\int }}\left[\left(\sqrt[]{16-y^2}-2\sqrt[]{3}\right)-\frac{\left(y^2-4\right)}{8}\right]dy}$              

  $=\frac{1}{3}\left(8\pi +4-12\sqrt[]{3}\right)$