Ncert Solutions Maths class 11th

$\begin{array}{cc}\hfill 2x-3y=\gamma +5\hfill & \hfill \alpha x+5y=\left(\beta +1\right)\hfill \end{array}\}infinitelymanysolution$

$\therefore \frac{2}{\alpha }=\frac{-3}{5}=\left(\frac{\gamma +5}{\beta +1}\right)$

(i) $\frac{2}{\alpha }=\frac{-3}{5}=\frac{\gamma +5}{\beta +1}$

$\alpha =\frac{5*2}{-3}$ 5x + 25 = -3β - 3

^{$5\gamma +3\beta =-28$}

^{$\left|9\alpha +5\gamma +3\beta \right|=\left|9*\frac{-10}{3}-28\right|$}

= ^{$\left|-30-28\right|=58$}

$S_n=\frac{n^2}{1-\frac{1}{{\left(n+1\right)}^2}}=\frac{n{\left(n+1\right)}^2}{\left(n+2\right)}={\left(n+1\right)}^2-2\left(n+1\right)+2-\frac{2}{\left(n+2\right)}$

$\frac{1}{26}+{\displaystyle \sum _{n=1}^{50}\left(S_n+\frac{2}{\left(n+1\right)}-\left(n+1\right)\right)=\frac{1}{26}+{\displaystyle \sum _{n=1}^{50}{\left(n+1\right)}^2}-3\left(n+1\right)+2+2\left(\frac{1}{\left(n+1\right)}-\frac{1}{\left(n+2\right)}\right)}$

$=\frac{1}{26}+45525-3*1325+2*50+2\left(\frac{1}{2}-\frac{1}{52}\right)$

$=\frac{1}{26}+41550+100+1-\frac{1}{26}=41651$

$\underset{x\to 1}{lim}\frac{sin\left(3x^2-4x+1\right)-x^2+1}{2x^3-7x^2+ax+b}=-2$

For this limit to be defined 2x³ – 7x² + ax + b should also trend to 0 or x ® 1.

$\therefore 2.{\left(1\right)}^3-7{\left(1\right)}^2+a\cdot 1+b=0$

 2 – 7 + (a + b) = 0

(a + b) = 5 …………….(i)

Now this becomes % form  $\therefore$ we apply L'lopital rule

$\underset{x\to 1}{lim}\frac{\left(3x^2-4x+1\right)-x^2+1}{2x^3-7x^2+ax+b}=\underset{x\to 1}{lim}\frac{cos\left(3x^2-4x+1\right)\left(6x-4\right)-2x}{6x^2-14x+a}$

Now the numerator again ® 0 as x = 1

$\therefore$  6x² – 14x + a ® 0 as x = 1

6 . (1)² – 14 + a = 0

a = 8 …………….(ii)

a + b = 5  $\therefore a-b=8-\left(-3\right)=11$       

(b = -3) ® from (i) & (ii)

  $x^2+y^2-2\sqrt[]{2}x-6\sqrt[]{2}y+14=0$

 centre $\left(\sqrt[]{2},3\sqrt[]{2}\right)$

radius  ${\left({\left(\sqrt[]{2}\right)}^2+{\left(3\sqrt[]{2}\right)}^2-14\right)}^{1/2}$

$={\left(2+18-14\right)}^{1/2}=\left(\sqrt[]{6}\right)$

${\left(x-2\sqrt[]{2}\right)}^2+{\left(y-2\sqrt[]{2}\right)}^2=r^2$

centre  $\left(2\sqrt[]{2},2\sqrt[]{2}\right)$

OA = $\sqrt[]{{\left(2\sqrt[]{2}-\sqrt[]{2}\right)}^2+{\left(2\sqrt[]{2}-3\sqrt[]{2}\right)}^2}=\sqrt[]{2+2}=2$                  

r² = ${\left(\sqrt[]{6}\right)}^2+{\left(2\right)}^2=6+4=10$  

$\underset{n\to \infty }{lim}6tan\left\{{\displaystyle \sum _{r=1}^{n}ta{n}^{-1}}\left(\frac{1}{r^2+3r+3}\right)\right\}$

$=\underset{n\to \infty }{lim}6tan\left\{{\displaystyle \sum _{r=1}^{n}ta{n}^{-1}}\left(\frac{\left(r+2\right)-\left(r+1\right)}{1+\left(r+2\right)\left(r+1\right)}\right)\right\}$

$=\underset{n\to \infty }{lim}6tan\left\{\frac{r}{2}-ta{n}^{-1}\left(2\right)\right\}$

$6*\frac{1}{2}=3$

$2y{e}^{x/y^2}dx+\left(y^2-4x{e}^{x/y^2}\right)dy=0$

  $\Rightarrow 2e^{2/y^2}\left(ydx-2xdy\right)+y^{2}dy=0$            

$2e^{x/y^2}d\left(\frac{x}{y^2}\right)+\frac{dy}{y}=0$

Integrating   $2e^{x/y^2}+Iny=c$

y = 1, n = 0  c = 2

  ^{$2e^{x/y^2}+Iny=2$}  

$\therefore y=e\Rightarrow 2e^{x/e^2}+1=2$

x = -e² In2

  $y=3?\left|x?\frac{1}{2}\right|=\left|x+1\right|$

Graph

Area =   $\left(\frac{1}{2}*\frac{3}{2}*\frac{3}{4}\right)*2+\frac{3}{2}*\frac{3}{2}=\frac{3}{2}*\frac{3}{2}*\frac{3}{2}=\frac{27}{8}$

  a + a₁……a_n, 100               n arithmetic mean 100

a + n = 33 ……. (i)

$\left(\frac{a_1}{a_n}=\frac{1}{7}\right)$

$\frac{\left(a+d\right)}{\left(100-d\right)}=\frac{7}{7}$

7a + 8d = 100…………… (ii)

a + (n + 1)d = 100………………. (iiI)

Solving these equations (i), (ii) & (iii), we get

n = 23 & d =    $\frac{15}{4}$

a = 10

Let the number of chocolates given to C₁, C₂, C₃ & C₄ be a, b, c, d respectively.

Given 4   $\le b\le 7$

$2\le c\le 6$

Now using these the maximum number of chocolates that can be given to C₁ or C₄ is 24 (where b & c are given 2 & 4 chocolates).

$\therefore 0\le a\le 24$

$0\le d\le 24$

& a + b + c + d = 30

So, total possible solution to the above equation.

Coefficient of x³⁰ in.

$\left(x^0+x+x^2+....+x^{24}\right)\left(x^4+x^5+....+x^7\right)\left(x^2+x^3+....+x^6\right)\left(x^0+x+x^2+....+x^{24}\right)$

=  ${\left(1+....+x^{24}\right)}^2\left(x^4\right)\left(1+x+....+x^3\right)*x^2\left(1+x+....+x^4\right)$

$=\left(x^{56}-2x^{31}+x^6\right)*\left(x^9-x^4-x^5+1\right)*{\left(x-1\right)}^{-4}$

x⁵⁶ & x³¹ can never give x³⁰ so we discard them.

$x^6*x^9*{\left(x-1\right)}^{-4}{\to }^{15+4-1}\text{?}{C}_{4-1}{=}^{18}\text{?}{C}_3$

Coefficient x³⁰ ® ¹⁸C₃ – ²³C₃ – ²²C₃ + ²⁷C₃

=   $\frac{18*17*16}{6}-\frac{23*22*21}{6}-\frac{22*21*20}{6}+\frac{27*26*25}{6}$

= 430