Maths

I = ∫? ² (x³+|x|)/ (e|x|+1) dx . (i)
I = ∫? ² (x³+|x|)/ (e? |x|+1) dx . (ii)
= ∫? ² |x| dx = 2∫? ² x dx
= [x²/2]? ² = (16/4 + 4/2) - 0
= 4+2=6

I = ∫ (e? (x²+1)/ (x+1)² dx = f (x)e? + c
I = ∫ (e? (x²-1+1+1)/ (x+1)² dx
I = ∫e? [ (x-1)/ (x+1) + 2/ (x+1)² ] dx
for x = 1
f' (1) = 12/24 - 12/16 = 3/4

cos? ¹ (y/2) = log? (x/5)? , |y| < 2
Differentiating on both side
(1/√ (1- (y/2)²) * (-y'/2) = 5 * (5/x) * (1/5)
(-xy')/ (2√ (1- (y²/4) = 5
Square on both side
(x²y'²)/4 = 25 * (4-y²)/4
Diff on both side
xy' + y'x² + 25y = 0

CD = √ (10+x²)² – (10–x²)² = 2√10|x|
Area
= 1/2 * CD * AB = 1/2 * 2√10|x| (20–2x²)
=> 10 – x² = 2x
3x² = 10
x = k
3k² = 10

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