Class 11th

Let S = 1 $+\frac{5}{6}+\frac{12}{6^2}+\frac{22}{6^3}+....$  

$\underset{\_}{-\frac{S}{6}=\frac{1}{6}+\frac{5}{6^2}+....}$  

$\frac{5S}{6}=1+\frac{4}{6}+\frac{7}{6^2}+\frac{10}{6^3}+....$  

$\underset{\_}{-\frac{5S}{36}=\frac{1}{6}+\frac{4}{6^2}+\frac{7}{6^3}+.....}$  

$\frac{25S}{36}=1+\frac{3}{6}+\frac{3}{6^2}+\frac{3}{6^3}+......$  

$\frac{25S}{36}=1+\frac{3/6}{1-1/6}$  

$\frac{25S}{36}=1+\frac{3/5}{1}$  

$S=\frac{288}{125}$

According to question, we can write

10 = $\frac{1}{2}a{t}^2...............\left(1\right)and$

$10+x=\frac{1}{2}a{\left(2t\right)}^2................\left(2\right)$

$\Rightarrow X=\frac{1}{2}A\left[{\left(2t\right)}^2-t^2\right]=3\left(\frac{1}{2}a{t}^2\right)=30m$

This is the case of fluids or gases. In such scenarios, the material does not have any limit to its flexibility and can be deformed to any extent. Such materials do not possess any stiffness and cannot regain their original form as they show no resistance towards any kind of pressure.

$\left|z\right|=3\Rightarrow$ circle with radius = 3

arg $\left(\frac{z-1}{z+1}\right)=\frac{\pi }{4},$ part of a circle (with radius $\sqrt[]{2}$ ). no common points

x⁴ + x² + 1 = 0

x⁴ + 2x² + 1 – x² = 0

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

$x=\pm \omega ,?{\omega }^2$

Now, =  ${\alpha }^{1011}+{\alpha }^{2022}-{\alpha }^{3033}$

=  ${\omega }^{1011}+{\omega }^{2022}-{\omega }^{3033}$

= 1 + 1 – 1 = 1

A diamond crystal with young modulus of more than 1050 Gpa is known to be the element which has the highest elastic moduli found in nature. This is possible due to the high electron density as well as strong covalent bonds between the carbon atoms.

$y+2x=\sqrt[]{11}+7\sqrt[]{7}$  ………(i)

              $2y+x=2\sqrt[]{11}+6\sqrt[]{7}$        ………(ii)

              $\Rightarrow x+y=\sqrt[]{11}+\frac{13}{3}\sqrt[]{7}$    ………(iii)

              Centre of the circle given by solving (i) & (ii)

              $as\left(\frac{8\sqrt[]{7}}{3},\sqrt[]{11}+\frac{5\sqrt[]{7}}{3}\right)$  

              Again $\sqrt[]{11}y-3x=\frac{5\sqrt[]{77}}{3}$ is tangent to the circle.

              $\therefore r=\left|\frac{\sqrt[]{11}\left(\sqrt[]{11}+\frac{5\sqrt[]{7}}{3}\right)-8\sqrt[]{7}-\frac{5\sqrt[]{77}}{3}}{\sqrt[]{11+9}}\right|=\left|\frac{11-8\sqrt[]{7}}{\sqrt[]{20}}\right|$  

              $\therefore {\left(5h-8k\right)}^2+5r^2=816$  

${\left(2x^3+\frac{3}{x^k}\right)}^{12}$

gen term = 

${=}^{12}{C}_r{2}^{12-r}.3^r.x^{36-3r-rk}$  

For constant term

36 – 3r – rk = 0

$\Rightarrow k=\frac{36-3r}{r}$  

for r = 1, 2, 4   

Possible values of k = 3, 1