Maths

Let A (α, 0,0), B (0, β, 0), C (0,0, γ), then the centroid is G (α/3, β/3, γ/3) = (1,1,2).
α = 3, β = 3, γ = 6
∴ Equation of plane is x/α + y/β + z/γ = 1
⇒ x/3 + y/3 + z/6 = 1
⇒ 2x + 2y + z = 6
∴ Required line passing through G (1,1,2) and normal to the plane is (x-1)/2 = (y-1)/2 = (z-2)/1.

A = [cosθ, sinθ], [-sinθ, cosθ]
A² = [cos2θ, sin2θ], [-sin2θ, cos2θ]
⇒ A? = [cos4θ, sin4θ], [-sin4θ, cos4θ]
B = [cos4θ, sin4θ], [-sin4θ, cos4θ] + [cosθ, sinθ], [-sinθ, cosθ]
= [cos4θ + cosθ, sin4θ + sinθ], [- (sin4θ + sinθ), cos4θ + cosθ]
det (B) = (cos4θ + cosθ)² + (sin4θ + sinθ)²
= (cos²4θ + sin²4θ) + (cos²θ + sin²θ) + 2 (cos4θcosθ + sin4θsinθ)
= 1 + 1 + 2cos (4θ - θ)
= 2 + 2cos3θ
Given 3θ = 3π/5
|B| = 2 + 2cos (3π/5)
= 2 + 2 (- (√5-1)/4) = 2 - (√5-1)/2 = (4-√5+1)/2 = (5-√5)/2 ∈ (1,2)

Kindly consider the following Image

f' (c) = 1 + lnc = e/ (e-1)
lnc = e/ (e-1) - 1 = (e - (e-1)/ (e-1) = 1/ (e-1)
c = e^ (1/ (e-1)

Equation of line is x/3 + y/1 = 1
⇒ x + 3y - 3 = 0
The image (x? , y? ) of point (-1, -4) is given by:
(x? - (-1)/1 = (y? - (-4)/3 = -2 (1 (-1) + 3 (-4) - 3) / (1² + 3²)
(x? + 1)/1 = (y? + 4)/3 = -2 (-1 - 12 - 3)/10 = -2 (-16)/10 = 16/5
x? + 1 = 16/5 ⇒ x? = 11/5
(y? + 4)/3 = 16/5 ⇒ y? + 4 = 48/5 ⇒ y? = 28/5

y = (2/π x - 1)cosec x
dy/dx = (2/π)cosec x - (2/π x - 1)cosec x cot x
⇒ dy/dx + (2/π x - 1)cosec x cot x = 2/π cosec x
⇒ dy/dx + ycot x = 2/π cosec x
This is a linear differential equation. The integrating factor P (x) is the coefficient of y.
⇒ P (x) = cot x

Applying Rolle's theorem in for function f (x), there exists c such that f' (c) = 0, c ∈ (0,1).
Again applying Rolle's theorem in [0, c] for function f' (x), there exists c? such that f' (c? ) = 0, c? ∈ (0, c).
Option A is correct.

Given equation is 2x (2x + 1) = 1 ⇒ 4x² + 2x - 1 = 0. Roots of the equation are α and β.
∴ α + β = -2/4 = -1/2 ⇒ β = -1/2 - α
and
4α² + 2α - 1 = 0 ⇒ α² = (1-2α)/4 = 1/4 - α/2
Now
α = 1/2 - 2α²
Substituting into the expression for β:
β = -1/2 - (1/2 - 2α²) = -1 + 2α²

f (x) = sin²x (λ + sinx)
f' (x) = 2sinxcosx (λ + sinx) + sin²x (cosx) = sinxcosx (2λ + 3sinx)
For extrema, f' (x) = 0
sinx = 0, cosx = 0, or sinx = -2λ/3
For more than 2 points of extrema in the interval, sinx = -2λ/3 must have solutions other than where sinx=0 or cosx=0.
-1 < -2/3 < 1 and -2/3 0
This gives λ ∈ (-3/2, 3/2) - {0}