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New answer posted
2 months agoContributor-Level 10
2x=tan (π/9)+tan (7π/18)
=sin (π/9+7π/18) / cos (π/9)cos (7π/18)
=sin (π/2) / cos (π/9)cos (7π/18)
=1 / cos (π/9)cos (7π/18)
=1 / cos (π/9)sin (π/2−7π/18)
=1 / cos (π/9)sin (π/9)
⇒x=1 / 2cos (π/9)sin (π/9)
=1 / sin (2π/9)=cosec (2π/9)
Again 2y=tan (π/9)+tan (5π/18)
⇒2y=sin (π/9+5π/18) / cos (π/9)cos (5π/18)
=sin (7π/18) / sin (π/2−π/9)sin (π/2−5π/18)
=sin (7π/18) / sin (7π/18)sin (4π/18) = cosec (2π/9)
⇒|x−2y|=0
New answer posted
2 months agoContributor-Level 10
f (x)= {sinx, 0≤x2; 1, /2x 2+cosx, x>π}
f' (x)= {cosx, 0
f' (π/2? ) = 0
f' (π/2? ) = 0
f' (π? ) = 0
f' (π? ) = 0
⇒ f (x) is differentiable in (0, ∞)
New answer posted
2 months agoContributor-Level 10
. xdy - ydx - x² (xdy + ydx) + 3x? dx = 0
⇒ (xdy - ydx)/x² - (xdy + ydx) + 3x²dx = 0 ⇒ d (y/x) - d (xy) + d (x³) = 0
Integrate both side, we get
y/x - xy + x³ = c
Put x = 3, y = 3
⇒ 1 - 9 + 27 = c
c = 19
Put x = 4
y/4 - 4y = 19 - 64
⇒ y = 12
New answer posted
2 months agoContributor-Level 10
for reflexive (x, x)
x³ - 3x²x + 3x³ = 0
. reflexive
For symmetric
(x, y) ∈ R
x³ - 3x²y - xy² + 3y³ = 0
⇒ (x - 3y) (x² - y²) = 0
For (y, x)
(y - 3x) (y² - x²) = 0
⇒ (3x - y) (x² - y²) = 0
Not symmetric
New answer posted
2 months agoContributor-Level 10
(x-a)/3 = (y-b)/ (-4) = (z-c)/12 = -2 (3a-4b+12c+19)/ (3²+ (-4)²+12²)
(x-a)/3 = (y-b)/ (-4) = (z-c)/12 = (-6a+8b-24c-38)/169
(x, y, z) = (a–6, β, γ)
(a-b)-a)/3 = (β-b)/ (-4) = (γ-c)/12 = (-6a+8b-24c-38)/169
(β-b)/ (-4) = -2
=> β = 8+b
=> 3a – 4b + 12c = 150 . (i)
a + b + c = 5
=> 3a + 3b + 3c = 15 . (ii)
Applying (i) – (ii), we get :
= 56 + 216 + 7b – 9c = 56 + 216 – 135 = 137
New answer posted
2 months agoContributor-Level 10
Let the equation of circle be
x (x-1/2) + y² + λy = 0
=> x² + y² - x/2 + λy = 0
Radius = √ (1/16 + λ²/4) = 2
=> λ² = 63/4 => (x-1/4)² + (y+λ/2)² = 4
∴ This circle and parabola
y-α = (x-1/4)² touch each other, so
α = -λ/2 + 2 => α-2 = -λ/2 => (α-2)² = λ²/4 = 63/16
(4α–8)² = 63
New answer posted
2 months agoContributor-Level 10
As slope of line joining (1, 2) and (3, 6) is 2 given diameter is parallel to side
∴ a = √ (3-1)²+ (6-2)² = √20 and
b/ (a/2) = 4/a => b = 8/√5
Area
ab = 2√5 * 8/√5 = 16
New answer posted
2 months agoContributor-Level 10
T? = r/ (2r²)²+1)
= r/ (2r²+1)²- (2r)²)
= (1/4) * (4r)/ (2r²+2r+1) (2r²-2r+1)
S? = 1/4 Σ? ¹? [ 1/ (2r²-2r+1) - 1/ (2r²+2r+1) ]
=> S? = (1/4) * (220/221) = 55/221 = m/n
∴ m+n = 276
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