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New answer posted

a year ago

0 Follower 21 Views

A
alok kumar singh

Contributor-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

a year ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

f (x)= {sinx, 0? x? }
f' (x)= {cosx, 0

New answer posted

a year ago

0 Follower 13 Views

A
alok kumar singh

Contributor-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

a year ago

0 Follower 14 Views

A
alok kumar singh

Contributor-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

a year ago

0 Follower 6 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

a year ago

0 Follower 15 Views

A
alok kumar singh

Contributor-Level 10

⇒3αβ−2αβ=−1
⇒2αβ=4⇒αβ=2 . (i)
b.c=10
⇒−3α−2β−α=10
⇒4α+2β=−10
⇒2α+β=−5 . (ii)
From (i) and (ii)
α=−1/2, α=−2
β=−4, β=−1
a=i−2j−k
b=3i−2j+2k
c=2i−2j+k
a. (b*c)=9

New answer posted

a year ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

e? dy = e? /α dx
⇒−e? =e? /α+c
y (ln2)=ln2 and y (0)=−ln2
⇒−2=−1/α+c
⇒c=−2−1/α
⇒e? = 1/α e? −2−1/α
⇒−e? ² = 1/α e? ²−2−1/α
⇒2? ¹=3/α
⇒α=2

New answer posted

a year ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

Equation of the ellipse
(x−3)²/a² + (y+4)²/b² = 1
a=2
ae=1⇒e=1/2
⇒b²=3
Equation of tangent
y+4=m (x−3)±√4m²+3
⇒mx−y=4+3m±√4m²+3
⇒3m±√4m²+3=0
⇒9m²=4m²+3
⇒5m²=3

New answer posted

a year ago

0 Follower 9 Views

A
alok kumar singh

Contributor-Level 10

A² = [1 2 3; 0 1 2; 0 1]
A³=A².A= [1 3 6; 0 1 3; 0 1]
A²? = [1 20 1+2+3.20; 0 1 20; 0 1] = [1 20 210; 0 1 20; 0 1]
M= [20 210 520; 0 20 210; 0 20]
M (a? )=T? =n (n+1)/2
S? = 1/2 [ n (n+1) (2n+1)/6 + n (n+1)/2 ]
⇒S? =1540
⇒M=2020

New answer posted

a year ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

∫? ^ (π/2) sin³x e? sin²? dx
=∫? ^ (π/2) (1−cos²x)sinx e? (¹? cos²? )dx
=2∫? ^ (π/2) (1−cos²x)sinx e? (¹? cos²? )dx
Let cos²x=t⇒sin2xdx=−dt
=−2∫? (1−t)e? (¹? ) dt/ (−2cosx)
=1/e ∫? e? dt −∫? te? dt
=1/e [e? ]? − [te? −e? ]?
=2e−3∫? ¹ √t e? dt
⇒α=2, β=3
α+β=5

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