Maths

Note D = |3, 4, 5; 1, 2, 3; 4, 4| = 0 (R? → R? - 2R? + 3R? )
Now let P? = 4x+4y+4z-δ=0. If the system has solutions it will have infinite solution, so P? = αP? + βP?
Hence 3α+β=4 and 4α+2β=4 ⇒ α=2 and β=-2
So for infinite solution 2µ-2=δ ⇒ for 2µ ≠ δ+2 System inconsistent

I = ∫ (cosθ / (2sin²θ + 7sinθ + 3) dθ
sinθ = t ⇒ cosθdθ = dt
= (1/2) ∫ (1 / (t² + (7/2)t + 3/2) dt
= (1/2) ∫ (1 / (t+7/4)² - (5/4)²) dt
= (1/5) ln | (2t+1)/ (t+3)| + c
= (1/5) ln | (2sinθ+1)/ (sinθ+3)| + C
So A = 1/5
B (θ) = 5 (2sinθ+1)/ (sinθ+3)

y = |x| (x-1)
= { 0,   0 ≤ x < 1
{ x-1, 1 ≤ x < 2
Area = ∫? ² 2√x dx - ∫? ² (1) (1)
= [ (4x³/²)/3]? ² - 1/2 = (8√2)/3 - 1/2

Let a, ar, ar² . G.P.
T? + T? + T? = 3 ⇒ ar (1+r+r²) = 3
T? + T? + T? = 243 ⇒ ar? (1+r+r²) = 243
by (i) and (ii)
r? = 81 ⇒ r=3
∴ a = 1/13
S? = a (r? -1)/ (r-1) = (3? -1)/26

Let x = tanθ
y? = tan? ¹ (secθ-1)/tanθ) = tan? ¹ (tan (θ/2) = θ/2 = (1/2)tan? ¹x
x = sinφ, y? = tan? ¹ (2sinφcosφ)/cos2φ)
= tan? ¹ (tan2φ) = 2φ = 2sin? ¹x
dy? /dy? = (dy? /dx)/ (dy? /dx) = (1/2) (1/ (1+x²) / (2/√ (1-x²)
= (√ (1-x²) / (4 (1+x²) = (√ (1-1/4) / (4 (1+1/4) = √3 / 10

e? y'x? + 4x³e? + 2y' / (2√ (y+1) = 0 at (1,0)
y' + 4 + y' = 0 ⇒ y' = -2
equation of tangent at (1,0) is 2x + y - 2 = 0
So option (C) is correct.

Kindly consider the following Image

L = sin (3π/16)sin (-π/16)
= (1/2) (cos (π/4) - cos (π/8)
= (1/2) (1/√2 - cos (π/8)
M = cos (3π/16)cos (-π/16)
= (1/2) (cos (π/4) + cos (π/8)
= (1/2) (1/√2 + cos (π/8)

dy/dx + 2tanx · y = 2sinx
I.F. = e^ (∫2tanxdx) = sec²x
Solution is y·sec²x = ∫2sinx·sec²xdx + C
ysec²x = 2secx + C
0 = 2·2 + c ⇒ c = -4
ysec²x = 2secx - 4
y (π/4) = √2 - 2

Line are coplanar
so | [α, 5-α, 1], [2, -1, 1], | = 0
−5α + (α – 5)3 + 7 = 0
-2α = 8 ⇒ α = −4
⇒ L? : (x+2)/-4 = (y+1)/9 = (z+1)/1
Now by cross checking option (A) is correct.