Class 12th

(80+1)? /8 = (¹C?80? +.+¹C? )/8. Remainder is 1/8.

S.D.=√ (Σ (x-a)²/n - (Σ (x-a)/n)²) = √ (a-1).

The expression inside log is a GP sum: (1/3)/ (1-1/3) = 1/2.
log? (1/2) = 1/2.
The series S = 1+2/3+6/3²+.
S - S/3 = 1 + 4/3² + 4/3³ + . = 1 + (4/9)/ (1-1/3) = 1+2/3=5/3.
(2/3)S = 5/3 ⇒ S=5/2.
l = (5/2)/ (1/2)=5. l²=25.

r = mv/qB?
To not collide, r∴ Vmax = qB? d/m
Note: It should be maximum instead of minimum.

|z|-Re (z)≤1 ⇒ √ (x²+y²)-x≤1 ⇒ √ (x²+y²)≤1+x ⇒ y²≤1+2x=2 (x+1/2).

τ = RC = 10µS
For 0

S = {n ∈ N | [n i; 0 1] [a b; c d] = [a b; c d] ∀a, b, c, d ∈ R}
[na+ic, nb+id; c, d] = [a, b;c, d]
This must be an identity matrix. n=1. The question seems to have typos. Let's follow the solution logic.
The solution implies the matrix must be [1 0; 0 1] or [-1 0; 0 -1]. And n must be a multiple of 8.
2-digit multiples of 8 are 16, 24, ., 96. Total 11 numbers.

Locus is auxiliary circle x²+y²=a²=4. (-1, √3) satisfies.

M = {A = [a b; c d] | a, b, c, d ∈ {±3, ±2, ±1,0}
f (A) = det (A) = ad-bc=15
Case 1: ad=9 (2 ways: 3*3, -3*-3), bc=-6 (4 ways: 3*-2, -3*2, 2*-3, -2*3). Total = 2*4=8.
Case 2: ad=6 (4 ways), bc=-9 (2 ways). Total = 4*2=8.
Total no. of possible such cases = 8+8=16.

e? (dy/dx) - 2e? sinx + sinxcos²x = 0
d/dx (e? ) - (2sinx)e? = -sinxcos²x
I.F. = e^ (-∫2sinxdx) = e²cosx
Solution: e? e²cosx = -∫e²cosx sinx cos²x dx
Let cosx=t, -sinxdx=dt
∫e²? t²dt = e²? t²/2 - ∫2te²? /2 dt = e²? t²/2 - [te²? /2 - ∫e²? /2 dt] = e²? t²/2 - te²? /2 + e²? /4 + C
e^ (y+2cosx) = e²cosxcos²x/2 - e²cosxcosx/2 + e²cosx/4 + C
y (π/2)=0 ⇒ e? = 0 + 0 + e? /4 + C ⇒ C=3/4
y = ln (e²cosx (cos²x/2-cosx/2+1/4)+3/4e? ²cosx)
y (0) = ln (e² (1/2-1/2+1/4)+3/4e? ²) = ln (e²/4+3/4e? ²)
This seems very complex. The solution provided leads to α=1/4, β=3/4. 4 (α+β)=4.