Post mathematic related questions here

OP
rst

rst

Youngling
rst said:
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NEW OBJECTIVE QUESTION

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Plz explain your answer

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you can use following mathematic symbols(just do copy and paste)
£ ω ∴ ∂ Φ γ δ μ σ Є Ø Ω ∩ ≈ ≡ ≈ ≅ ≠ ≤ ≥ • ~ ± ∓ ∤ ◅∈ ∉ ⊆ ⊂ ∪ ⊥ ô ∫ Σ → ∞ Π Δ Ψ Γ ∮ ∇∂ √ °α β γ δ ε ζ η θ ι κ λ ν ξ ο π ρ σ τ υ φ χ ψ
x⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁿ
x₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋
Click to expand...

Domain of sin⁻¹x is [-1,1]

So -1≤log₂(x/2) ≤ 1

=> 2⁻¹≤(x/2) ≤ 2¹

=> 1≤ x ≤ 4
 
OP
rst

rst

Youngling
nisargshah95 said:
Couldn't solve this one -
∫ (tan(2x)/√(cos[SUP]6[/SUP]x+sin[SUP]6[/SUP]x))dx
Click to expand...

∫ (tan(2x)/√(cos[SUP]6[/SUP]x+sin[SUP]6[/SUP]x))dx
∫ (sin(2x)/cos2x √(cos[SUP]6[/SUP]x+sin[SUP]6[/SUP]x))dx---------------------------(1)

Now cos[SUP]6[/SUP]x+sin[SUP]6[/SUP]x=(cos² x+sin² x)³ -3cos² xsin² x(cos² x+sin² x)
=1-(3cos² xsin² x)
=1-(3/4)(sin2x)² =1-3/4(1-cos² (2x)
=[1+3cos² (2x)]/4

put cos 2x =t
using in (1) ,we get

∫ dt/t√(1+3t² )

Now it is easy to solve
 
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OP
rst

rst

Youngling
----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION

*img51.imageshack.us/img51/1486/fpt9.png

Plz explain your answer

(Also you can ask your mathematics related questions or queries
we will try to solve them)
----------------------------------------------------------------------------------------------------------------------
you can use following mathematic symbols(just do copy and paste)
£ ω ∴ ∂ Φ γ δ μ σ Є Ø Ω ∩ ≈ ≡ ≈ ≅ ≠ ≤ ≥ • ~ ± ∓ ∤ ◅∈ ∉ ⊆ ⊂ ∪ ⊥ ô ∫ Σ → ∞ Π Δ Ψ Γ ∮ ∇∂ √ °α β γ δ ε ζ η θ ι κ λ ν ξ ο π ρ σ τ υ φ χ ψ
x⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁿ
x₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋
 
Chaitanya

Chaitanya

Cyborg Agent
then there seems 2 be a problem how can 587 be remainder for any possible division of 167 ??

Edit : may be x is negative??

Edit 2 : I am an idiot.. there is no -ve no in option...
 
OP
rst

rst

Youngling
rst said:
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NEW OBJECTIVE QUESTION

*img51.imageshack.us/img51/1486/fpt9.png

Plz explain your answer

(Also you can ask your mathematics related questions or queries
we will try to solve them)
----------------------------------------------------------------------------------------------------------------------
you can use following mathematic symbols(just do copy and paste)
£ ω ∴ ∂ Φ γ δ μ σ Є Ø Ω ∩ ≈ ≡ ≈ ≅ ≠ ≤ ≥ • ~ ± ∓ ∤ ◅∈ ∉ ⊆ ⊂ ∪ ⊥ ô ∫ Σ → ∞ Π Δ Ψ Γ ∮ ∇∂ √ °α β γ δ ε ζ η θ ι κ λ ν ξ ο π ρ σ τ υ φ χ ψ
x⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁿ
x₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋
Click to expand...

a≡ b(mod x) means x divides a-b (or b-a)
In terms of remainder, a≡ b(mod x) means remainder will be same if we divide "a by x" or "b by x"

According to the question,
x divides 385 - 21 =364 and x divides 587 - 167 =420
H.C.F of 364 and 420 is 28

So 28 is the highest number which divides both 364 and 420
 
OP
rst

rst

Youngling
----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION

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Plz explain your answer

(Also you can ask your mathematics related questions or queries
we will try to solve them)
----------------------------------------------------------------------------------------------------------------------
you can use following mathematic symbols(just do copy and paste)
£ ω ∴ ∂ Φ γ δ μ σ Є Ø Ω ∩ ≈ ≡ ≈ ≅ ≠ ≤ ≥ • ~ ± ∓ ∤ ◅∈ ∉ ⊆ ⊂ ∪ ⊥ ô ∫ Σ → ∞ Π Δ Ψ Γ ∮ ∇∂ √ °α β γ δ ε ζ η θ ι κ λ ν ξ ο π ρ σ τ υ φ χ ψ
x⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁿ
x₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋
 
OP
rst

rst

Youngling
rst said:
----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION

*img16.imageshack.us/img16/7392/o2p8.png

Plz explain your answer

(Also you can ask your mathematics related questions or queries
we will try to solve them)
----------------------------------------------------------------------------------------------------------------------
you can use following mathematic symbols(just do copy and paste)
£ ω ∴ ∂ Φ γ δ μ σ Є Ø Ω ∩ ≈ ≡ ≈ ≅ ≠ ≤ ≥ • ~ ± ∓ ∤ ◅∈ ∉ ⊆ ⊂ ∪ ⊥ ô ∫ Σ → ∞ Π Δ Ψ Γ ∮ ∇∂ √ °α β γ δ ε ζ η θ ι κ λ ν ξ ο π ρ σ τ υ φ χ ψ
x⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁿ
x₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋
Click to expand...

Diff. e^(ax) cosbx w.r.t x we get
ae^(ax) cosbx - be^(ax) sinbx
e^(ax)[a cosbx - bsinbx]-----------------(1)

Now r e^(ax) cos[bx+tan⁻¹(b/a)] =r e^(ax) [cosbx sin{tan⁻¹(b/a)} - sinbx cos{tan⁻¹(b/a)} ]
=r e^(ax) [cosbx {a /√ (a² +b²)} - sinbx {b /√ (a² +b²)} ]
=r e^(ax) /√ (a² +b²) [acosbx - bsinbx ]-------------(2)

Comparing (1) and (2)
r=√ (a² +b²)
 
OP
rst

rst

Youngling
----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION

*img33.imageshack.us/img33/8869/vtgv.png

Plz explain your answer

(Also you can ask your mathematics related questions or queries
we will try to solve them)
----------------------------------------------------------------------------------------------------------------------
you can use following mathematic symbols(just do copy and paste)
£ ω ∴ ∂ Φ γ δ μ σ Є Ø Ω ∩ ≈ ≡ ≈ ≅ ≠ ≤ ≥ • ~ ± ∓ ∤ ◅∈ ∉ ⊆ ⊂ ∪ ⊥ ô ∫ Σ → ∞ Π Δ Ψ Γ ∮ ∇∂ √ °α β γ δ ε ζ η θ ι κ λ ν ξ ο π ρ σ τ υ φ χ ψ
x⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁿ
x₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋
 
OP
rst

rst

Youngling
rst said:
----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION

*img33.imageshack.us/img33/8869/vtgv.png

Plz explain your answer

(Also you can ask your mathematics related questions or queries
we will try to solve them)
----------------------------------------------------------------------------------------------------------------------
you can use following mathematic symbols(just do copy and paste)
£ ω ∴ ∂ Φ γ δ μ σ Є Ø Ω ∩ ≈ ≡ ≈ ≅ ≠ ≤ ≥ • ~ ± ∓ ∤ ◅∈ ∉ ⊆ ⊂ ∪ ⊥ ô ∫ Σ → ∞ Π Δ Ψ Γ ∮ ∇∂ √ °α β γ δ ε ζ η θ ι κ λ ν ξ ο π ρ σ τ υ φ χ ψ
x⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁿ
x₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋
Click to expand...

Let f(x)=y,then we have
∫ y sinx cosx dx =[1/ 2(b²-a²)] logy +c

Differentiating both sides we,get
y sinx cosx =[y'/ 2(b²-a²)y]

2y (b²-a²)sinx cosx = y'/ y

2 (b²-a²)sinx cosx =[y'/ y²]

(b²-a²)sin2x dx =[dy/ y²]
integrating we get
-[(b²-a²)cos2x] /2 = -(1/ y )

y=2/[(b²-a²)cos2x]
 
Last edited:
OP
rst

rst

Youngling
----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION

*img441.imageshack.us/img441/6261/cgyq.png

Plz explain your answer

(Also you can ask your mathematics related questions or queries
we will try to solve them)
----------------------------------------------------------------------------------------------------------------------
you can use following mathematic symbols(just do copy and paste)
£ ω ∴ ∂ Φ γ δ μ σ Є Ø Ω ∩ ≈ ≡ ≈ ≅ ≠ ≤ ≥ • ~ ± ∓ ∤ ◅∈ ∉ ⊆ ⊂ ∪ ⊥ ô ∫ Σ → ∞ Π Δ Ψ Γ ∮ ∇∂ √ °α β γ δ ε ζ η θ ι κ λ ν ξ ο π ρ σ τ υ φ χ ψ
x⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁿ
x₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋
 
Niilesh

Niilesh

Padawan
^ cot⁻¹(√2+x/2-x) = cos⁻¹[(√2+x)/2)]
so the original expression becomes (2+x)/4 [ cos(cos⁻¹x) = x ]
so ans is 1/4

or one can always apply chain rule...
 
OP
rst

rst

Youngling
----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION

*img10.imageshack.us/img10/4633/dvr7.png

Plz explain your answer

(Also you can ask your mathematics related questions or queries
we will try to solve them)
----------------------------------------------------------------------------------------------------------------------
you can use following mathematic symbols(just do copy and paste)
£ ω ∴ ∂ Φ γ δ μ σ Є Ø Ω ∩ ≈ ≡ ≈ ≅ ≠ ≤ ≥ • ~ ± ∓ ∤ ◅∈ ∉ ⊆ ⊂ ∪ ⊥ ô ∫ Σ → ∞ Π Δ Ψ Γ ∮ ∇∂ √ °α β γ δ ε ζ η θ ι κ λ ν ξ ο π ρ σ τ υ φ χ ψ
x⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁿ
x₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋
 
OP
rst

rst

Youngling
----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION

*img202.imageshack.us/img202/149/oj3b.png

Plz explain your answer

(Also you can ask your mathematics related questions or queries
we will try to solve them)
----------------------------------------------------------------------------------------------------------------------
you can use following mathematic symbols(just do copy and paste)
£ ω ∴ ∂ Φ γ δ μ σ Є Ø Ω ∩ ≈ ≡ ≈ ≅ ≠ ≤ ≥ • ~ ± ∓ ∤ ◅∈ ∉ ⊆ ⊂ ∪ ⊥ ô ∫ Σ → ∞ Π Δ Ψ Γ ∮ ∇∂ √ °α β γ δ ε ζ η θ ι κ λ ν ξ ο π ρ σ τ υ φ χ ψ
x⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁿ
x₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋
 
OP
rst

rst

Youngling
rst said:
----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION

*img202.imageshack.us/img202/149/oj3b.png

Plz explain your answer

(Also you can ask your mathematics related questions or queries
we will try to solve them)
----------------------------------------------------------------------------------------------------------------------
you can use following mathematic symbols(just do copy and paste)
£ ω ∴ ∂ Φ γ δ μ σ Є Ø Ω ∩ ≈ ≡ ≈ ≅ ≠ ≤ ≥ • ~ ± ∓ ∤ ◅∈ ∉ ⊆ ⊂ ∪ ⊥ ô ∫ Σ → ∞ Π Δ Ψ Γ ∮ ∇∂ √ °α β γ δ ε ζ η θ ι κ λ ν ξ ο π ρ σ τ υ φ χ ψ
x⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁿ
x₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋
Click to expand...

principal value of cos⁻¹ x lies in [0,π ]

principal value of sin⁻¹ x lies in [-π/2,π/2 ]

Intersection of region is [0,π/2 ]

So 0≤ 2cos⁻¹ x ≤ π/2
0≤ cos⁻¹ x ≤ π/4
cos0≥ x ≥ cosπ/4
1≥ x ≥ 1/√2
 
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