Class 12th
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New answer posted
a year agoContributor-Level 9
f(3)=f(4) ⇒ α=12
f'(x) = (x²-12)/(x(x²+12))
∴ f'(c)=0 ⇒ c=√12
∴ f''(c) = 1/12
New answer posted
a year agoContributor-Level 10
2(cos²θ/sin²θ) - 5/sinθ + 4 = 0
(2sinθ - 1)(sinθ - 2) = 0
sinθ = 1/2 only
∴ θ = π/6, 5π/6
↓↓
θ?, θ?
∫(π/6 to 5π/6) cos²3θdθ = ∫(π/6 to 5π/6) (1+cos6θ)/2 dθ = π/3
New answer posted
a year agoContributor-Level 10
6.00
b·a = c·a
|a+b-c|² = |a|²+|b|²+|c|²+2(a·b - b·c - a·c)
= 4+16+16+2(a·b - 0 - a·b) = 36
⇒ |a+b-c| = 6
New answer posted
a year agoContributor-Level 10
Yes, the phase of the reactant (solid, liquid or gas) actually affects the rate of a reaction. Gases and Liquids tend to move more freely and effectively as compared to solids, which increaes the frequency of particles colliding with each other.
New answer posted
a year agoContributor-Level 10
0.50
y = x²-3x+2, x+y=a, x-y=b
x?=2 x?=1
y? = 4-6+2 = 0 y?=0
(2,0)
(1,0)
b=2
a=1
∴ a/b = 1/2 = 0.5
New answer posted
a year agoContributor-Level 10
This is possible because these catalysts have a lower activation energy, which allows a comparatively higher number of particles to react at a faster rate. More number of particles will mean more frequent collisions which speeds up the process.
New answer posted
a year agoContributor-Level 10
Probability (at most two machines will be out of service) = (3/4)³ . k
⇒ ?C?(1/4)?(3/4)? + ?C?(1/4)¹(3/4)? + ?C?(1/4)²(3/4)³ = (3/4)³ . k
⇒ 17/8 (3/4)³ = (3/4)³ . k
⇒ k = 17/8
New answer posted
a year agoContributor-Level 10
f(x) = (3x²+ax-2-a)e?
f'(x) = (3x²+ax-2-a)e? + e?(6x+a)
= e?(3x²+(a+6)x-2)
∴ x=1 is a critical point
∴ f'(1)=0
∴ 3+a+6-2=0
a = -7
∴ f'(x) = e?(3x²-x-2)
= e?(3x+2)(x-1)
∴ maxima at x = -2/3
∴ minima at x = 1
New answer posted
a year agoContributor-Level 9
y = (8²?-8?²?)/(8²?+8?²?) ⇒ (1+y)/(1-y) = 8? ⇒ 4x = log?((1+y)/(1-y))
x = (1/4)log?((1+y)/(1-y)), f?¹(x) = (1/4)log?((1+x)/(1-x))
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