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New answer posted

10 months ago

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V
Vishal Baghel

Contributor-Level 10

While approaching: v = v? (c / (c - vcosθ)
While receding: v = v? (c / (c + vcosθ)

New answer posted

10 months ago

0 Follower 34 Views

A
alok kumar singh

Contributor-Level 10

Required area = 4 [∫? ¹/² 2y²dy + ½*½*1] = 5/6.

 

New answer posted

10 months ago

0 Follower 11 Views

A
alok kumar singh

Contributor-Level 10

α+β=64; αβ=256=2?
(α³/β? )¹/? + (β³/α? )¹/? = (α+β)/ (αβ)? /? = 64/32=2.

New answer posted

10 months ago

0 Follower 4 Views

R
Raj Pandey

Contributor-Level 9

I = ∫ (π/24 to 5π/24) dx/ (1+³√tan2x). Using King's rule.
2I = ∫ (π/24 to 5π/24) dx = 4π/24=π/6. I=π/12.

New answer posted

10 months ago

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R
Raj Pandey

Contributor-Level 9

dy/dx - 1 = xe^ (y-x). Let y-x=t. dt/dx = xe? e? dt=xdx.
-e? = x²/2+C. y (0)=0⇒t=0⇒-1=C.
-e^ (x-y) = x²/2-1. y=x-ln (1-x²/2).
y'=1+x/ (1-x²/2)=0 ⇒ x=-1. Min value at x=-1.
y (-1)=-1-ln (1/2) = -1+ln2. This differs from solution.

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10 months ago

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New answer posted

10 months ago

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R
Raj Pandey

Contributor-Level 9

f' (x) = 12sin³xcosx+30sin²xcosx+12sinxcosx = 3sin2x (2sin²x+5sinx+2) = 3sin2x (2sinx+1) (sinx+2).
In [-π/6, π/2], sinx+2>0. 2sinx+1>0 except at x=-π/6. sin2x>0 for x∈ (0, π/2), <0 for x (-/6,0).
So f' (x)<0 on (-/6,0) (decreasing) and f' (x)>0 on (0, π/2) (increasing).

New answer posted

10 months ago

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A
Aadit Singh Uppal

Contributor-Level 10

Successful collisions help in the formation of new products, whereas in unsuccessful collisions the molecules just bounce back without causing any reaction. In such cases, the reaction fails and no new product or bond is formed.

New answer posted

10 months ago

0 Follower 2 Views

A
Aadit Singh Uppal

Contributor-Level 10

For a collision to happen, the molecules must collide with each other at a high frequency. But the result can still lead to disappointment. This is because even if molecules collide effectively, they also need to be aligned properly in symmetry. Only this can help in the formation of new bonds. Hence, the molecular orientation is considered important in he concept of collision theory.

New answer posted

10 months ago

0 Follower 31 Views

R
Raj Pandey

Contributor-Level 9

Vectors are coplanar. Determinant is zero. Row operations.
This leads to 2a=b+c.

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