Ncert Solutions Maths class 11th

$tan\theta =\frac{3x}{18}=\frac{x}{6}........\left(i\right)$ and

$tan2\theta =\frac{10x}{8}=\frac{5x}{9}..........\left(ii\right)$

Solving (i) and (ii)  $\frac{2tan\theta }{1-ta{n}^2\theta }=\frac{5x}{9}wegetx=\sqrt[]{\frac{72}{5}}$  

Height of pole = 10x =  $12\sqrt[]{10}$

q = 18° Þ 2q + 3q = 90° Þ sin 3q = 1 – sin 2q Þ cos q Þ cosq (4 sin² q + 2sinq - 1) = 0

$?cos\theta \ne 0\therefore 4si{n}^2\theta +2sin\theta -1=0\Rightarrow cose{c}^{2}18°-2cosec18°-4=0$

? x² – 2x – 4 = 0

${\left|2\overrightarrow{a}+3\overrightarrow{b}\right|}^2={\left|3\overrightarrow{a}+\overrightarrow{b}\right|}^2\Rightarrow 4{\left|\overrightarrow{a}\right|}^2+9{\left|\overrightarrow{b}\right|}^2+12\overrightarrow{a}.\overrightarrow{b}=9{\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2+6\overrightarrow{a}.\overrightarrow{b}$

$\Rightarrow -5{\left|\overrightarrow{a}\right|}^2+8{\left|\overrightarrow{b}\right|}^2+6\overrightarrow{a}.\overrightarrow{b}=0.........\left(i\right)$

$cos60°=\frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}=\frac{1}{2}\therefore \overrightarrow{a}.\overrightarrow{b}=4\left|\overrightarrow{b}\right|and\left|\overrightarrow{a}\right|=8$

from (i) we get $\left|\overrightarrow{b}\right|=5$

$?\frac{3}{1^2*2^2}+\frac{5}{2^2*3^2}+\frac{7}{3^2*4^2}+........$

$\therefore t_r=\frac{2r+1}{r^2{\left(r+1\right)}^2}=\frac{1}{r^2}-\frac{1}{{\left(r+1\right)}^2}$

$S_{10}={\displaystyle \sum _{r=1}^{10}{t}_r=}{\displaystyle \sum _{r=1}^{10}\left(\frac{1}{r^2}-\frac{1}{{\left(r+1\right)}^2}\right)=1-\frac{1}{11^2}=\frac{120}{121}}$

Equation of plane is 3x – 2y + 4z – 7 + $\lambda$ (x + 5y – 2z + 9) = 0. (i)

It passes through (1, 4, -3) and we get $\lambda =\frac{2}{3}$

$\therefore$ from (i) we get 11x + 4y + 8z – 3 = 0 Þ -11x – 4y – 8z + 3 = 0

$\therefore \alpha +\beta +\lambda =-11-4-8=-23$  

Since A (sec θ, 2 tanθ) & B $\left(sec?,2tan?\right)$  lies on 2x² – y² = 2 then

2sec² θ - 4tan^2_θ = 2 or sec² θ - 2 tan² θ = 1

-> $ta{n}^2\theta =0so\theta =0$

Similarly    $?=0but\theta +?=\frac{\pi }{2}$ (given) so not possible

Hence question is not correct

Let $l={\displaystyle \underset{6}{\overset{16}{\int }}\frac{lo{g}_e{x}^2}{lo{g}_e{x}^2+lo{g}_e\left(x^2-44x+484\right)}dx..........\left(i\right)}$

By property ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx={\displaystyle \underset{a}{\overset{b}{\int }}f\left(a+b-x\right)dx}}$

$\left(i\right)\Rightarrow l={\displaystyle \underset{6}{\overset{16}{\int }}\frac{lo{g}_e{\left(22-x\right)}^2}{lo{g}_e{\left(22-x\right)}^2+lo{g}_e{x}^2}dx}..........\left(ii\right)$      

(i) + (ii) 2l = ${\displaystyle \underset{6}{\overset{16}{\int }}1dx=10}$

l = 5

Kindly consider the following figure

$3*7^{22}+2*10^{22}-44=3*{\left(1+6\right)}^{22}+2{\left(1+9\right)}^{22}-44$

$=3\left[1{+}^{22}{C}_1*6{+}^{22}{C}_2*6^2{+}^{22}{C}_3*6^3+....{.}^{22}{C}_{22}{6}^{22}\right]$                

= -39 on division by 18

= (-54 + 15) on division by 18 = 15

 Slope of C₁C₂ = $\frac{3}{4}=tan\theta$  

By parametric form C₁ (1 + 5 cosθ, 2 + 5sinθ)

& C₂ (1 – 5 cos θ, 2 – 5 sinθ)

C_{1 $\left(1+5*\frac{4}{5}.2+5*\frac{3}{5}\right)\&C_2\left(1-5*\frac{4}{5}.2-5*\frac{3}{5}\right)$}

So, $\left|\left(\alpha +\beta \right)\left(r+\delta \right)\right|=40$