Ncert Solutions Maths class 11th

$S=\frac{7}{5}+\frac{9}{5^2}+\frac{13}{5^3}+\frac{19}{5^4}..........\infty$

$\frac{S}{5}=\frac{7}{5^2}+\frac{9}{5^2}+\frac{13}{5^4}+------\infty$

$Let\frac{4S-7}{5}=t$

$\frac{4t}{5}=\frac{2}{25}\left\{\frac{1}{1-\frac{1}{5}}\right\}=\frac{1}{10}$

$t=\frac{1}{8}$

$\frac{4S-7}{5}=\frac{1}{8}$

$S=\frac{61}{32}$

$\therefore 160S=305$

The parabola : ${\left(x-\frac{1}{2}\right)}^2=y-\frac{3}{4}$  

-> y = x² – x + 1 . (i)

$P\left(-\frac{1}{2},\frac{7}{4}\right)$                

$N_P:y=\frac{x}{2}+2.........\left(ii\right)$                

(i) & (ii) -> Q (2, 3)

$P{Q}^2=\frac{125}{16}$                

Total 4 digit number

4 digit divisible by 7

1001, 1008, -9996

9996 = 1001 + (n₁ – 1) 7

n₁ = 1286

4 digit no divisible by 3

1002, 1005, -9999

9999 = 1002 + (n₂ – 1)3

n₂ = 3000

4 digit number visible by 21

1008, 1031, -9996

n₃ = 429

4 digit number divisible by 7 or 3

= 9000 – 1286 – 3000 + 429

= 5143

$AP=\sqrt[]{9+1-6+4+1}$

AP = 3 = AQ

$r=\sqrt[]{1+4-1}=2$        

$tan\theta =\frac{3}{2}$  

$\frac{Areaof\Delta APQ}{Areaof\Delta BPQ}=\frac{AR}{RB}=\frac{3sin\theta }{2cos\theta }=\frac{9}{4}$

$f\left(2\right)=f\left(4\right)=0andf\left(x\right)=x^3-6x^2+ax+b$  

$\Rightarrow f\left(x\right)=\left(x-2\right)\left(x-4\right)x=x^3-6x^2+8x$               

->a = 8, b = 0

$f\text{'}\left(x\right)=0\Rightarrow x=2\pm \frac{2}{\sqrt[]{3}},x_4=2+\frac{2}{\sqrt[]{3}},f\left(x_4\right)=\frac{-16}{3\sqrt[]{3}}$           

$f\text{'}\left(x_3\right)=\frac{\sqrt[]{3}}{2}f\left(x_4\right)=\frac{-8}{3}$               

The tangent to the parabola

$y^2=8xat\left(2,-4\right)is-4y=4\left(x+2\right)$  

x + y + 2 = 0

$OA=\sqrt[]{a}$         

$\Rightarrow \left|\frac{0+0+2}{\sqrt[]{2}}\right|=\sqrt[]{a}$  

$\sqrt[]{2}=\sqrt[]{a}$      

a = 2

Required number =   $\frac{4*2+24*3+36*4}{2}=112$

Let point on ellipse (2sinθ, 3cosθ) and the mid point of line segment joining (-3, -5) and

(2 sin θ, 3cosθ) will be (h, k)

$\frac{2sin\theta -3}{2}=h\frac{3cos\theta -5}{2}=k$

sin² θ + cos² θ = 1

${\left(\frac{2h+3}{2}\right)}^2+{\left(\frac{2k+5}{3}\right)}^2$ = 1

$36x^2+16y^2+108x+80y+145=0$

$\frac{dy}{dx}=\frac{2^{x}y+2^y.2^x}{2^x+2^{x+y}lo{g}_{e}2}$

${\displaystyle \int \frac{1+2^{y}lo{g}_{e}2}{y+2^y}dy={\displaystyle \int dx}}$        

=> ln|y + 2^y| = x + c

y (0) = 0

ln |y + 2^y| = x

y = 1

x = ln 3

$x\in \left(1,2\right)$

Angle bisectors are

$\frac{x-2y-2z+1}{3}=\pm \frac{2x-3y-6z+1}{7}$              

->x – 5y + 4z + 4 = 0, (i)

3x – 23y – 32z + 10 = 0 . (ii)

As distance of a point (-1, 0,0) on x – 2y – 2z + 1 = 0

from (i) is greater than that form (ii)

(ii) is the acute angle bisector.