Ncert Solutions Maths class 11th

$Area=\frac{1}{2}\left(\sqrt[]{5}-1\right)\frac{9-\sqrt[]{5}}{4}-{\displaystyle \underset{1}{\overset{\sqrt[]{5}}{\int }}\frac{1}{2}\sqrt[]{5-x^2}dx}$

$=\frac{1}{8}\left(-14+10\sqrt[]{5}\right)-\frac{1}{2}{\displaystyle \underset{co{s}^{-1}\frac{1}{\sqrt[]{5}}}{\overset{0}{\int }}\left(-si{n}^2\theta \right)d\theta }$

$A=\frac{-14}{8}+\frac{5}{4}\sqrt[]{5}-\left(\frac{5}{4}co{s}^{-1}\frac{1}{\sqrt[]{5}}-\frac{1}{2}\right)$

$=\frac{5}{4}\sqrt[]{5}-\frac{5}{4}-\frac{5}{4}co{s}^{-1}\frac{1}{\sqrt[]{5}}$

$\alpha =\frac{5}{4},\beta =\frac{-5}{4},\gamma =\frac{-5}{4}$

$\left|\alpha +\beta +\gamma \right|=\frac{5}{4}$

$R_1\to R_1-R_2,R_2\to R_2-R_3$

$\Delta =\left|\begin{array}{ccc}\hfill a-c+1\hfill & \hfill b-c\hfill & \hfill a-b\hfill \\ \hfill b-d-1\hfill & \hfill c-d\hfill & \hfill b-c\hfill \\ \hfill x-b+d\hfill & \hfill x+d\hfill & \hfill x+c\hfill \end{array}\right|$

$C_1\to C_1-C_2\&C_2\to C_2-C_3$

$\Delta =\left|\begin{array}{ccc}\hfill a+1-b\hfill & \hfill 2b-c-a\hfill & \hfill a-b\hfill \\ \hfill b-1-c\hfill & \hfill 2c-b-d\hfill & \hfill b-c\hfill \\ \hfill -b\hfill & \hfill d-c\hfill & \hfill x+c\hfill \end{array}\right|=\left|\begin{array}{ccc}\hfill -\lambda +1\hfill & \hfill 0\hfill & \hfill -\lambda \hfill \\ \hfill -\lambda -1\hfill & \hfill 0\hfill & \hfill -\lambda \hfill \\ \hfill -b\hfill & \hfill \lambda \hfill & \hfill x+c\hfill \end{array}\right|,R_1\to R{}_1-R_2$

$=\left|\begin{array}{ccc}\hfill 2\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill -\lambda -1\hfill & \hfill 0\hfill & \hfill -\lambda \hfill \\ \hfill -b\hfill & \hfill \lambda \hfill & \hfill x+c\hfill \end{array}\right|=2{\lambda }^2=2\Rightarrow {\lambda }^2=1$

0 = -16 m + c . (i)

$\frac{\left|m\left(-10\right)+C\right|}{\sqrt[]{m^2+1}}=2.............\left(ii\right)$

(i) & (ii) Þ 36 m² = 4 (m² + 1)

$m=\frac{1}{2\sqrt[]{2}},C=\frac{8}{\sqrt[]{2}}$

$4\sqrt[]{2}\left(m+c\right)=34$

0 = 16 m + c . (i)

 $\frac{\left|m\left(?10\right)+C\right|}{\sqrt[]{m^2+1}}=2.............\left(ii\right)$

(i) & (ii) 36 m2 = 4 (m2 + 1)

$m=\frac{1}{2\sqrt[]{2}},C=\frac{8}{\sqrt[]{2}}$

$4\sqrt[]{2}\left(m+c\right)=34$

$x?\left(x\right)={\displaystyle \underset{}{\overset{}{\int }}\left(3t^2-2?\left(t\right)\right)dt,x>-2,By}$  Leibniz theorem

$?\left(x\right)+x?\text{'}\left(x\right)=3x^2-2?\text{'}\left(x\right)\Rightarrow ?\text{'}\left(x\right)+\frac{?\left(x\right)}{x+2}=\frac{3x^2}{x+2}..............\left(i\right)$

$\therefore I.F.=e^{{\displaystyle \int \frac{dx}{x+2}}}=x+2$

Multiplying (i) by I.F. we get,

$\therefore ?\left(x\right)=\frac{x^3+8}{x+2}$

$\therefore ?\left(2\right)=4$

VOWELS

vowels – 2, constants – 4

all the consonants never come together = 6! – 3! 4! =720 – 144 = 576

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

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