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New answer posted

2 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

∫? ^ (π/2) sin³x e? sin²? dx
=∫? ^ (π/2) (1−cos²x)sinx e? (¹? cos²? )dx
=2∫? ^ (π/2) (1−cos²x)sinx e? (¹? cos²? )dx
Let cos²x=t⇒sin2xdx=−dt
=−2∫? (1−t)e? (¹? ) dt/ (−2cosx)
=1/e ∫? e? dt −∫? te? dt
=1/e [e? ]? − [te? −e? ]?
=2e−3∫? ¹ √t e? dt
⇒α=2, β=3
α+β=5

New answer posted

2 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

A= {1,2,3,4,5.100}
B= {4,7,10,13,16,19.}
C= {2,4,6,8,10.}
B−C= {7,13,19, .97}
A∩ (B−C)= {7,13,19, .97}
sum of elements=832

New answer posted

2 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

Direction ratio of line (1, -1, -6)
Equation of line (x−3)/1 = (y+4)/-1 = (z+5)/-6 =k
x=k+3, y=−k−4, z=−6k−5
Solving with plane k=−2
⇒x=1, y=−2, z=7
⇒Distance=√ (3−1)²+6²+3²=√49=7

New answer posted

2 months ago

0 Follower 73 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 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 4 Views

A
alok kumar singh

Contributor-Level 10

f' (x)=g (x)
f" (x)=g' (x)
⇒g (x).g' (x)=f' (x).f" (x)
f (x) has five roots
⇒f' (x) has atleast 4 roots.
And f" (x) has atleast of 3 roots
⇒g (x).g' (x)=0 has atleast 7 roots in (a, b)

New answer posted

2 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

General form Σ? ? C? (1/2)? < 1/2
Σ? ? C? < 2?
⇒? C? +? C? +.? C? < 128
⇒256− (? C? +? C? +.? C? ) < 128
⇒? C? +? C? +.? C? > 128
n≥5

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

 

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