Class 12th

${\displaystyle \underset{0}{\overset{1}{\int }}\left[-8x^2+6x-1\right]dx}$  

Let f (x) = -8x² + 6x – 1

f (0) = -1, f (1) = -3

max f (x) =  $-\frac{D}{4a}=-\frac{36-32}{-32}=\frac{1}{8}$  

= $-\frac{1}{4}-1\left(\frac{3}{4}-\frac{\lambda }{2}\right)-2\left(\frac{\sqrt[]{17}-3}{8}-\frac{3}{4}\right)-3\left(1-\frac{\sqrt[]{17}-3}{8}\right)$  

= $\frac{\sqrt[]{17}-13}{8}$  

|A| = 2

$\left|\left|A\right|adj\left(5adj{A}^3\right)\right|$

= $\left|A{P}^3\right|\left|adj\left(5adj\left(A^3\right)\right)\right|$

$\begin{array}{l}={\left|A\right|}^{15}.5^6\\ =2^{15}*5^6=2^9*10^6\end{array}$

System of equation is

$\left(\begin{array}{ccc}\hfill 2\hfill & \hfill 3\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill \left|\lambda \right|\hfill \end{array}\right)\left(\begin{array}{l}x\\ y\\ z\end{array}\right)=\left[\begin{array}{l}-2\\ 4\\ 4\lambda -4\end{array}\right]$

R₁ – 2 R₂, R₃ – R₂

$\left(\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill -3\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill -2\hfill & \hfill \left|\lambda \right|-1\hfill \end{array}\right)\left(\begin{array}{l}x\\ y\\ z\end{array}\right)=\left(\begin{array}{l}-10\\ 4\\ 4\lambda -8\end{array}\right)$

System of equation will have no solution for = -7.

$f\left(x\right)=[\begin{array}{ccc}\hfill 2n\hfill & \hfill \hfill & \hfill n=2,4,6....\hfill \\ \hfill n-1\hfill & \hfill \hfill & \hfill n=3,7,11,15....\hfill \\ \hfill \frac{n+1}{2}\hfill & \hfill \hfill & \hfill n=1,5,9,13....\hfill \end{array}$

for n = 2, 4, 6 ……

f (n) = 4, 8, 12, ….4 (n) form        

for n = 3, 7, 11, 15, ….

f (n) = 1, 3, 5, 7, …. (4n + 1) or, (4n + 3) from

f is one and onto.

Organic compound → AgBr (s)

  0.5 g                    0.40 g

Here; Moles of Br in organic compound = Moles of Br in Ag Br

= Moles of AgBr

=  $\frac{0.40}{188}mol$

Mass of Br in organic compound =  $\frac{0.4}{188}*80g$  

$\begin{array}{l}\mathrm{\%}ofBr=\frac{0.4*80}{188*0.5}*100\mathrm{\%}\\ =34\mathrm{\%}\end{array}$

$f\left(x\right)=a{x}^2+bx+c$

g (x) = px + q

$f\left(g\left(x\right)\right)=a{\left(px+q\right)}^2+b\left(p\right)\left(+q\right)+c$  

$\Rightarrow 8x^2-2x=a{\left(px+q\right)}^2+b\left(px+q\right)+c$  

Compare 8 = ap² …………… (i)

-2 = a (2pq) + bp

0 = aq² + bq + c

$9\left(f\left(x\right)\right)=p\left(a{x}^2+bx+c\right)+q$  

->4x² + 6x + 1 = apx² + bpx + cp + q

->ap = 4 ……………. (ii)

6 = bp

1 = cp + q

From (i) & (ii), p = 2, q = -1

->b = 3, c = 1, a = 2

f (x) = 2x² + 3x + 1

f (2) = 8 + 6 + 1 = 15

g (x) = 2x – 1

g (2) = 3

$N=M^2+M^4+.....+M^{98}$

$=\left(-{\alpha }^{2}I\right)+{\left(-{\alpha }^{2}I\right)}^2+....+{\left(-{\alpha }^{2}I\right)}^{49}$

$=I\left(-{\alpha }^2+{\alpha }^4-{\alpha }^6+....-{\alpha }^{98}\right)$

N =  $-I\left({\alpha }^2-{\alpha }^4+{\alpha }^6.......+{\alpha }^{98}\right)$

$=-I\frac{{\alpha }^2\left(1-{\left(-{\alpha }^2\right)}^{49}\right)}{1-\left(-{\alpha }^2\right)}$

N =  $\frac{-I{\alpha }^2\left(1+{\alpha }^{98}\right)}{1+{\alpha }^2}$  

Now  $\left(I-m^2\right)N=-2I$

$\left(I+{\alpha }^{2}I\right)\frac{\left(-I{\alpha }^2\cdot (1+{\alpha }^{98}\right)}{1+{\alpha }^2}=-2I$  

-> a¹⁰⁰ + a² = 2

->a = $\pm 1$

Fix the unit place, find the chances for the first three digits

unit digit as 1, total ways = 9.10²

unit digit as 2, total ways = 4.5²

unit digit as 3 total ways = 3.4²

unit digit as 4 total ways = 2.3²

unit digit as 5 total ways = 1.2²

unit digit as 6 total ways = 1.2²

unit digit as 7 total ways = 1.2²

unit digit as 8 total ways = 1.2²

unit digit as 9 total ways = 1.2²

All metal carbonyls have synergic bonds

Rate constant, k = 5.5 * 10^-14 s^-1.

$t=\frac{2.303}{k}log\frac{{\left[R\right]}_0}{\left[R\right]}$  

 = $\frac{2.303}{k}log3-(i)$  

$t_{50\mathrm{\%}}=\frac{2.303}{k}log2-(ii)$  

 From (i) & (ii)

$\frac{t_{67\mathrm{\%}}}{t_{50\mathrm{\%}}}=\frac{log3}{log2}$  

= 1.58 t_50%

So;         t_67% is 15.8 * 10^-1 times half life.

X = 16 (the nearest integer)