Ncert Solutions Maths class 11th

${\displaystyle \underset{-1}{\overset{1}{\int }}{x}^2{e}^{\left[x^3\right]}dx=}{\displaystyle \underset{-1}{\overset{0}{\int }}{x}^2.e^{-1}dx+}{\displaystyle \underset{0}{\overset{1}{\int }}{x}^2.e^{0}dx=\frac{1}{e}}{\displaystyle \underset{-1}{\overset{0}{\int }}{x}^{2}dx+{\displaystyle \underset{0}{\overset{1}{\int }}{x}^{2}dx}}$

$=\frac{1}{e}{\left[\frac{x^3}{3}\right]}_{-1}^0+{\left[\frac{x^3}{3}\right]}_0^1=\frac{1}{e}\left[0-\left(\frac{-1}{3}\right)\right]+\left[\frac{1}{3}-0\right]=\frac{1}{3e}+\frac{1}{3}=\frac{e+1}{3e}$

$\frac{x^2}{25}+\frac{y^2}{16}=1$

16 = 25 (1 – e²)

$\Rightarrow e=\frac{3}{5}$

OF₁ = 5 $\left(\frac{3}{5}\right)=3$

For Hyperbola : e' $=\frac{5}{3}$

a = 3

$Hyperbola,\frac{x^2}{9}-\frac{y^2}{16}=1$

 $x,y\in \left(0,\pi \right)$

cos x + cos y – cos (x + y) = $\frac{3}{2}$

$2cos\frac{x+y}{2}.cos\frac{x-y}{2}-\left(2co{s}^2\frac{x+y}{2}-1\right)=\frac{3}{2}$

$\Rightarrow 2co{s}^2\frac{x+y}{2}-2cos\frac{x-y}{2}.cos\frac{x+y}{2}+\frac{1}{2}=0$

As,

$x,y\in \left(0,\pi \right)$

$\frac{-\pi }{2}<\frac{x-y}{2}<\frac{\pi }{2}$

$sin\frac{x-y}{2}=0\Rightarrow \frac{x-y}{2}=0\Rightarrow x=y$

then equation : cos x + cos y – cos (x + y) = $\frac{3}{2}$

$cosx=\frac{2\pm \sqrt[]{4-4}}{4}=\frac{1}{2}$

$x=y=\frac{\pi }{3}\Rightarrow sinx+cosy=\frac{\sqrt[]{3}}{2}+\frac{1}{2}=\frac{\sqrt[]{3}+1}{2}$

$z^2+\alpha z+\beta =0,\alpha ,\beta \in R$

roots : 1 – 2i, 1 + 2i

Sum of roots = 2 = -α and product of roots = 5 = β

α - β = -2 - 5 = -7

Combined equation of pair of lines OP and OQ is

$x^2+2y^2=2{\left(x+y\right)}^2$

$\Rightarrow x^2+4xy=0\Rightarrow x\left(x+4y\right)=0\Rightarrow \{\begin{array}{l}x=0\left(lineOP\right)\\ y=-\frac{x}{4}\left(lineOQ\right)\end{array}$

$tan\left(90°+\theta \right)=-\frac{1}{4}$

$-cot\theta =-\frac{1}{4}\Rightarrow tan\theta =4$

$\theta =ta{n}^{-1}4=co{t}^{-1}\frac{1}{4}=\frac{\pi }{2}-ta{n}^{-1}\frac{1}{4}$

$\angle POQ=\pi -\theta =\frac{\pi }{2}+ta{n}^{-1}\frac{1}{4}$

${\alpha }^2-6\alpha -2=0,{\beta }^2-6\beta -2=0$

${\alpha }^2-2=6\alpha ,{\beta }^2-2=6\beta$

$\frac{a_{10}-2a_8}{3a_9}=\frac{{\alpha }^{10}-{\beta }^{10}-2\left({\alpha }^8-{\beta }^8\right)}{3\left({\alpha }^9-{\beta }^9\right)}=\frac{{\alpha }^8\left({\alpha }^2-2\right)-{\beta }^8\left({\beta }^2-2\right)}{3\left({\alpha }^9-{\beta }^9\right)}$

$\frac{{\alpha }^8.6\alpha -{\beta }^8.6\beta }{3\left({\alpha }^9-{\beta }^9\right)}=2$

Using the standard equations of a hyperbola:

$9\left({\mathrm{e}}^2+l\right)$ and directrix $focus\mathrm{a}\mathrm{e}=\sqrt[]{10}$

By multiplying both focus and directrix, we get
$\frac{\mathrm{a}}{\mathrm{e}}=\frac{9}{\sqrt[]{10}}$ and $\Rightarrow {\mathrm{a}}^2=9$
Now $\mathrm{e}=\frac{\sqrt[]{10}}{3}$
$\mathrm{}(ae{)}^2=a^2+b^2$

$\frac{dy}{dx}=\frac{ax-by+a}{bx+cy+a}$

$=bxdy+cydy+ady=axdx-bydx+adx$

$=\frac{c{y}^2}{2}+ay-\frac{a{x}^2}{2}-ax+bxy=k$

$a{x}^2+a{y}^2+2ax-2ay=k$

$\Rightarrow x^2+y^2+2x-2y=\lambda$

Short distance of (11,6)

= $\sqrt[]{12^2+5^2}-5$ $\text{exception 25:}$

= 13 – 5

= 8

$x={\displaystyle \sum _{n=0}^{\infty }{a}^n}=\frac{1}{1-a}y={\displaystyle \sum _{n=0}^{\infty }{b}^n=\frac{1}{1-b}{\displaystyle \sum _{n=0}^{\infty }{c}^n=\frac{1}{1-c}}}$

Now,

a, b, c AP

1 – a, 1 – b, 1 – c AP

$\frac{1}{1-a},\frac{1}{1-b},\frac{1}{1-c}\to HP$

x, y, z HP

$\overline{z}=i{z}^2$

Let z = x + iy

x – iy = I (x2 – y2 + 2xiy)

Case-I

x = 0

y2 = y

y = 0, 1

Case – II

$y=-\frac{1}{2}$

$\Rightarrow x^2-\frac{1}{4}=\frac{1}{2}\Rightarrow x=\pm \frac{\sqrt[]{3}}{2}$

Area of polygon

$=\frac{1}{2}\left|\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill \frac{\sqrt[]{3}}{2}\hfill & \hfill \frac{-1}{2}\hfill & \hfill 1\hfill \\ \hfill \frac{-\sqrt[]{3}}{2}\hfill & \hfill \frac{-1}{2}\hfill & \hfill 1\hfill \end{array}\right|=\frac{1}{2}\left|-\sqrt[]{3}-\frac{\sqrt[]{3}}{2}\right|=\frac{3\sqrt[]{3}}{4}$