Maths

Get insights from 6.5k questions on Maths, answered by students, alumni, and experts. You may also ask and answer any question you like about Maths

Follow Ask Question
6.5k

Questions

0

Discussions

14

Active Users

0

Followers

New answer posted

2 months ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

Any point on line (1)
x=α+k
y=1+2k
z=1+3k
Any point on line (2)
x=4+Kβ
y=6+3K
Z=7+3K?
⇒1+2k=6+3K, as the intersect
∴1+3k=7+3K?
⇒K=1, K? =−1
x=α+1; x=4−β
⇒y=3; y=3
z=4; z=4
Equation of plane
x+2y−z=8
⇒α+1+6−4=8 . (i)
and 4−β+6−4=8 . (ii)
Adding (i) and (ii)
α+5−β+12−8=16
α−β+17=24
⇒α−β=7

New answer posted

2 months ago

0 Follower 6 Views

A
alok kumar singh

Contributor-Level 10

Lt? →? x/ (1−sinx)¹/? − (1+sinx)¹/? )
= 2x/ (1−sinx)¹/? − (1+sinx)¹/? ) Multiply by conjugate
= 4x/ (1−sinx)¹/²− (1+sinx)¹/²) Multiply by conjugate
= 8x/ (1−sinx−1−sinx) Multiply by conjugate
= 4x/sinx = −4

New answer posted

2 months ago

0 Follower 6 Views

A
alok kumar singh

Contributor-Level 10

[3¹/²]^ (log? (25? ¹+7)+ [3¹/? ]^ (log? (5? ¹+1) = 180
⇒ 45 (5²? ²+7) / (5? ¹+1) = 180
⇒ (5²? ²+7)/ (5? ¹+1) = 4
Put 5? ¹=t
⇒ (t²+7)/ (t+1) = 4
⇒ t²−4t+3=0
⇒t=1,3
⇒5? ¹=1 or 5? ¹=3
⇒x=1 or x−1=log?3

New answer posted

2 months ago

0 Follower 4 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

2 months ago

0 Follower 5 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

2 months ago

0 Follower 6 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

2 months ago

0 Follower 4 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

2 months ago

0 Follower 6 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

2 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

2 months ago

0 Follower 13 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

Get authentic answers from experts, students and alumni that you won't find anywhere else

Sign Up on Shiksha

On Shiksha, get access to

  • 65k Colleges
  • 1.2k Exams
  • 679k Reviews
  • 1800k Answers

Share Your College Life Experience

×
×

This website uses Cookies and related technologies for the site to function correctly and securely, improve & personalise your browsing experience, analyse traffic, and support our marketing efforts and serve the Core Purpose. By continuing to browse the site, you agree to Privacy Policy and Cookie Policy.