Post mathematic related questions here
darkv0id
Journeyman
Niilesh
Padawan
For competitive exam we can use thisNice thread guys! A good way to recall mathematics
Given f''(x)= -f(x) .., consider f(x)= sin x
and then f(x)= - g(x) .., so g(x)= -sin x limits?
So.., as h(x) = {f(x)^2} + {g(x)^2} = (sinx)^2 + (-sinx)^2 = 2[(sinx)^2]
and given h(5)=3 and req is h(10) i.e h(2x)
So the new eq will be h(2x) = 2(sin2x)^2..
we need to write 2[sin²(2x)] in terms of 2[sin²x] an we will get the ans..
Im hopeless from here..
enough brain excercise for today
Let f(x) = a * sinx (where a ∈ R )
=>h(x) = 2a²sin²x
Now, h(5) = 3
=>3/sin²5 = 2a²
Now h(x) becomes = (3/sin²5)*sin²x
=>h(10) = (3/sin²5) * sin²10 = 3*4*cos²5 = 0.96557
So ans is (D)
rohanz
Journeyman
Seems right but we won't get calculators during the exam :O
Niilesh
Padawan
You know that cos5 should be a little greater than cos(3π/2) so their is no way 3*4*cos²5 will be equal to an integer so.......
Last edited:
OP
rst
Youngling
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NEW OBJECTIVE QUESTION
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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
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NEW OBJECTIVE QUESTION
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Plz explain your answer
(Also you can ask your mathematics related questions or queries
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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
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x₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋
axelzdly1
mR.coNfusEd
Assume the equation to be a+b=0 for a= |2x²-5x+3| and b= x-1
to satisfy a+b=0 there are two conditions.. either < a=b=0 > or < a= -b >
case 1: a= |2x²-5x+3| = 0
=> 2x²-5x+3=0 x=(1.5,1)
case 2: b= x-1 =0 => x=1
common solution i.e, x=1 ∩ x=1,1.5 is 1..
for second condition apply
| 2x²-5x+3 | => ± (2x²-5x+3) = (1-x) and you will get the same result i.e, x= 1
so the solution is 1.., unique solution
Corrections? any other simple methods?
to satisfy a+b=0 there are two conditions.. either < a=b=0 > or < a= -b >
case 1: a= |2x²-5x+3| = 0
=> 2x²-5x+3=0 x=(1.5,1)
case 2: b= x-1 =0 => x=1
common solution i.e, x=1 ∩ x=1,1.5 is 1..
for second condition apply
| 2x²-5x+3 | => ± (2x²-5x+3) = (1-x) and you will get the same result i.e, x= 1
so the solution is 1.., unique solution
Corrections? any other simple methods?
Niilesh
Padawan
Ummm... Sometimes your questions are too easy
Case 1: 2x²-5x+3≥0 =>x∉(1,3/2)
=>2x²-5x+3 + x-1 = 0
=>x = 1,1
Common - x=1
Case 2: 2x²-5x+3≤0 x∈(1,3/2)
=> -2x²+5x-3 + x-1 =0
=> x² -3x + 2 = 0
x= 1,2
Common - x∈Φ
So only solution is 1(only one solution)
Assume the equation to be a+b=0 for a= |2x²-5x+3| and b= x-1
to satisfy a+b=0 there are two conditions.. either < a=b=0 > or < a= -b > There is no need of the first condition, 2nd already covers it
case 1: a= |2x²-5x+3| = 0
=> 2x²-5x+3=0 x=(1.5,1)
case 2: b= x-1 =0 => x=1
common solution i.e, x=1 ∩ x=1,1.5 is 1..
for second condition apply
| 2x²-5x+3 | => ± (2x²-5x+3) = (1-x) and you will get the same result i.e, x= 1
so the solution is 1.., unique solution
Corrections? any other simple methods?
Last edited:
OP
rst
Youngling
Ummm... Sometimes your questions are too easy
Case 1: 2x²-5x+3≥0 =>x∉(1,3/2)
=>2x²-5x+3 + x-1 = 0
=>x = 1,1
Common - x=1
Case 2: 2x²-5x+3≤0 x∈(1,3/2)
=> -2x²+5x-3 + x-1 =0
=> x² -3x + 2 = 0
x= -1,2
Common - x∈Φ
So only solution is 2(only one solution)
How is it -1 ??
OP
rst
Youngling
So only one solution is 1 or 2
Ummm... Sometimes your questions are too easy
Case 1: 2x²-5x+3≥0 =>x∉(1,3/2)
=>2x²-5x+3 + x-1 = 0
=>x = 1,1
Common - x=1
Case 2: 2x²-5x+3≤0 x∈(1,3/2)
=> -2x²+5x-3 + x-1 =0
=> x² -3x + 2 = 0
x= 1,2
Common - x∈Φ
So only solution is 2(only one solution)
I mean only solution is not x=2
as at x=2, 2x²-5x+3=1>0 which contradicts case (2)
OP
rst
Youngling
----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION
*img843.imageshack.us/img843/3268/lmuo.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 ₊ ₋
NEW OBJECTIVE QUESTION
*img843.imageshack.us/img843/3268/lmuo.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
Padawan
I think this was the first question given in our class to tell how to solve this kind of questions
BTW i know two methods i will use the general one
f(x+y) = f(x) . f(y)
Now partially differentiating wrt x we get
f'(x+y) = f'(x) . f(y)
Now at x = 0
f'(y)=2f(y)
=>f'(x)=2f(x)
we can also get f(x) by integration.....
OP
rst
Youngling
I think this was the first question given in our class to tell how to solve this kind of questions
BTW i know two methods i will use the general one
f(x+y) = f(x) . f(y)
Now partially differentiating wrt x we get
f'(x+y) = f'(x) . f(y)
Now at x = 0
f'(y)=2f(y)
=>f'(x)=2f(x)
we can also get f(x) by integration.....
f '(x) represent ordinary diff.
But you used it for partial diff.
the_conqueror
Ambassador of Buzz
Niilesh
Padawan
ohhh... diffrentiation is not taught to us till now on maths so.....
i think he objected to only the use of f'(x) for patial diff.
OP
rst
Youngling
f(x+y)=f(x)f(y)
f(0+0)=f(0)f(0)
=>f(0)=f²(0)
=> f(0)=0,1
but f(x)≠0 for any xЄ R => f(0)=1
Then, f'(x)= lim(h → 0) [f(x+h)-f(x)]/h
f'(x)= lim(h → 0) [f(x)f(h)-f(x)]/h
f'(x)= f(x)lim(h → 0) [f(h)-1]/h --------(1)
putting x=0 in (1),we get
f'(0)= f(0)lim(h → 0) [f(h)-1]/h
f'(0)/ f(0)=lim(h → 0) [f(h)-1]/h
2=lim(h → 0) [f(h)-1]/h--------------------(2)
using (2) in (1),we get
f'(x)= 2f(x)
OP
rst
Youngling
----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION
*img266.imageshack.us/img266/5720/y32v.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 ₊ ₋
NEW OBJECTIVE QUESTION
*img266.imageshack.us/img266/5720/y32v.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 ₊ ₋