Maths

Since student guesses only two wrong. So there are three possibilities

(i) both wrong in section A

(ii) both wrong in section B

(iii) one wrong in each section A and B.

$\therefore$ Required possibilities =

${=}^4{C}_4{*}^6{C}_4{\left(\frac{3}{4}\right)}^4*{\left(\frac{1}{4}\right)}^4{\left(\frac{3}{4}\right)}^2{+}^4{C}_3{*}^6{C}_5{\left(\frac{3}{4}\right)}^3{\left(\frac{1}{4}\right)}^5*\frac{1}{4}*\frac{3}{4}$ ${+}^4{C}_2{*}^6{C}_6*{\left(\frac{3}{4}\right)}^2{\left(\frac{1}{4}\right)}^2*{\left(\frac{1}{4}\right)}^6$

$=\frac{27}{4^{10}}\left[15*27+24*3+2\right]=\frac{27*479}{4^{10}}$

OM² = OP² = PM²

^{$\left|\frac{1+r}{\sqrt[]{2}}\right|=r^2-1$}

^{$\therefore r=3$}

^$\therefore$ equation of circle is ${\left(x-1\right)}^2+{\left(y-3\right)}^2=3^2$

$\therefore h+k+r=7$

Required area

= ${\displaystyle {\int }_{-4}^2\left(4-y-\frac{y^2}{2}\right)dy}$

= ${\left(4y-\frac{y^2}{2}-\frac{y^3}{6}\right)}_{-4}^2=18squnits$

S = 1 + 3 + 3² + 3³ + ….+ 3²⁰²¹   $=\frac{3^{2022}-1}{2}=\frac{1}{2}\left[a^{1011}-1\right]$

^{$=\frac{1}{2}\left[9^{1011}-1\right]=\frac{1}{2}\left[100k+10110-1-1\right]$}                          

= 50k₁ + 4

Remainder = 4

$\left\{a\in \left(1,2,3,......,100\right):HCF\left(a,24\right)=1\right\}$

HCF of (a, 24) = 1 $\therefore$ a = 1, 5, 7, 11, 13, 17, 19, 23 sum of these numbers = 96

$\therefore$ There are four such blocks and a number 97 is there upto 100.

$\therefore$ complete sum = 96 + (24 * 8 + 96) + (48 * 8 + 96) + (72 * 8 + 96) + 97 = 1633

Sum of all given numbers = 31

Difference between odd and even positions must be 0,11 or 22 but 0 and 22 are not possible.

$\therefore$ Hence 11 is possible.

This is possible only when either 1, 2, 3, 4 if filled in odd places in order and remaining in other order.

Hence 2, 3, 5 or 7, 2, 1 or 4, 5, 1 at even places.

$\therefore$ Total possible ways = (4! * 3!) * 4 = 576

 $S=\{\left(\begin{array}{cc}\hfill 1\hfill & \hfill a\hfill \\ \hfill 0\hfill & \hfill b\hfill \end{array}\right),a,b\in 1,2,3,.....100$

$\therefore A=\left(\begin{array}{cc}\hfill -1\hfill & \hfill a\hfill \\ \hfill 0\hfill & \hfill b\hfill \end{array}\right)$ then even power of A as A = $\left(\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right).$

If b = 1 & $a\in \left\{1,2,3.....100\right\}$ and n (n + 1) is always even

$\therefore T_1,T_2,T_3,........,T_n$ are all 1 for b = 1 and each value of a.

$\therefore {\displaystyle \underset{n=1}{\overset{1000}{\cap }}{T}_n=100}$

Circle $\left|z-3\right|\le 1\Rightarrow {\left(x-3\right)}^2+y^2\le 1$

and line $z\left(4+3i\right)+\overline{z}\left(4-3i\right)\le 24$

$\Rightarrow 4x-3y\le 12\therefore slope=tan\theta =\frac{4}{3}$

$?f_{\lambda }\left(x\right)=4\lambda x^3-36\lambda x^2+36x+48$

$f_{\lambda }\left(x\right)=12\left(\lambda x^2-6\lambda x+3\right)$

For increasing $f_{\lambda }\left(x\right)\ge 0$

$f_{\lambda }^{*}\left(x\right)=\frac{4}{3}{x}^3-12x^2+36x+48\therefore f_{\lambda }^{*}\left(1\right)+f_{\lambda }^{*}\left(-1\right)=73\frac{1}{2}-1\frac{1}{2}=72$

 $?-\frac{dx}{dy}=\frac{x^2}{xy-x^2{y}^2-1}$

$\therefore \frac{dy}{dx}=\frac{xy-x^2{y}^2-1}{x^2}$

$Letxy=v\Rightarrow x\frac{dy}{dx}+y=\frac{dv}{dx}$

Put x = 1, y = 1 $\Rightarrow ta{n}^{-1}=c\Rightarrow c=\frac{\pi }{4}$

$\therefore ta{n}^{-1}\left(xy\right)=\mathcal{l}nx=\frac{\pi }{4}$

$e\left(y\left(e\right)\right)=tan\left(1+\frac{\pi }{4}\right)=\frac{tan1+1}{1-tan1}$