Class 12th

| 1 -1 -1 |
| 1 -k | = 0
| k 2 1 |

⇒ 1 (1 + 2k) + 1 (1 + k²) – 1 (2 – k) = 0
2k + 1 + 1 + k² − 2 + k = 0
k² + 3k = 0
k = 0, -3

A = [1]
[0 1]
Now A² = [1] [1] = [1 2]
[0 1] [0 1] [0 1]
A³ = A².A = [1 2] [1] = [1 3]
[0 1] [0 1] [0 1]
Similarly A²? ¹¹ = [1 2011]
[0 1]

If charge (-Q) at origin is replaced by (+Q), then electric field at the centre of the cube is zero. Thus, electric field at the centre of the cube is as if only (-2Q) charge is present at the origin.

$\overrightarrow{E}=\frac{1}{4\pi {\epsilon }_0}\frac{\left(-2Q\right)}{{\left(\frac{\sqrt[]{3}}{2}a\right)}^2}.\frac{1}{\sqrt[]{3}}\left(\widehat{x}+\widehat{y}+\widehat{z}\right)=\frac{|-2Q}{3\sqrt[]{3}\pi {\epsilon }_0{a}^2}\left(\widehat{x}+\widehat{y}+\widehat{z}\right)$

Length of normal
k = y √ (1 + (dy/dx)²)

⇒ ∫ (y dy) / √ (k² - y²) = ∫ dx
Let k² – y² = t²
-2y dy = 2t dt
-√ (k² - y²) = x + c
It passes through (0, k)
x² + y² = k²

| 1+x² |
| x 1+x | = − (x - α? ) (x − α? ) (x – α? ) (x – α? )
| x² x 1+x |

Put x = 0
| 1 0 |
| 0 1 0 | = - (-α? ) (-α? ) (-α? ) (-α? )
| 0 1 |
α ⇒? α? α? α? = −1

$\frac{dq}{dt}=t=20t+8t^2$

$\Rightarrow {\displaystyle \underset{0}{\overset{q}{\int }}dq={\displaystyle \underset{0}{\overset{15}{\int }}\left(20t+8t^2\right)dt}}$

$\Rightarrow q=20*\frac{15^2}{2}+8.\frac{15^3}{3}=11250C$  

I = ∫? ² [log? x] dx
By using graph
I = ∫? 0 dx + ∫? ² 1 dx
I = 0 + e² - e

A : Series limit of Lyman series

B : Third line of Balmer series

C : Second line of Paschen series

When connected in series, equivalent capacitance,

$C_1=\frac{C*C}{C+C}=\frac{C}{2}$          

When connected in parallel, equivalent capacitance

C₂ = C + C = 2C

$\frac{C_1}{C_2}=\frac{C/2}{2C}=\frac{1}{4}$            

$i=\frac{6-4}{2+8}=0.2A$

$V_x+4+0.2*8=V_Y$

$\Rightarrow V_Y-V_X=5.6V$