Ncert Solutions Maths class 12th

Given : $\widehat{a}\cdot \widehat{b}=\widehat{b}\cdot \widehat{c}=\widehat{c}\cdot \widehat{a}=cos\theta \left(say\right)$

$\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\left|\overrightarrow{c}\right|=14$

$\left(\overrightarrow{a}\times \overrightarrow{b}\right)\cdot \left(\overrightarrow{b}\times \overrightarrow{c}\right)$

$=\overrightarrow{a}\cdot \left[\left(\overrightarrow{b}\cdot \overrightarrow{c}\right)\overrightarrow{b}-\left(\overrightarrow{b}\cdot \overrightarrow{b}\right)\overrightarrow{c}\right]$

$=\left(\overrightarrow{a}\cdot \overrightarrow{b}\right)\left(\overrightarrow{b}\cdot \overrightarrow{c}\right)-{\left|\overrightarrow{b}\right|}^2\overrightarrow{a}\cdot \overrightarrow{c}$

$=\left|\overrightarrow{a}\right|{\left|\overrightarrow{b}\right|}^2\left|\overrightarrow{c}\right|\left(co{s}^2\theta -cos\theta \right)=14\left|\overrightarrow{b}\right|\left(co{s}^2\theta -cos\theta \right)$

Similarly, $\left(\overrightarrow{b}\times \overrightarrow{c}\right)\cdot \left(\overrightarrow{c}\times \overrightarrow{a}\right)$

= $\left|\overrightarrow{b}\right|{\left|\overrightarrow{c}\right|}^2\left|\overrightarrow{a}\right|\left(co{s}^2\theta -sin\theta \right)=14\left|\overrightarrow{c}\right|\left(cos2\theta -sin\theta \right)\&\left(\overrightarrow{c}\times \overrightarrow{a}\right)\cdot \left(\overrightarrow{a}\times \overrightarrow{b}\right)$

$=\left|\overrightarrow{c}\right|{\left|\overrightarrow{a}\right|}^2\left|\overrightarrow{b}\right|\left(co{s}^2\theta -sin\theta \right)=14\left|\overrightarrow{a}\right|\left(co{s}^2\theta -sin\theta \right)$

Given : 14 $\left(co{s}^2\theta -cos\theta \right)\left(\left|\overrightarrow{a}\right|+\left|\overrightarrow{b}\right|+\left|\overrightarrow{c}\right|\right)=168$

$\left|\overrightarrow{a}\right|+\left|\overrightarrow{b}\right|+\left|\overrightarrow{c}\right|=\frac{12}{co{s}^2\theta -cos\theta }=\frac{12}{\frac{1}{4}-\left(-\frac{1}{2}\right)}=\frac{12}{\frac{3}{4}}=16$

Given : $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are coplanar & pair wise equal angle.

$\left|z-1+i\right|\ge \left|z\right|$

$\left|z+i\right|=\left|z-1\right|$

W = (2x, y) = (α, y)

Let S represent the line segment AB

For ‘B’

x² + y² = 4

x = y

x² = 2

$x=\pm \sqrt[]{2}$

$B\left(-\sqrt[]{2},\sqrt[]{2}\right)$

$A\left(\frac{1}{2},-\frac{1}{2}\right)$

W (2x, y) lies on AB

$-\sqrt[]{2}<2x\le \frac{1}{2}$

$\left(\left|z\right|<2\right)$

$-\frac{1}{\sqrt[]{2}}<x\le \frac{1}{4}$

 $\begin{array}{ccc}\hfill 3R\hfill & \hfill 2R\hfill & \hfill 4B\hfill \\ \hfill 5B\hfill & \hfill 3W\hfill & \hfill 2W\hfill \end{array}$

I        II

Let

E₁ : a red ball is transferred from I to II

E₂ : a black is transferred from I to II

E₃ :a white transferred from I to II

E : a black ball is drawn from 2nd bag after a ball from I to II was transferred.

$P\left(\frac{E_1}{E}\right)=\frac{P\left(E_1\cap E\right)}{P\left(E\right)}$

$P\left(E\right)=P\left(E_1\cap E\right)+P\left(E_2\cap E\right)+P\left(E_3\cap E\right)$

= $P\left(E_1\right)\cdot P\left(\frac{E}{E_1}\right)+....+.....$

= $\frac{3}{10}\cdot \frac{5}{10}+\frac{4}{10}\cdot \frac{6}{10}+\frac{3}{10}\cdot \frac{5}{10}=\frac{54}{100}$

$P\left(E_1/E\right)=\frac{15/100}{54/100}=\frac{5}{18}$

I(x) = ${\displaystyle \int \frac{se{c}^{2}x-2022}{si{n}^{2022}x}dx}$

$={\displaystyle \int si{n}^{-2022}x\cdot se{c}^{2}xdx-{\displaystyle \int 2022si{n}^{-2022}xdx}}$

$=si{n}^{-2022}x\cdot tanx-{\displaystyle \int \left(-2022\right)si{n}^{-2023}x\cdot cosx\cdot tanxdx-{\displaystyle \int 2022si{n}^{-2022}xdx}}$

$=tanx\cdot si{n}^{-2022}x+2022{\displaystyle \int si{n}^{-2022}-2022{\displaystyle \int si{n}^{-2022}xdx}}$

I(x) = $tanx\cdot si{n}^{-2022}x+c$

Given, $I\left(\frac{\pi }{4}\right)=2^{1011}$

$2^{1011}=\frac{1}{{\left(\frac{1}{\sqrt[]{2}}\right)}^{2022}}+c\Rightarrow c=0$

$I\left(x\right)=\frac{tanx}{si{n}^{2022}x},I\left(\frac{\pi }{3}\right)=\frac{\sqrt[]{3}}{\left(\frac{\sqrt[]{3}}{2}\right)}=\frac{2^{2022}}{{\left(\sqrt[]{3}\right)}^{2021}}$

${\mathcal{l}}_1:$

$\overrightarrow{r}=\frac{1}{8}\widehat{i}+\lambda \left(-\frac{1}{8},\frac{1}{4\sqrt[]{2}},0\right)$

$\overrightarrow{r}=-\frac{1}{8}\widehat{i}+\mu \left(\frac{1}{8},0,-\frac{-1}{6\sqrt[]{3}}\right)$

$\overrightarrow{n}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill \frac{-1}{8}\hfill & \hfill \frac{1}{4\sqrt[]{2}}\hfill & \hfill 0\hfill \\ \hfill \frac{1}{8}\hfill & \hfill 0\hfill & \hfill \frac{-1}{6\sqrt[]{3}}\hfill \end{array}\right|$

$=\left(-\frac{1}{24\sqrt[]{6}},\frac{-1}{48\sqrt[]{3}},-\frac{1}{32\sqrt[]{2}}\right)$

$d=projectionofA\overrightarrow{C}on\overrightarrow{n}=\left(-\frac{2}{8},0,0\right).\frac{\left(4,2\sqrt[]{2},3\sqrt[]{3}\right)}{\sqrt[]{16}+8+27}=\frac{-1}{\sqrt[]{16+8+27}}$

d²⁼ $\frac{1}{51}$                                       

$f\left(x\right)=\left|\left(x-1\right)\left(x+1\right)\left(x-3\right)\right|+x-3$  

For    $x\in \left[-1,1\right]\cup \left[3,\infty \right)$

y = x² (x – 3)

  $\frac{dy}{dx}=2x\left(x-3\right)+x^2=3x^2-6x=3x\left(x-2\right)$

Number of max = 2

Number of min = 1

  $\overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill a_1\hfill & \hfill a_2\hfill & \hfill a_3\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill \lambda \hfill \end{array}\right|$              

$=\left(\lambda a_2-a_3,a_3-a_1\lambda ,a_1-a_2\right)$

$=\left(13,-1,-4\right)$

$\overrightarrow{a}.\overrightarrow{b}=-21$

$\Rightarrow a_1+a_2+a_3\lambda =-21$

$\lambda a_2-a_3=13\Rightarrow \lambda =\frac{13+a_3}{a_2}=\frac{a_3+1}{a_1}$

$\overrightarrow{b}-\overrightarrow{a}=\left(3,-1,10\right)$

$\overrightarrow{b}+\overrightarrow{a}=\left(-1,3,-4\right)$

= 11 + 3 + 14               

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

$=\frac{1}{3}{\left[\frac{1}{4}\mathcal{l}n\frac{x-2}{x+2}-\frac{1}{2}\mathcal{l}n\frac{x-1}{x+1}\right]}_3^6$

$=\mathcal{l}n{\left(\frac{49}{40}\right)}^{\frac{1}{12}}$

$\frac{5\left(b-2\right)}{b+2}/\frac{4{\left(b-1\right)}^2}{{\left(b+1\right)}^2}=\frac{49}{40}$

$\frac{4b}{2\left(b^2+1\right)}=\frac{-b+198}{99b-2}$

->b³ – 3b – 198 = 0

b = 6

  $f\left(x\right)=\left[2x^2+1\right]$

  $g\left(x\right)=\{\begin{array}{l}2x-3,x<0\\ 2x+3,x\ge 0\end{array}$

  $f\left(g\left(x\right)\right)=\left[2{\left(g\left(x\right)\right)}^2\right]+1$              

 for 0 < x < 1

-3 < 2x – 3 < -1

2 < 2 (2x – 3)² < 18

31 + 15 = 46

  $ta{n}^{-1}\left(\frac{\frac{1}{\sqrt[]{2}-1}}{\frac{1}{\sqrt[]{2}}}\right)$

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

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

$=-\theta$

$tan\theta =\sqrt[]{2}-1$

$tan2\theta =\frac{2\left(\sqrt[]{2}-1\right)}{1-{\left(\sqrt[]{2}-1\right)}^2}$

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

tan 2 θ = 1

$\theta =\frac{\pi }{8}$