Ncert Solutions Maths class 11th

  $\Delta G=\Delta H-T\Delta S$

For spontaneity

$\begin{array}{l}\Delta G=-ve\\ \Delta S=+ve\\ \Delta H=-ve\end{array}$        

K $=\frac{4\sqrt[]{3}}{\sqrt[]{3}x+y}$ $\frac{\text{exception 25:}}{\text{exception 25:}}$

K = $\frac{\sqrt[]{3}x-y}{4\sqrt[]{3}}$ $\frac{\text{exception 25:}}{\text{exception 25:}}$

$\therefore \frac{4\sqrt[]{3}}{\sqrt[]{3}x+y}=\frac{\sqrt[]{3}x-y}{4\sqrt[]{3}}$

$\frac{x^2}{16}-\frac{y^2}{48}=1$

48 = 16 (e2 – 1)

$e=\sqrt[]{4}=2$

Let a_n be the side of square A_n

$a_n=\sqrt[]{2}{a}_{n+1}$

a₁ = 12

a_n = 12 $*{\left(\frac{1}{\sqrt[]{2}}\right)}^{n-1}$

${\left(a_n\right)}^2<1$

$\frac{144}{2^{\left(n-1\right)}}<1$

$\Rightarrow 2^{\left(n-1\right)}>144$

$\Rightarrow n-1\ge 8$

$\Rightarrow n\ge 9$

Kindly consider the following figure

$x^2-2\left(3k-1\right)x+8k^2-7>0\Rightarrow a>0,\&D<0,$

a = 1 > 0 and D < 0

4 (3k – 1)² – 4 (8k² – 7) < 0

  $\left(9k^2+1-6k-8k^2+7<0\right)$

$k(k-4)-2(k-4)<0$

$k\in \left(2,4\right)$

K = 3

sin 2 + tan 2 > 0

$\frac{2tan\theta }{1+ta{n}^2\theta }+\frac{2tan\theta }{1-ta{n}^2\theta }>0$

Let tan = x

$\frac{2x}{1+x^2}+\frac{2x}{1-x^2}>0$

$tan\theta <-1or0<tan\theta <1$

$\theta \in \left(0,\frac{\pi }{4}\right)\cup \left(\frac{\pi }{2},\frac{3\pi }{4}\right)\cup \left(\pi ,\frac{5\pi }{4}\right)\cup \left(\frac{3\pi }{2},\frac{7\pi }{4}\right)$

According to question,

$\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)=\frac{1}{4}$

$\frac{1}{a}+\frac{1}{b}=\frac{1}{2}........(i)$

Equation of required line is $\frac{x}{a}+\frac{y}{b}=1$ $\frac{}{}\frac{}{}$

Obviously B (2, 2) satisfying condition (i)

$Given\left|z+5\right|\le 4..........\left(i\right)$

->Represent a circle ${\left(x+5\right)}^2+y^2\le 16$  

$=z\left(1+i\right)+\overline{z}\left(1-i\right)\ge -10..........\left(ii\right)$           

->Represent a line X – y $\ge -5$  

So max |z + 1|² = AQ²

$={\left(-4-2\sqrt[]{2}\right)}^2+8$         

$\begin{array}{l}=32+16\sqrt[]{2}\\ =\alpha +\beta \sqrt[]{2}\end{array}$        

Hence a + b = 48

18 = 3² * 2

For G.C.D to be 3. no. of four digits should be only multiple of 3, but not multiple of 9 & also should not be even.

As we know no. of the form                         

9 k -> 1000

9 k + 1 -> 1000

9 k + 2 -> 1000

9 k + 3 -> 1000  -> Total no. = 2000

9 k + 4 -> 1000

9 k + 5 -> 1000

9 k + 6 ->1000

9 k + 7 -> 1000

9 k + 8 -> 1000

In which half will be even & half be odd so Required no. = 1000

Given sequence is -16, 8, -4, 2, .

are in G.P. with first term a = -16 & common ratio r =   $\frac{-1}{2}$

Now $t_p=a{r}^{P-1}=-16{\left(-\frac{1}{2}\right)}^{p-1}\&t_q=a{r}^{q-1}=-16{\left(-\frac{1}{2}\right)}^{q-1}$  

So A.M. = -8 $\left[{\left(-\frac{1}{2}\right)}^{p-1}+{\left(\frac{-1}{2}\right)}^{q-1}\right]=\alpha \left(let\right)$

Given equation is 4x² – 9x + 5 = 0 gives $x=1,\frac{5}{4}$  

From roots we get possible value of b = 1 so

$16{\left(-\frac{1}{2}\right)}^{\frac{P+q-2}{2}}=1OR{\left(-\frac{1}{2}\right)}^{\frac{p+q-2}{2}}=\frac{1}{16}-{\left(-\frac{1}{2}\right)}^4$

$\Rightarrow \frac{p+q-2}{2}=4\Rightarrow p+q=10$