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New answer posted

a year ago

0 Follower 29 Views

V
Vishal Baghel

Contributor-Level 10

The value of (1+sin (2π/9)+icos (2π/9)/ (1+sin (2π/9)-icos (2π/9)³
= (1+cos (5π/18)+isin (5π/18)/ (1+cos (5π/18)-isin (5π/18)³
= (2cos² (5π/36)+2isin (5π/36)cos (5π/36)/ (2cos² (5π/36)-2isin (5π/36)cos (5π/36)³
= (cos (5π/36)+isin (5π/36)/ (cos (5π/36)-isin (5π/36)³
= (e^ (i5π/36)/e^ (-i5π/36)³ = (e^ (i5π/18)³ = e^ (i5π/6) = cos (5π/6)+isin (5π/6)
= -√3/2 + i/2

New answer posted

a year ago

0 Follower 5 Views

V
Vishal Baghel

Contributor-Level 10

Let L be the common normal to parabola
y = x²+7x+2 and line y = 3x-3
⇒ slope of tangent of y=x²+7x+2 at P=3
⇒ dy/dx|for p = 3
⇒ 2x+7=3 ⇒ x=-2 ⇒ y=-8
So P (-2, -8)
Normal at P: x+3y+C=0
⇒ C=26 (satisfies the line)
Normal: x+3y+26=0

New answer posted

a year ago

0 Follower 8 Views

V
Vishal Baghel

Contributor-Level 10

Sol. Let t? denotes r+1th term of (αx? + βx? )¹?
t? = ¹? C? (αx? )¹? (βx? )? = ¹? C? α¹? β? x? ¹?
If t? is independent of x
90-15r=0 ⇒ r=6
This differs from the solution.
Let's follow the solution's powers.
(10-r)/9 - r/6 = 0 ⇒ r=4
maximum value of t? is 10K (given)
⇒ ¹? C? α? β? is maximum
By AM ≥ GM (for positive numbers)
(α³/2+α³/2+β²/2+β²/2)/4 ≥ (α? β? /16)¹/?
⇒ α? β? ≤ 16
So, 10K = ¹? C?16
⇒ K=336

New question posted

a year ago

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New question posted

a year ago

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New question posted

a year ago

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New answer posted

a year ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

x+z=h. x=80cos30, y=80sin30. tan75= (h-y)/z. h=80.

New answer posted

a year ago

0 Follower 8 Views

A
alok kumar singh

Contributor-Level 10

MD²+MC² is minimum when M is the midpoint of projection of CD on AB.

 

New answer posted

a year ago

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A
alok kumar singh

Contributor-Level 10

√3|a+b|+|a-b| ≤

2√ (3+1)=4.

New answer posted

a year ago

0 Follower 27 Views

A
alok kumar singh

Contributor-Level 10

f" (0)=10, f" (0+)=2λ. λ=5.

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