Ncert Solutions Maths class 12th

$\frac{h-cos\theta }{2}=\frac{k-sin\theta }{3}=\frac{w-0}{1}$

$=\frac{-1\left(2cos\theta +3sin\theta -6\right)}{14}\Rightarrow h=\frac{cos-2\left(2cos\theta +3sin\theta -6\right)}{14}$

$k=\frac{5sin\theta -6cos\theta +18}{14}$

${\left(5h+6k-12\right)}^2+4{\left(3h+5k-9\right)}^2=1$

  $S_1=\left\{z,\in c:\lfloor z_1-3\rfloor =\frac{1}{2}\right\}and{S}_2=\left\{z_2\in c:\left|z_2-\left|z_2+1\right|\right|=\left|z_2+\left|z_2-1\right|\right|\right\}$

Then, for $z_1\in S_1$  and $z_2\in S_2,$  the least value of |z₂ – z₁|

${\left|z_2+\left|z_2-1\right|\right|}^2={\left|z_2-\left|z_2+1\right|\right|}^2$      

$\Rightarrow \left(z_2+{\overline{z}}_2\right)\left(\left|z_2-1\right|+\left|z_2+1\right|-2\right)=0$      

$\therefore z_2+{\overline{z}}_2=0or\left|z_2-1\right|+\left|z_2+1\right|-2=0$          

$\therefore$ z₂ lies on imaginary axis or on real axis with in [-1, 1]

also $\left|z_1-3\right|=\frac{1}{2}$  lie on circle having centre 3 and radius   $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

Clearly $\left|z_1-z_2\right|min=\frac{5}{2}-1=\frac{3}{2}$

$7^{2022}+3^{2022}$

$={\left(49\right)}^{1011}+{\left(9\right)}^{1011}$

$={\left(50-1\right)}^{1011}+{\left(10-1\right)}^{1011}$

$=5\lambda -1+5k-1$

= 5m – 2

Remainder = 5 – 2 = 3

Probability that chosen candidate is female $=\frac{40}{60}=\frac{2}{3}$

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

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

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

$\left|\overrightarrow{a}\right|=9\&\left(x\overrightarrow{a}+y\overrightarrow{b}\right).\left(6y\overrightarrow{a}-18x\overrightarrow{b}\right)=0$

$\Rightarrow 6xy{\left|\overrightarrow{a}\right|}^2-18x^2\left(\overrightarrow{a}.\overrightarrow{b}\right)+6y^2\left(\overrightarrow{a}.\overrightarrow{b}\right)-18xy{\left|\overrightarrow{b}\right|}^2=0$

This should hold $\forall x,x,y\in R*R$

$\therefore {\left|\overrightarrow{a}\right|}^2=3{\left|\overrightarrow{b}\right|}^2\&\left(\overrightarrow{a}.\overrightarrow{b}\right)=0$

$\therefore \left|\overrightarrow{a}*\overrightarrow{b}\right|=\frac{{\left|\overrightarrow{a}\right|}^2}{\sqrt[]{3}}=\frac{81}{\sqrt[]{3}}=27\sqrt[]{3}$

a₀ = 0, a₁ = 0

a_n+2 = 3a_n+1 – 2a_n + 1

a₂₅ a₂₃ – 2a₂₅a₂₂ – a₂₃a₂₄ + 4a₂₂a₂₄ =?

a₂ = 3a₁ – 2a₀ + 1

a₃ = 3a₂ – 2a₁ + 1

a₄ = 3a₃ – 2a₂ + 1

a₅ = 3a₄ – 2a₃ + 1

a_n+2  = 3a_n+1 – 2a_n + 1

$\left(+\right)\Rightarrow \left(a_2+a_3+a_4+....+a_{n+1}+a_{n+2}\right)$

-> a_n+2 = 2 (a₂ + a₃ + …. + a_n + a_n+1) –2 (a₁ + a₂ + ….+ a_n) + n + 1

a_n+2 = 2a_n+1 + n + 1

a₂₅ a₂₃ -2a₂₅ a₂₂ -a₂₃ a₂₄ + 4a₂₂ a₂₄

= a₂₅ (a₂₃ – 2a₂₂) -2a₂₄ (a₂₃ – 2a₂₂)

As a_n+2 = 2a_n+1 + n + 1

-> a_n+2 – 2a_n+1 = n + 1

-> a_n+1 -2a_n = n

-> 24 * 22 = 528

By its given condition

$\stackrel{}{}\stackrel{}{}\stackrel{}{}$  $\therefore \overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are linearly independent vectors is $\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]\ne 0$ $\stackrel{}{}\stackrel{}{}\stackrel{}{}$

Now, $\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]$ $\stackrel{}{}\stackrel{}{}\stackrel{}{}$ = $\left|\begin{array}{ccc}\hfill 1+t\hfill & \hfill 1-t\hfill & \hfill 1\hfill \\ \hfill 1-t\hfill & \hfill 1+t\hfill & \hfill 2\hfill \\ \hfill t\hfill & \hfill -t\hfill & \hfill 1\hfill \end{array}\right|\ne 0$ $\begin{array}{ccc}\hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill \end{array}$

$R_1-R_2,R_2+R_3$

$R_1-R_2\Rightarrow \left|\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill -3\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 3\hfill \\ \hfill t\hfill & \hfill -t\hfill & \hfill \mathcal{l}\hfill \end{array}\right|\ne 0\Rightarrow t\ne 0$

$\underset{x?0}{lim}\frac{\mathcal{l}n\frac{1+5x}{1+?x}}{x}=10$

$\underset{x?0}{lim}\mathcal{l}n\left\{\frac{1+\left(5??\right)x}{\frac{\left(5??\right)x}{1+?x}?\left(\frac{1+?x}{5??}\right)}\right\}$

5 = 10

= 5

$l={\displaystyle {\int }_{-3}^{103}\left(\left[sin\left(\pi x\right)\right]+e^{\left[cos\left(2\pi x\right)\right]}\right)}dx$

$\left[sin\pi x\right]$ is periodic with period 2 and

$e^{\left[cos2\pi x\right]}$ is periodic with period 1.

So,

$I=52{\displaystyle {\int }_0^2\left(\left[sin\left(\pi x\right)\right]+e^{\left[cos2\pi x\right]}\right)}dx$

= $\frac{52}{e}$