@darkvoid.. h(x)= ( f ( x )² + g( x ) )² not ( f ( x )² ) + ( g ( x )² ) ??
aren't there 4 parenthesis around each of the f,g functions..?
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
For competitive exam we can use this
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)
Seems right but we won't get calculators during the exam :O
Ummm... Sometimes your questions are too easy----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION
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 ₊ ₋
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?
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)
It's 1, typoHow is it -1 ??
So only one solution is 1 or 2It's 1, typo
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)
wth happened to meSo only one solution is 1 or 2
I mean only solution is not x=2
as at x=2, 2x²-5x+3=1>0 which contradicts case (2)
I think this was the first question given in our class to tell how to solve this kind of questions----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION
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 ₊ ₋
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.
ohhh... diffrentiation is not taught to us till now on maths so.....f '(x) represent ordinary diff.
But you used it for partial diff.
i think he objected to only the use of f'(x) for patial diff.No, he did it correctly AFAIK.
----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION
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 ₊ ₋