Class 11th

Get insights from 8k questions on Class 11th, answered by students, alumni, and experts. You may also ask and answer any question you like about Class 11th

Follow Ask Question
8k

Questions

0

Discussions

38

Active Users

0

Followers

New answer posted

2 months ago

0 Follower 72 Views

A
alok kumar singh

Contributor-Level 10

N=2¹? 5¹? 11¹¹ 13¹¹
5→4n+1 type→number of choice=11
11→4n+3 type→number of choice=6
13→4n+1 type→number of choice=12
. Number of divisor of 4n+1 type=11*6*12=792

New answer posted

2 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Z = (3+2icosθ)/ (1-3icosθ) = (3−6cos²θ)+i (11cosθ) / (1+9cos²θ)
Real part = 0
⇒3−6cos²θ=0
⇒θ=45?
⇒sin²3θ+cos²θ=1/2+1/2=1

New answer posted

2 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

A * A = A A s i n θ n ^

= A A s i n 0 ° n ^

= 0 [since Angle between the vectors are zero degree]

A * A = 0

New answer posted

2 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

e? +e³? −4e²? −e? +1=0
Divide by e²?
⇒ (e²? +e? ²? )− (e? +e? )−4=0
Put e? +e? =t>0
e²? +e? ²? +2=t²
t²−2−t−4=0
⇒t²−t−6=0
⇒t=3, t=−2 but t≠−2
⇒t=3
⇒e? +e? =3
Number of solution=2

New answer posted

2 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

p: there exist M>0
Such that x≥M for all x∈S
Obviously ~p: M>0 such that x. Negation of 'there exists' is 'for all'.

New answer posted

2 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

2 months ago

0 Follower 4 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

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.