Class 11th

Given series $\left\{3*1\right\},\left\{3*2,3*3,3*4\right\},\left\{3*5,3*6,3*7,3*8,3*9\right\}.........$  

$\therefore$  11^th set will have 1 + (10)2 = 21 terms

Also up to 10^th set total 3 * k type terms will be 1 + 3 + 5 + ……… +19 = 100 terms

$\therefore Set11=\left\{3*101,3*102,......3*121\right\}$ $\therefore$ Sum of elements = 3 * (101 + 102 + ….+121)

$=\frac{3*222*21}{2}=6993.$   

Factors of 36 = 2².3².1

Five-digit combinations can be 

(1, 2, 3, 3), (1, 4, 3, 1), (1, 9, 2, 1), (1, 4, 9, 11), (1, 2, 3, 6, 1), (1, 6, 1, 1)

i.e., total numbers  $\frac{5!5!}{2!2!}+\frac{5!}{2!2!}+\frac{5!}{2!2!}+\frac{5!}{3!}+\frac{5!}{2!}+\frac{5!}{3!2!}=\left(30*3\right)+20+60+10=180.$  

Process-AB $\Rightarrow$ Isobaric,                                                           

Process-AC  $\Rightarrow$ Isothermal, and

Process-ADÞAdiabatic

W₂ < W₁ < W₃

$\sim \left(p\iff \sim q\right)\wedge q=\left(p\iff \sim q\right)\wedge qis:$

$\therefore \left(\sim \left(P\iff \sim Q\right)\right)\wedge$ q is equivalent to $\left(p\Rightarrow q\right)\wedge$

According to question, we can write

$d=d_0\left(1+\alpha \Delta T\right)\Rightarrow 6.241=6.230\left(1+1.4*10^{-5}*\Delta T\right)$

$\Rightarrow T=126.18+27=153.18°C$

Any tangent to y² = 24x at (a, b) is by = 12 (x + a) therefore   Slope = $\frac{12}{\beta }$  

and perpendicular to 2x + 2y = 5 Þ 12 = b and a = 6 Hence hyperbola is $\frac{x^2}{6^2}-\frac{y^2}{12^2}$  = 1 and normal is drawn at (10, 16)

therefore equation of normal  $\frac{36x}{10}+\frac{144y}{16}=36+144\Rightarrow \frac{x}{50}+\frac{y}{20}=1$  This does not pass through (15, 13) out of given option.

According to question, we can write

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

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

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

Let point P : (h, k)

Therefore according to question,  ${\left(h-1\right)}^2+{\left(k-2\right)}^2+{\left(h+2\right)}^2+{\left(k-1\right)}^2=14$

$\therefore$ locus of P(h, k) is $x^2+y^2+x-3y-2=0$  

Now intersection with x – axis are  $x^2+x-2=0\Rightarrow x=-2,1$  

Now intersection with y – axis are  $y^2-3y-2=0\Rightarrow y=\frac{3\pm \sqrt[]{17}}{2}$  

Therefore are of the quadrilateral ABCD is =  $\frac{1}{2}\left(\left|x_1\right|+\left|x_2\right|\right)\left(\left|y_1\right|+\left|y_2\right|\right)=\frac{1}{2}*3*\sqrt[]{17}=\frac{3\sqrt[]{17}}{2}$  

$\frac{dT}{dt}=-k\left(T-T_0\right)$  

$\frac{dT}{\left(T-T_0\right)}=-kdt$  

$\Rightarrow k=-\frac{1}{6}\mathcal{l}n\left(\frac{2}{3}\right)$             - (1)

Now $\frac{dT}{dt}=-k\left(T-T_0\right)$

t = 10.257

t₂ = 6 + 10.257 = 16.257 minutes

16.25 and or 16

Given G.P's 2, 2², 2³, …60 term and 4, 4², 4³, … of 60

Now G.M. =  ${\left(2\right)}^{\frac{225}{8}}\Rightarrow {\left(2,2^2,2^3,....\right)}^{\frac{1}{60+n}}={\left(2\right)}^{\frac{225}{8}}\Rightarrow n=\frac{57}{8},20son=20$

$\therefore {\displaystyle \sum _{k=1}^{n}k(n-k)}20*\frac{20*21}{2}-\frac{20*21*41}{6}=1330$