Ncert Solutions Maths class 11th

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

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

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

Let a, ar, ar², . an increasing G.P then r > 1 & a > 0

given : ar + ar5 = $\frac{25}{2}$ .(i)

and $a{r}^{2}a{r}^4=25\Rightarrow a^2{r}^6=25\Rightarrow {\left(a{r}^3\right)}^2=25$

$\Rightarrow a{r}^3=5.......\left(ii\right)$

From (i) and (ii), $\frac{ar\left(1+r^4\right)}{a{r}^3}=\frac{25}{2*5}$

$\Rightarrow 2+2r^4=5r^2\Rightarrow 2r^4-5r^2+2=0$

$\Rightarrow \left(2r^2-1\right)\left(r^2-2\right)=0$

$\Rightarrow r^2=\frac{1}{2},2\therefore r^2=2\Rightarrow r=\sqrt[]{2}$

$\therefore t_4+t_6+t_8=5+10+20=35$

Put x = $\frac{1}{3}$

S = 1 + 2x + 7x² + 12x³ + 17x⁴ + 22x⁵ + .

$\underset{\_}{xS=x+2x^2+7x^3+12x^4+17x^5+.........}$

Subtracting,

$\left(1-x\right)S=1+x+5x^2+5x^3+5x^4+5x^5+......$

= $1-4x+5x+5x^2+5x^3+.........$

$=\frac{1-x-4x+4x^2+5x}{1-x}=\frac{4x^2+1}{1-x}$

$S=\frac{4x^2+1}{{\left(1-x\right)}^2}putx=\frac{1}{3}wegets=\frac{\frac{4}{9}+1}{\frac{4}{9}}=\frac{13}{4}$

x – y = 0 . (i)

x + 2y = 3. (ii)

2x + y = 6 . (iii)

Solving (i) & (ii), we get (1, 1)

Solving (ii) & (iii), we get (3, 0)

Solving (iii) & (i), we get (2, 2)     

$AB=\sqrt[]{5,}BC=\sqrt[]{5},CA=\sqrt[]{2}$

$\therefore AB=BC$ Isosceles triangle

Hence, option (B) is correct answer.

By properties,

OA.AB = PA.AQ, OA=1, AB = 12

12 = a²

$a=2\sqrt[]{3}$

$\therefore AreaofPQB=\frac{1}{2}*12*2.2\sqrt[]{3}=24\sqrt[]{3}$ sq. units

Let direction ratio of the normal to the required plane are l, m, n

$\begin{array}{l}\therefore 3l+m-2n=0\\ \therefore 2l-5m-n=0\\ \therefore \frac{l}{-11}=\frac{m}{-1}=\frac{n}{-17}\end{array}$             

Equation of required plane

11 (x – 1) + 1 (y – 2) + 17 (z + 3) = 0

Any point on line  $\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}=r$ is P (r + 3, 2r + 4, 2r + 5) lies on x + y + z = 17,

5r + 12 = 17

r = 1

$P\left(4,6,7\right)A\left(1,1,9\right)distAP=\sqrt[]{9+25+4}=\sqrt[]{38}$  

x² + y² = 36         ……(i)

y² = 9x      .(ii)

Solving (i) & (ii)

x² + 9x – 36 = 0

(x + 12) (x – 3) = 0

x = 3

Let $A={\displaystyle \underset{0}{\overset{3}{\int }}\left(\sqrt[]{36-x^2}-3\sqrt[]{x}\right)dx}$  

$={\left[\frac{x\sqrt[]{36-x^2}}{2}+18si{n}^{-1}\frac{x}{6}\right]}_0^3-3.{\frac{x^{3/2}}{3/2}]}_0^3$       

$=3\pi -\frac{3\sqrt[]{3}}{2}$        

Required area = $=\frac{1}{2}\pi {\left(6\right)}^2+2\left(3\pi -\frac{3\sqrt[]{3}}{2}\right)=\left(24\pi -3\sqrt[]{3}\right)sq.unit$

BC = CN = x = 75

$cot\theta =\frac{3h}{75}$

$tan\theta =\frac{h}{75}$            

$\frac{3h^2}{75*75}=1\Rightarrow h=\frac{75}{\sqrt[]{3}}=25\sqrt[]{3}$

p + q = 2

 p⁴ + q⁴ = 272

${\left(p^2+q^2\right)}^2-2p^2{q}^2=272$          

  ${\left[{\left(p+q\right)}^2-2pq\right]}^2-2p^2{q}^2=272$          

Let pq = t Þ (4 – 2t)² – 2t² = 272

2t² – 16 t – 256 = 0

->t = pq = 16

Required equation x² – 2x + 16 = 0