Maths

$\Delta =p!\left(p+1\right)!\left(p+2\right)!{\Delta }_1=2p!\left(p+1\right);\left(p+2\right)!=2{\left(p!\right)}^3.{\left(p+1\right)}^2.{\left(p+2\right)}^1\to \alpha +\beta =3+1=4$

${\Delta }_1=\left|\begin{array}{ccc}\hfill 1\hfill & \hfill p+1\hfill & \hfill \left(p+2\right)\left(p+1\right)\hfill \\ \hfill 1\hfill & \hfill p+2\hfill & \hfill \left(p+3\right)\left(p+2\right)\hfill \\ \hfill 1\hfill & \hfill p+3\hfill & \hfill \left(p+4\right)\left(p+3\right)\hfill \end{array}\right|=\left|\begin{array}{ccc}\hfill 0\hfill & \hfill -1\hfill & \hfill \left(p+2\right)\left(-2\right)\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill \left(p+3\right)\left(-2\right)\hfill \\ \hfill 1\hfill & \hfill p+3\hfill & \hfill \left(p+4\right)\left(p+3\right)\hfill \end{array}\right|$

= 2(p + 3) – 2 (p + 2) = 2

 $A=\left[\begin{array}{ccc}\hfill a_1\hfill & \hfill a_2\hfill & \hfill a_3\hfill \\ \hfill b_1\hfill & \hfill b_2\hfill & \hfill b_3\hfill \\ \hfill c_1\hfill & \hfill c_2\hfill & \hfill c_3\hfill \end{array}\right],a_2,b_2,c_2\in \left\{0,1\right\}$

S = $a_1+a_2+a_3+b_1+b_2+b_3+c_1+c_2+c_3$ is prime

$0\le s\le 9$

Prime value = 2,3, 5, 7

$ForS=2,1+1+0+0+0+0+0+0+0\to \frac{9!}{2!7!}=36$

Total number of matrices

= 36 + 84 + 126 + 36 = 282

 $T_5{=}^n{C}_4{\left(2^{\frac{1}{4}}\right)}^{n-4}.{\left({\left(\frac{1}{3}\right)}^{\frac{1}{4}}\right)}^4{=}^n{C}_4.2^{\frac{n-4}{4}}.\frac{1}{3}$

$T_6{=}^9{C}_5{\left(2^{\frac{1}{2}}\right)}^4.{\left({\left(\frac{1}{3}\right)}^{\frac{1}{4}}\right)}^5{=}^9{C}_5.2.\frac{1}{3^{5/4}}=\frac{2}{3}.\frac{1}{3^{\frac{1}{4}}}.\frac{9*8*7*6}{4*3*2*1}$

 ${\displaystyle \sum _{r=1}^{\infty }\frac{a+\left(r-1\right)d}{2^r}=4,4\left(a+d\right)=?}$

$\Rightarrow a{\displaystyle \sum _{r=1}^{\infty }{\left(\frac{1}{2}\right)}^r+d{\displaystyle \sum _{r=1}^{\infty }\left(r-1\right){\left(\frac{1}{2}\right)}^r=4}}$

$={\left(\frac{1}{2}\right)}^2{\displaystyle \sum _{k=0}^{\infty }k{\left(\frac{1}{2}\right)}^{k-1}}$

$=\frac{1}{4}.\frac{1}{{\left(1-\frac{1}{2}\right)}^2}$

= 1

a + d = 4

4 (a + d) = 16

(a, 4a, 7). (3, 1, 2b)= 0

3a + 4a – 14b = 0 a – 2b = 0 ……. (i)

(a, 4a, 7) . (b, a, 2) = 0

ab – 4a² + 14 = 0……… (ii)

(i) & (ii) =>2b² – 16b²+ 14 = 0

$\Rightarrow b^2=1\Rightarrow a=2b=\pm 2$

Plane : x – y + z = 0

P (4, 5, 1)

4 + 5 + 1 = 10

 $\overline{x}=15\Rightarrow {\displaystyle \sum _{i=1}^{20}{x}_i=300}$

$\frac{{\displaystyle \sum _{i=1}^{20}{\left(x_i-15\right)}^2}}{20}=9\Rightarrow {\displaystyle \sum _{i=1}^{20}{\left(x_i-15\right)}^2=180}$

${\displaystyle \sum _{i=1}^{20}{\left(x_i+\alpha \right)}^2=178*20=3560}$

$4680+2\alpha \left(300\right)+20{\alpha }^2=3560$

${\alpha }^2+30\alpha +234-178=0$

= 2, 28

${\alpha }_{max}2={\left(-2\right)}^2=4$

 $\frac{7}{2}\left(1+cos2\theta \right)-\frac{3}{2}\left(1-cos2\theta \right)-2co{s}^{2}2\theta =2$

Put cos 2 $\theta$= t

Equation 2t²– 5t = 0, t (2t – 5) = 0

$t=0,\frac{5}{2}$

cos 20 = 0, 0 < 2 < 4

$x_1+x_2=2\left(ta{n}^2\theta +co{t}^2\theta \right)\Rightarrow ?$

$=2\left(1+1\right)+2\left(1+1\right)+2\left(1+1\right)+2\left(1+1\right)=16$

$f\left(x\right)=3^{{\left(x^2-2\right)}^3+4}=81.3^{{\left(x^2-2\right)}^3}$

$f\text{'}\left(x\right)=81.3^{{\left(x^2-2\right)}^3}.\mathcal{l}n3.3{\left(x^2-2\right)}^2.2x$

From graph : P, Q, R

S = {1,2, 3, …., n, 2022}

HCF (n, 2022) = 1

2022 = 2 * 1011 ->3 * 337

2022 = 2 * 3 * 337 (prime factorization)

Let n (A) = no members divisible by 2 = 1011

Let n (B) = no members divisible by 3 = 674

Let n (C) = no members divisible by 337 = 6

$n\left(A\cup B\cup C\right)=n\left(A\right)+n\left(B\right)+n\left(C\right)-n\left(A\cap B\right)-n\left(B\cap C\right)-n\left(C\cap A\right)+n\left(A\cap B\cap C\right)$

= 1011 + 674 + 6 – 337 – 2 – 3 + 1

= 1350

$n\left(A\cap B\right)=337$

$n\left(\left(A\cup B\cup C\right)\text{'}\right)=2022-1350=672$

Prob. $=\frac{672}{2022}=\frac{336}{1011}=\frac{112}{337}$

 $f\left(x\right)=\left|x-1\right|cos\left|x-2\right|sin\left|x-1\right|+\left(x-3\right)\left|x^2-5x+4\right|$

$f\left(x\right)=\left|x-1\right|cos\left(x-2\right)sin\left|x-1\right|+\left(x-3\right)\left|x-1\right|\left|x-4\right|$

x = 1, 4 (doubtful points)

Diff. at x = 1

$\underset{x\to 1}{lim}\frac{f\left(x\right)-f\left(1\right)}{x-1}=\underset{x\to 1}{lim}\frac{\left|x-1\right|\left(sin\left|x-1\right|cos\left(x-2\right)+\left(x-3\right)\left|x-4\right|\right)}{x-1}$

$\begin{array}{l}RHD=\underset{x\to 4^{+}}{lim}\frac{3\left(sin3cos2\right)}{0^{+}}=+\infty \\ LHD=\underset{x\to 4^{-}}{lim}\frac{3\left(sin3cos2\right)}{0^{-}}=-\infty \end{array}\}\Rightarrow$ Not diff. at x = 4