Maths NCERT Exemplar Solutions Class 11th Chapter Two

$x^2+(y-1{)}^2=y^2$

$\Rightarrow x^2-2y+1+y^2=y^2\Rightarrow x^2=2y-1\Rightarrow x^2=2\left(y-\frac{1}{2}\right)$

Foci (0,1)

α=0, β=1;α+β=1

Kindly consider the following Image 

$1+2x+3x^2+4x^3\dots =(1-x{)}^{-2}$

${\left(1+2x+3x^2+4x^3\dots \right)}^{-2}=(1-x{)}^4=1-4x+6x^2-4x^3+x^4$

$a_0=1,\mathrm{}{a}_1=-4$

 $a_0-a_1=5$

$\text{27.(32)}\begin{array}{rr}& I=2{\int }_0^{\pi /2}??\left({\mathrm{c}\mathrm{o}\mathrm{s}}^3?x+{\mathrm{c}\mathrm{o}\mathrm{s}}^4?x+{\mathrm{s}\mathrm{i}\mathrm{n}}^4?x\right)dx\\ & I=2\left[{\int }_0^{\pi /2}??{\mathrm{c}\mathrm{o}\mathrm{s}}^3?xdx+{\int }_0^{\pi /2}??{\mathrm{c}\mathrm{o}\mathrm{s}}^4?xdx+{\int }_0^{\pi /2}??{\mathrm{s}\mathrm{i}\mathrm{n}}^4?xdx\right]\\ & I=2\left[\frac{2}{3}*1+\frac{3*1}{4*2}*\frac{\pi }{2}+\frac{3*1}{4*2}*\frac{\pi }{2}\right]\\ & I=2\left[\frac{2}{3}+\frac{3\pi }{8}\right]\Rightarrow I=\frac{4}{3}+\frac{6\pi }{8}\Rightarrow 24I=32+18\pi \\ & 24I-18\pi =32\end{array}$

$9*9*3=81*3=243$

Clearly all equations can be $\lambda x-2y-6z=0$  form where $\lambda =1,\frac{4}{3},\frac{6}{5}$  

So for $x=0y=-3z$

So $x^2+y^2+z^2=10z^2$  (as  ) $x=0\mathrm{}y=-3z$

ATQ $10\le 10z^2\le 200$

$1\le z^2\le 20\Rightarrow z=\pm 1,\pm 2,\pm 3,\pm 4$

By property

If first and Last term of Andhra Pradesh and HP are equal (of nn  terms) then $a_r{H}_S=$  first *  x last term

If $r+s=n+1$ r+s=n+1

Equation of tangent of at $y=\mathrm{l}\mathrm{o}\mathrm{g}?x$  at $\left(e^{\lambda },\lambda \right)$  is $y-\lambda =\frac{1}{e^{\lambda }}\left(x-e^{\lambda }\right)$

It cuts y   axis at $(0,\lambda ,-1)$

Also normal to $y^2=8x$    at $\left(\mathrm{7,4}\right)$    is $y-4=-(x-2)$  

  $y=-x+6\mathrm{}$  Cuts y axis at $\left(\mathrm{0,6}\right)\mathrm{}$  SO ATQ $6=-1+\lambda \Rightarrow \lambda =7$

Plane containing both line $x=1$

Image of point $(\mathrm{2,1},3)$  in the plane $\frac{\alpha -2}{1}=\frac{\beta -1}{0}=\frac{\gamma -3}{0}=-2\left(1\right)$

$\alpha =0,\beta =1,\gamma =3;\alpha +\beta +\gamma =4$

Single digit factor of $12\to \mathrm{1,2},\mathrm{3,4},6$

  $(x,y,z)$  where , $xyz=12$

$(\mathrm{6,2},1)(\mathrm{4,3},1)(\mathrm{3,2},2)$

$\begin{array}{ccc}\downarrow & \downarrow & \downarrow \\ \text{6Ways}& \text{6 Ways}& \text{3Ways}\end{array}$

 Total 15 Ways