Class 11th

$a^2\left(e^2-1\right)=b^2$

$e=\sqrt[]{\frac{5}{2}}\Rightarrow b^2=\frac{3a^2}{2}$

Length of latus rectum $\frac{2b^2}{a}=6\sqrt[]{2}$  

$3a=6\sqrt[]{2}\Rightarrow a=2\sqrt[]{2}$  

$b=2\sqrt[]{3}$  

y = 2x + c is tangent to hyperbola

$\therefore c^2=a^2{m}^2-b^2=20$  

$f\left(x\right)=x^7+5x^3+3x+1$

$f\text{'}\left(x\right)=7x^6+15x^4+3>0\forall x\in R$

$\therefore f\left(x\right)$ is increasing

 for x $\infty$ - , f (x) $\infty$

x = 0, f (x) = 1

$\therefore f\left(x\right)$ = 0 has only one real root.

Let $\frac{A_2}{A_1}=\frac{A_3}{A_2}=...=r$  

$\therefore A_1{A}_3{A}_5{A}_7=\frac{1}{1296}$

$\Rightarrow A_1{r}^3=\frac{1}{6}.......\left(i\right)$  

Again, A₂ + A₄ = $\frac{7}{36}$  

$A_{1}r=\frac{7}{36}-\frac{1}{6}=\frac{1}{36}........\left(ii\right)$

$\therefore A_6+A_8+A_{10}=A_1{r}^5\left(1+r^2+r^4\right)=43$  

  $\therefore A_6+A_8+A_{10}=A_1{r}^5\left(1+r^2+r^4\right)=43$  

Sum of digits

1 + 2 + 3 + 5 + 6 + 7 = 24

So, either 3 or 6 rejected at a time

Case 1 Last digit is 2

……….2

no. of cases = 

  Case 2 Last digit is 6

……….6

= 4! = 24

Total cases = 72

$T_{r+1}{=}^{15}\text{?}{C}_r\cdot {\left(2x^{1/5}\right)}^{15-r}{\left(-x^{-1/5}\right)}^r$  

= ¹⁵C_{r $\cdot 2^{15-r}\cdot x^{\frac{15-2r}{5}}\cdot {\left(-1\right)}^r$}

Coefficient of x^-1 -> r = 10 ->m = ¹⁵C_{10 $\cdot 2^5$}

$x^{-3}\Rightarrow r=15\Rightarrow n{=}^{15}\text{?}{C}_{15}\cdot 2^0=1$  

now mn² = ¹⁵C_{r $\cdot 2^r$}  

r = 5

For expansion in vacuum, workdone, w = 0

For isothermal process,   $\Delta U=0$  

According to first law of thermodynamics,

$\begin{array}{l}\Delta U=q+w\\ q=0\end{array}$  

sin x = 1 – sin² x

sin x =  $\frac{-1+\sqrt[]{5}}{2},\frac{-1-\sqrt[]{5}}{2}\left(rejected\right)$  

draw y = sin x

y =  $\frac{\sqrt[]{5}-1}{2},$ find their pt. of intersection.