Maths

f (x) = x? – 4x + 1 = 0
f' (x) = 4x³ – 4
= 4 (x–1) (x²+1+x)
=> Two solution

lim? (x→7) (18- [1-x])/ ( [x-3a])
exist & a∈I.
= lim? (x→7) (17- [-x])/ ( [x]-3a)
exist
RHL = lim? (x→7? ) (17- [-x])/ ( [x]-3a) = 25/ (7-3a) [a ≠ 7/3]
LHL = lim? (x→7? ) (17- [-x])/ ( [x]-3a) = 24/ (6-3a) [a ≠ 2]
LHL = RHL
25/ (7-3a) = 8/ (2-a)
∴ a = -6

dy/dx = (ax-by+a)/ (bx+cy+a)
=> bxdy + cydy + ady = axdx – bydx + adx
cy²/2 + ay – ax²/2 – ax + bxy = k
ax² + ay² + 2ax – 2ay = k
=> x² + y² + 2x – 2y = λ
Short distance of (11,6)
= √12²+5² – 5
= 13 – 5
= 8

x = Σ a? = 1/ (1-a); y = Σ b? = 1/ (1-b); z = Σ c? = 1/ (1-c)

Now,
a, b, c → AP
1-a, 1-b, 1-c → AP
1/ (1-a), 1/ (1-b), 1/ (1-c) → HP
x, y, z → HP

x + 2y + z = 2
αx + 3y – z = α
–αx + y + 2z = –α

Δ = | (1, 2, 1), (α, 3, -1), (-α, 1, 2) | = 1 (6+1) – 2 (2α–α) + 1 (α+3α) = 7+2α
α = –7/2

z? = iz²
Let z = x + iy
x – iy = I (x² – y² + 2xiy)

Case-I
x = 0
–y² = –y
y = 0, 1

Case - II
y = – 1/2
=> x² – 1/4 = 1/2 => x = ±√3/2

Area of polygon
= 1/2 | (0, 1, 1), (√3/2, -1/2, 1), (-√3/2, -1/2, 1) |
= 1/2 | -√3/2 - √3/2 | = 3√3/4

Kindly consider the following Image 

(P? ¹AP - I)²
= (P? ¹AP - I) (P? ¹AP - I)
= P? ¹A (PP? ¹)AP - P? ¹AP - P? ¹AP + I
= P? ¹A²P - 2P? ¹AP + I
= P? ¹ (A² - 2A + I)P = P? ¹ (A - I)²P
| (P? ¹AP - I)²| = |P? ¹ (A - I)²P| = |P? ¹| | (A - I)²| |P| = | (A - I)²| = |A - I|²
A - I = [1, 7, w²], [-1, w², 1], [0, -w, -w]
|A - I| = 1 (-w³ + w) - 7 (w) + w² (w) = -w³ + w - 7w + w³ = -6w.
|A - I|² = (-6w)² = 36w².

Let A = [a? ]? Sum of diagonal elements of A.A? is Tr (A.A? ) = ∑? ∑? a? ² = 9.
where each a? ∈ {0, 1, 2, 3}.
Case I: One of a? = 3 and rest are 0. (3²=9). There are? C? = 9 ways.
Case II: Two of a? are 2, one is 1, and rest are 0. (2² + 2² + 1² = 9). There are? C? *? C? = 36 * 7 = 252 ways.
Case III: One of a? = 2, five are 1, and rest are 0. (2² + 1²+1²+1²+1²+1² = 9). There are? C? *? C? = 9 * 56 = 504 ways.
Case IV: All nine a? = 1. (1² * 9 = 9). There is 1 way.
Total = 9 + 252 + 504 + 1 = 766.