Ncert Solutions Maths class 11th

$si{n}^{7}x=co{s}^{7}x=1,x\in \left[0,4\pi \right]$            

will satisfy for sin x = 1, cos x = 0

$\Rightarrow x=\frac{\pi }{2}\&\frac{5\pi }{2}.$              

or, cos x = 1, sin x = 0

x = 0, 2π, 4π              total 5 solutions

f : R -> R.

$f\left(x\right)=[\begin{array}{l}\frac{x^3}{{\left(1-cos2x\right)}^2}log\left(\frac{1+2x{e}^{-2x}}{{\left(1-x{e}^{-x}\right)}^2}\right)x\ne 0\\ \alpha x=0\end{array}$                

As f is continuous at x = 0

$\therefore \alpha =\underset{x\to 0}{Lim}f\left(x\right)$              

$=\underset{x\to 0}{Lim}\frac{1}{2}\left[e^{-2x}+e^{-x}\right]=\frac{1}{2}*2=1$            

$\therefore \alpha =1$              

  $S_{10}=530\Rightarrow 5\left[2a+9d\right]=530$

2a + 9d = 106 . (i)

$S_5=140\Rightarrow \frac{5}{2}\left[2a+4d\right]=140$              

a + 2d = 28 . (ii)

Solving (i) & (ii) a = 88 d = 10

$\therefore S_{20}-S_6=14a+175d=1862$            

L : 2x + y = k.

$y=-2x+k.istangent\frac{x^2}{3}-\frac{y^2}{3}=1$                

$\Rightarrow k^2=3{\left(-2\right)}^2-3=9\Rightarrow k=3ask>0$              

y = -2x + 3 is also tangent to y² = $4\left(\frac{\alpha }{4}\right)x$  

$\Rightarrow 3=\frac{\alpha /4}{-2}$ -> a = -24

$36x^2+36y^2-108x+120y+c=0$

$\Rightarrow x^2+y^2-3x+\frac{10}{3}y+\frac{c}{36}=0$              

The circle not interesting either axes

$\therefore g^2<c\&f^2<c$               

  $\Rightarrow \frac{9}{4}<\frac{c}{36}\Rightarrow c>81-\left(i\right)\&\frac{25}{9}<\frac{c}{36}\Rightarrow c>100......\left(ii\right)$            

(i)  $\cap$ (ii) -> c > 100 .(iii)

Point of intersection between x – 2y – 4 = 0 & 2x – y – 5 = 0 is (2 - 1) lies inside the circle.

$\therefore 4+1-6-\frac{10}{3}+\frac{c}{36}<0\Rightarrow c<156.......\left(iv\right)$         

-> $\left(iii\right)\cap \left(iv\right)$ 100 < c < 156

${\left(\sqrt[]{50}\right)}^2={\left(\sqrt[]{45}\right)}^2+{\left(\sqrt[]{5}\right)}^2$

$\therefore \angle B=90°$ circum centre = $O\left(\frac{1}{2},\frac{11}{2}\right)$

Mid point of BC = $D\left(2,\frac{17}{2}\right)$

Equation of OD is y = 2x + $\frac{9}{2}.$ This line passes through

$\left(0,\frac{\alpha }{2}\right)\therefore \alpha =9$

$a_{n+1}=\frac{a_{n+2}-a_n}{2}letp={\displaystyle \sum _{n=1}^{\infty }\frac{a_n}{8^n}}$

$\Rightarrow 64\frac{a_{n+2}}{8^{n+2}}=\frac{16a_{n+1}}{8^{n+1}}+\frac{a_n}{8^n}$

$\Rightarrow 64{\displaystyle \sum _{n=1}^{\infty }\frac{a_{n+2}}{8^{n+2}}=}{\displaystyle \sum _{n=1}^{\infty }\frac{16a_{n+1}}{8^{n+1}}+}{\displaystyle \sum _{n=1}^{\infty }\frac{a_n}{8^n}}\Rightarrow 64\left(p-\frac{a_1}{8}-\frac{a_2}{8^2}\right)=16\left(p-\frac{a_1}{8}\right)+p$

$\Rightarrow 64\left(p-\frac{1}{8}-\frac{1}{8^2}\right)=16\left(p-\frac{1}{8}\right)+p\Rightarrow 64p-8-1=16p-2+p$

$?f\left(x\right)=ln\left(x+\sqrt[]{1+x^2}\right)\therefore f\left(-x\right)=-f\left(x\right)$ .

Hence f(x) is an odd function.

$Nowg\left(t\right)={\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}cos\left(\frac{\pi }{4}t+f\left(x\right)\right)dx}$

Put t = 0, g(0) = ${\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}cos\left(f\left(x\right)\right)dx=2}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}cos\left(f\left(x\right)\right)dx}...............\left(i\right)$

where cos(f(x)) is an even function.

Now again put t = 1,

$\therefore g\left(1\right)=\frac{1}{\sqrt[]{2}}*g\left(0\right),from\left(i\right)$

$\therefore g\left(0\right)=\sqrt[]{2}g\left(1\right)$

Centre of the smallest circle is A

Centre of the largest circle is B

$r_1=\left|CP+CB\right|=3\sqrt[]{2}+3$ and

$r_2=\left|CP-CB\right|=3\sqrt[]{2}-3$    

$\frac{r_1}{r_2}=\frac{3\sqrt[]{2}+3}{3\sqrt[]{2}-3}=3+2\sqrt[]{2}$