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New answer posted

a year ago

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A
alok kumar singh

Contributor-Level 10

 |z−i|=|z+5i| So, z lies on perpendicular bisector of (0, 1) and (0, 5) i.e., line y = 2 as |z| = 2 z = 2i x = 0 and y = 2 so, x + 2y + 4 = 0

New answer posted

a year ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

Let a and b be the roots of the equation  x 2 + ( 3 − a ) x + 1 = 2 a

Therefore a + b = a – 3, ab = 1 – 2a Þ a2 + b2 = (a – 3)2 – 2 (1 – 2a) = a2 – 6a + 9 – 2 + 4a = a2 – 2a + 7 = (a – 1)2 + 6 Þ So,   α 2 + β 2 ≥ 6

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 3 Views

V
Vishal Baghel

Contributor-Level 10

 l=60∫0π2 (sin6x−sin4xsinx+sin4x−sin2xsinx+sin2xsinx)dx

l=60∫0π2 (2cos5x+2cos3x+2cosx)dx

l=60 (25sin5x+23sin3x+2sinx)|0π/2 = 104

New answer posted

a year ago

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V
Vishal Baghel

Contributor-Level 10

X  0 1 2 3

P (X) 16 12 310 130

σ2=Σx2P (x)− (Σ*P (x)2=56100)

100σ2=56

New answer posted

a year ago

0 Follower 60 Views

V
Vishal Baghel

Contributor-Level 10

 |z2|=|z¯|.21−|z|⇒|z|=1

z2=z¯⇒z3=1

∴z=w  or  w2

wn= (1+w)n= (−w2)n

New answer posted

a year ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

 tan? +5tan2? tan? =5? tan2?

tan3? =5

? =n? 3+? 3tan? =5

Five solution.

New answer posted

a year ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

 6312+10 (1311+2310+2239+2338+....+2103)

6312=10311 (611−16−1)

=212.1m.n=12

New answer posted

a year ago

0 Follower 7 Views

V
Vishal Baghel

Contributor-Level 10

 y2=−x2

y=mx−18m

This tangent pass through (2, 0)

m=±14i.e.,   one  tangent  is  x−4y−2=0, 17r=9

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