Maths

Critical point of function are x = $\frac{-1}{2},$ 1 and -2 but x = -2 is making zero .

$\therefore nondifferentiableatx=\frac{-1}{2},1$      

$A={\displaystyle \underset{\pi /4}{\overset{5\pi /4}{\int }}\left(sinx-cosx\right)dx={\left[-cosx-sinx\right]}_{\pi /4}^{5\pi /4}}=\left(\frac{1}{\sqrt[]{2}}+\frac{1}{\sqrt[]{2}}\right)-\left(-\frac{1}{\sqrt[]{2}}-\frac{1}{\sqrt[]{2}}\right)$

A =    $2\sqrt[]{2}$

A⁴ = 64

 $\underset{n\to \infty }{lim}{\displaystyle \sum _{r=0}^{n-1}\frac{n}{{\left(n+r\right)}^2}}$

$\underset{n\to \infty }{lim}{\displaystyle \sum _{r=0}^{n-1}\frac{1}{n}.\frac{1}{{\left(1+\frac{r}{n}\right)}^2}={\displaystyle \underset{0}{\overset{1}{\int }}\frac{1}{{\left(1+x\right)}^2}={\left[-\frac{1}{1+x}\right]}_0^1}=-\frac{1}{2}-\left(-1\right)=\frac{1}{2}}$

Number divisible by 3;

(a) sum of the digit must be divisible by 3

(i)  $\boxed{1}\boxed{2}\boxed{3}$  3! = 6

(ii) 1, 3, 5 -> 3! = 6

(iii) 2, 3, 4 -> 3! = 6

(iv) 3, 4, 5 = 3! = 6

Total = 24

(b) Divisible by

$\boxed{}\boxed{}\boxed{5}$  4 * 3 = 12

(c) Now common divisible by both

$\boxed{1}\boxed{3}\boxed{5}$ 2! = 2

For 3, 4  $\boxed{}\boxed{}\boxed{5}$  2! = 2

$\therefore$ Total ways = 24 + 12 – 4 = 32

$A_{3*3}$

$det\left(A\right)=42A=\left[\begin{array}{ccc}\hfill 2a_1\hfill & \hfill 2a_2\hfill & \hfill 2a_3\hfill \\ \hfill 2b_1\hfill & \hfill 2b_2\hfill & \hfill 2b_3\hfill \\ \hfill 2c_1\hfill & \hfill 2c_2\hfill & \hfill 2c_3\hfill \end{array}\right]$

For R₂ →2R₂ + 5R₃

$det\left(B\right)=\left|\begin{array}{ccc}\hfill 2a_1\hfill & \hfill 2a_2\hfill & \hfill 2a_3\hfill \\ \hfill 4b_1+10c_1\hfill & \hfill 4b_2+10c_2\hfill & \hfill 4b_3+10c_3\hfill \\ \hfill 2c_1\hfill & \hfill 2c_2\hfill & \hfill 2c_3\hfill \end{array}\right|=\left|\begin{array}{ccc}\hfill 2a_1\hfill & \hfill 2a_2\hfill & \hfill 2a_3\hfill \\ \hfill 4b_1\hfill & \hfill 4b_2\hfill & \hfill 4b_3\hfill \\ \hfill 2c_1\hfill & \hfill 2c_2\hfill & \hfill 2c_3\hfill \end{array}\right|=16det\left(A\right)=64$

$\frac{a^{a^x}+a^{1-a^x}}{2}\ge \sqrt[]{a^{a^x}.a^{1-a^x}}$ $\left(AM\ge GM\right)$

$\Rightarrow a^{a^x}+a^{1-a^x}\ge 2\sqrt[]{a}$

Equation of given line x – y + 1 = 0 .(i),

equation of per perpendicular line PP' is

-x – y + $\lambda =0$  

As line passing through (3, 5)

$\therefore -3-5+\lambda =0$   

$\lambda =8$  

Equation of line PP' is –x – y + 8 = 0 .(ii)

Solving (i) and (ii)

$\begin{array}{l}-2y=-9\\ y=\frac{9}{2}\end{array}\}Q=\left(\frac{7}{2},\frac{9}{2}\right)$

$y=\frac{9}{2}$

$\therefore x-\frac{9}{2}+1=0$

$P\text{'}=\left(4,4\right)$           

$\frac{5+y_1}{2}=\frac{9}{2}$           

y₁ = 4

P {a person chosen from the group has chest disorder}

$=\frac{160}{400}*0.35+\frac{100}{400}*0.20+\frac{140}{400}*0.10$

$P\left(\frac{Personissmoker\&non-vegetarian}{Personhaschestdisorder}\right)$

$=\frac{\frac{160}{400}*0.35}{\frac{160}{400}*0.35+\frac{100}{400}*0.20+\frac{140}{400}*0.10}$

$=\frac{40*0.7}{40*0.7+25*0.4+35*0.2}=\frac{23}{23+10+7}=\frac{28}{45}$

$x^2-2\left(3k-1\right)x+8k^2-7>0\Rightarrow a>0,\&D<0,$       

a = 1 > 0 and D < 0

4 (3k – 1)² – 4 (8k² – 7) < 0

$\left(9k^2+1-6k-8k^2+7<0\right)$            

$k(k-4)-2(k-4)<0$    

$k\in \left(2,4\right)$           

K = 3