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New answer posted

6 months ago

0 Follower 27 Views

A
alok kumar singh

Contributor-Level 10

S? : |z−2|≤1 is circle with centre (2,0) and radius less than equal to 1.
S? : z (1+i)+z? (1−i)≥4
Put z=x+iy
y≤x−2
Solving with S1
⇒x=2−1/√2, y=-1/√2
Point of intersection P= (2−1/√2, −i/√2)
|z−5/2|² = | (2−1/√2)−i (1/√2)−5/2|² = (5√2+4)/4√2 = (5+2√2)/4

New answer posted

6 months ago

0 Follower 7 Views

A
alok kumar singh

Contributor-Level 10

α=max {2? sin³?2? cos³? }
=max {2? sin³? 2? cos³? }=2¹?
β=min {2? sin³?2? cos³? }=2? ¹?
α¹/? +β¹/? = b/8
⇒4+1/4 = b/8
⇒17/4 = b/8 ⇒ b=-34
Again α¹/? β¹/? =c/8
⇒4*1/4 = c/8
⇒c=8
⇒c−b=8+34=42

New answer posted

6 months ago

0 Follower 12 Views

A
alok kumar singh

Contributor-Level 10

A? =B? (i)
A³B²=A²B³. (ii)
Subtract (i) & (ii)
⇒A³ (A²−B²)=B³ (B²−A²)
⇒ (A²−B²) (A³+B³)=0
A²−B² is invertible matrix
∴A²−B²≠0
⇒A³+B³=0
∴? A³+B³? =0

New answer posted

6 months ago

0 Follower 11 Views

A
alok kumar singh

Contributor-Level 10

Co-ordinate of Q (b+2, a)
⇒ 1/√2 + 7i/√2 = (b+2+ai)e^ (iπ/4)
= (b+2+ai) (cos (π/4)+isin (π/4)
⇒ b−a+2=−1
b+2+a=7
⇒a=4
b=1
⇒2a+b=9

New answer posted

6 months ago

0 Follower 6 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

6 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

mean = Σx? f? /Σf? = (32+8α+9β)/ (8+α+β)=6
⇒2α+3β=16 . (i)
d? =x? −x? =−4,0,2,3
f? d? ²=64,0,4α,9β
Variance σ²=Σf? d? ²/Σf? =6.8
⇒ (64+4α+9β)/ (8+α+β)=6.8
⇒2.8α+ (−2.2β)=9.6
⇒28α−22β=96
14α−11β=48 . (ii)
Solving (i) and (ii),
⇒β=2, α=5
New mean=Σx? f? /Σf? =85/15=17/3

New answer posted

6 months ago

0 Follower 29 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

6 months ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

6 months ago

0 Follower 12 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

6 months ago

0 Follower 8 Views

A
alok kumar singh

Contributor-Level 10

f (x)= {sinx, 0≤xπ}
f' (x)= {cosx, 0π}
f' (π/2? ) = 0
f' (π/2? ) = 0
f' (π? ) = 0
f' (π? ) = 0
⇒ f (x) is differentiable in (0, ∞)

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