Ncert Solutions Maths class 12th

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

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

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)

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

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

Let f(x) = 8 sin x - sin 2x

$f\text{'}\left(x\right)=8cosx-2cos2x$              

$f\text{'}\text{'}\left(x\right)=-8sinx+4sin2x$               

= -8 (sin x) (1 – cos x)

f''(x) < 0, f'(x) is a decreasing function

  ^{$f\text{'}\left(\frac{\pi }{3}\right)<f\text{'}\left(x\right)<f\text{'}\left(\frac{\pi }{a}\right)$}               

  $5<f\text{'}\left(x\right)<\frac{8}{\sqrt[]{2}}$

${\displaystyle \underset{\pi /4}{\overset{\pi /3}{\int }}dx<{\displaystyle \underset{\pi /4}{\overset{\pi /3}{\int }}\frac{f\left(x\right)}{x}<{\displaystyle \underset{\pi /4}{\overset{\pi /3}{\int }}\frac{8}{\sqrt[]{2}}}}}$

$5\pi 12<l<\frac{8}{\sqrt[]{2}}.\frac{\pi }{\sqrt[]{2}}$             

Find RHL, x = -1 + h

$\underset{h\to 0}{lim}a.sin\left(\frac{\pi \left[-1+h\right]}{2}\right)+\left[2+1-h\right]=-a+2$               

 Find LHL, x = -1 – h

$\underset{h\to 0}{lim}asin\frac{\left(\pi \left[-1-h\right]\right)}{2}=\left[2+1+h\right]=3$               

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

$V={\left|z\right|}^2+{\left|z-3\right|}^2+{\left|z-6i\right|}^2$  

Put z = x + iy

$V=3\left[{\left(x-1\right)}^2+{\left(y-2\right)}^2+10\right]$                

For minimum x = 1, y = 2, V₀ = 30

So, z₀ = 1 + 2i

f (a) = a, a is max of powers of prime P such that P^a divides a.

f (2) = 1; g (2) = 3

f (3) = 1; g (3) = 4

f (4) = 2; g (4) = 5

f (5) = 1; g (5) = 6

f (2) + g (2) = 4

f (3) + g (3) = 5

f (4) + g (4) = 7

f (5) + g (5) = 7

So, f (x) + g (x) is many-one-and an into function. We won't get 'I' in the range.

  $\frac{x-a}{3}=\frac{y-b}{-4}=\frac{z-c}{12}=\frac{-2\left(3a-4b+12c+19\right)}{3^2+{\left(-4\right)}^2+12^2}$

$\frac{x-a}{3}=\frac{y-b}{-4}=\frac{z-c}{12}=\frac{-6a+8b-24c-38}{169}$               

$\left(x,y,z\right)\equiv \left(a-6,\beta ,\gamma \right)$               

  $\frac{\left(a-b\right)-a}{3}=\frac{\beta -b}{-4}=\frac{\gamma -c}{12}=\frac{-6a+8b-24c-38}{169}$              

  $\frac{\beta -b}{-4}=-2$              

$\Rightarrow \beta =8+b$               

$\Rightarrow 3a-4b+12c=150$                   ….(i)

a + b + c = 5

$\Rightarrow 3a+3b+3c=15$ ….(ii)

Applying (i) – (ii), we get :

= 56 + 216 + 7b – 9c = 56 + 216 – 135 = 137

First we arrange 5 red cubes in a row and assume  number of blue cubes between them

$Here,x_1+x_2+x_3+x_4+x_5+x_6=11$            

and    $x_2,x_3,x_4,x_5\ge 2$

so $x_1+x_2+x_3+x_4+x_5+x_6=3$  

No. of solutions = ⁸C₅ = 56

$f\left(x\right)=\frac{2e^{2x}}{e^{2x}+e^x}andf\left(1-x\right)=\frac{2e^{2-2x}}{e^{2-2x}+e^{1-x}}$

$\therefore \frac{f\left(x\right)+f\left(1-x\right)}{2}=1$

i.e. f (x) + f (1 – x) = 2

$\therefore f\left(\frac{1}{100}\right)+f\left(\frac{2}{100}\right)+....+f\left(\frac{99}{100}\right)$

${\displaystyle \sum _{x=1}^{49}f\left(\frac{x}{100}\right)+f\left(1-\frac{x}{100}\right)+f\left(\frac{1}{2}\right)}$

= 49 * 2 + 1 = 99