Ncert Solutions Maths class 11th

Since centres C₁ (1, 1) and C₂ (9, 1) lies opposite sides of the line3x + 4y = α

$((3+4-\alpha )((27+4-\alpha )<0\Rightarrow \alpha \in \left(7,31\right)..........\left(i\right)$

Also length of perpendicular from centre of the circle is greater than radius of the circle.

$\frac{\left|3+4-\alpha \right|}{5}\ge 1\Rightarrow \alpha \le 2or\alpha \ge 12............\left(ii\right)$

and $\frac{\left|27+4-\alpha \right|}{5}\ge 2\Rightarrow \alpha \le 21or\alpha \ge 41...........\left(iii\right)$

from (i), (ii) and (iii) we get $$ $\in$ [12, 21]

$\therefore$ sum of all integers = 165 

$l={\displaystyle \underset{-1/2}{\overset{1}{\int }}\left[2x\right]dx+}{\displaystyle \underset{-1/2}{\overset{1}{\int }}\left|2x\right|dx}$

let  $l_1={\displaystyle \underset{-1/2}{\overset{1}{\int }}\left[2x\right]dxput2x=t\Rightarrow dx=\frac{dt}{2}}$

$\therefore l=l_1+l_2=0+\frac{5}{8}=\frac{5}{8}\Rightarrow 8l=5$

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