Class 12th

$\alpha =\frac{\Delta l_C}{\Delta l_E}=\frac{3.5}{4}=\frac{7}{8}$

$\beta =\frac{\alpha }{1-\alpha }=\frac{7/8}{1-7/8}=7$

lim (x→1) [f (1)g (x)-f (1)-g (1)f (x)+g (1)] / [f (1)g (x)-f (x)g (1)], form: 0/0
lim (x→1) [f (1)g' (x)-g (1)f' (x)] / [f (1)g' (x)-f' (x)g (1)] = 1

Focus of a spherical convex mirror is in the same side of centre of curvature. Thus, f = + $\frac{1}{2}r.$

? + b? = λ? c? (i)
b? + c? = λ? (ii)
Form (i) – (ii),
? – c? = λ? c? – λ?
(1 + λ? )? = (1 + λ? )c?
? ? and c? are non – collinear
⇒ 1 + λ? = 0, 1 + λ? = 0
λ? = λ? = -1
⇒? + b? + c? = 0

I = ∫ (cos 4x-1)/ (cot x-tan x) dx = ∫ (2 cos² 2x-1-1)/ (cos²x-sin²x)/ (sin x cos x) dx
= 2 ∫ (cos² 2x-1) sin x cos x) / (cos 2x) dx = ∫ (cos² 2x-1) sin 2x) / (cos 2x) dx
Put cos 2x = t
-2 sin 2x dx = dt
I = -1/2 ∫ (t²-1)/t dt = 1/2 ∫ (1/t - t) dt
= 1/2 (ln|t| - t²/2) + c
= 1/2 ln|cos 2x| - 1/4 cos² (2x) + c

| 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