Post mathematic related questions here

[rst]

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NEW OBJECTIVE QUESTION

*img827.imageshack.us/img827/8467/rangeb.png

Plz explain your answer

(Also you can ask your mathematics related questions or queries
we will try to solve them)
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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 ₊ ₋

Let f(x)=y
Then y= 9^x-3^x+1
y= 3^2x-3^x+1

Put 3^x =u
Then y= u²-u+1
y= u²-u+(1/4)+1-(1/4) {for making it perfect square}
y= [u-(1/2)]²+ (3/4)
As [u-(1/2)]² can't be negative,so its least value is 0
Hence y≥ (3/4)

Ans is option (4)

 

[rst]

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NEW OBJECTIVE QUESTION

*img42.imageshack.us/img42/9306/z0l.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]

^ using all the conditions(for log to be defined) from the left we will get
18x-x²-77 >3
x²-18x+8*10< 0
So 8< x <10

 

[rst]

^ using all the conditions(for log to be defined) from the left we will get
18x-x²-77 >3
x²-18x+8*10< 0
So 8< x <10

Great explanation
Thanks a lot, Niilesh

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NEW OBJECTIVE QUESTION

*img42.imageshack.us/img42/9306/z0l.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 ₊ ₋

log_4(log_5(log_3(18x-x^2-77))) is defined if log_5(log_3(18x-x^2-77))>0

log_3(18x-x^2-77)>5^0
log_3(18x-x^2-77)>1
(18x-x^2-77)>3^1
18x-x²-77 >3

 

[rst]

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NEW OBJECTIVE QUESTION

*img4.imageshack.us/img4/7743/h3x.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]

^
We know that always in the above mentioned functions
f(g(x)) = x
Differentiating both sides wrt x
f'(g(x)) * g'(x) = 1
g'(x)=1/f'(g(x))
Now since f'(x) = 1/1+x³
So, g'(x)= 1+(g(x))³

 

[rst]

^
We know that always in the above mentioned functions
f(g(x)) = x
Differentiating both sides wrt x
f'(g(x)) * g'(x) = 1
g'(x)=1/f'(g(x))
Now since f'(x) = 1/1+x³
So, g'(x)= 1+(g(x))³

Nice method

Can you prove f(g(x)) = x if g is inverse function of f

 

[Niilesh]

umm....
i can explain it in words but it would be better to prove it mathematically. Give me any relation of a function and its inverse that you know(if any) and i will try to prove it .

 

[rst]

umm....
i can explain it in words but it would be better to prove it mathematically. Give me any relation of a function and its inverse that you know(if any) and i will try to prove it .

If g is inverse function of f
Then f(x)=y and g(y)=x

So g(f(x))=g(y)=x
---------------------------------------------------

If g is inverse function of f

Then f(x)=y and g(y)=x

So f(g(y))=f(x)=y



or f(g(x))=x

 

[Niilesh]

If g is inverse function of f
Then f(x)=y and g(y)=x not valid always

So g(f(x))=g(y)=x not valid always
---------------------------------------------------

If g is inverse function of f

Then f(x)=y and g(y)=x Not valid always

So f(g(y))=f(x)=y Valid always



or f(g(x))=x Valid always

Reason - If f(x) is not one-one then there will be more than one value of g(f(x)) but since g(x) is a function we take a interval in f(x) in which it is bijective

 

[rst]

Reason - If f(x) is not one-one then there will be more than one value of g(f(x)) but since g(x) is a function we take a interval in f(x) in which it is bijective

g is inverse of function f
So f is invertible function

Inverse exists only for bijective function
Hence f is bijective (one -one and onto)
So f is always one one

 

[Niilesh]

g is inverse of function f
So f is invertible function

Inverse exists only for bijective function
Hence f is bijective (one -one and onto)
So f is always one one

Technically that is right but sometimes we just make non bijective function( f(x) ) bijective by adjusting their domain and co-domain so that it becomes invertible and find its inverse( g(x) ) but when we refer to f(x) we mean the original function
Example - Sin x and Sin⁻¹x

Sin(Sin⁻¹(x)) = x

Sin⁻¹(Sin (x)) = many different values including 'x' depending upon its value

*www.goiit.com/upload/2009/7/18/ab59e6078e1f4a31a4c0e6e644cde7f7_1408226.gif

 

[rst]

Technically that is right but sometimes we just make non bijective function( f(x) ) bijective by adjusting their domain and co-domain so that it becomes invertible and find its inverse( g(x) ) but when we refer to f(x) we mean the original function
Example - Sin x and Sin⁻¹x

Sin(Sin⁻¹(x)) = x

Sin⁻¹(Sin (x)) = many different values including 'x' depending upon its value

*www.goiit.com/upload/2009/7/18/ab59e6078e1f4a31a4c0e6e644cde7f7_1408226.gif

For sine function
domain = R
range=[-1,1]
So sine function is many one function
That why sine is not invertible function

sinx becomes invertible if its domain is [- π/2, π/2] {its range is [-1,1]}
For Sin⁻¹(x) domain becomes [-1,1] and range [- π/2, π/2]

now Sin⁻¹(Sin (x)) =x (always)

 

[Niilesh]

For sine function
domain = R
range=[-1,1]
So sine function is many one function
That why sine is not invertible function

sinx becomes invertible if its domain is [- π/2, π/2] {its range is [-1,1]}
For Sin⁻¹(x) domain becomes [-1,1] and range [- π/2, π/2]

now Sin⁻¹(Sin (x)) =x (always)

Try to read and understand my post again -

Technically that is right but sometimes we just make non bijective function( f(x) ) bijective by adjusting their domain and co-domain so that it becomes invertible and find its inverse( g(x) ) but generally when we refer to f(x) we mean the original function

 

[rst]

Try to read and understand my post again -

I understand your point

But as f is invertible function and g is its inverse
so f(x)=y and g(y)=x will always be true (according to the given hypothesis)

Sin⁻¹(Sin (x)) = many different values including 'x' depending upon its value
Above statement is true in general
But in the above question or proof we are talking about invertible function
So we have to put values in interval [- π/2, π/2]
otherwise it will not be invertible (this will be contradiction to our hypothesis]

 

[rst]

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

*img515.imageshack.us/img515/5681/cxn.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]

I understand your point

But as f is invertible function and g is its inverse
so f(x)=y and g(y)=x will always be true (according to the given hypothesis)

Sin⁻¹(Sin (x)) = many different values including 'x' depending upon its value
Above statement is true in general
But in the above question or proof we are talking about invertible function
So we have to put values in interval [- π/2, π/2]
otherwise it will not be invertible (this will be contradiction to our hypothesis]

Fair enough
BTW sin x will not be only invertible in [- π/2, π/2]

 

[axelzdly1]

Nice 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

 

[darkv0id]

^ Such a solution is okay for competitive exam purposes, but how do you "assume" that f(x) = sinx ?

EDIT: I think you read the question wrong... h(x)= (f(x)^2 + g(x))^2 and not f(x)^2 + g(x)^2

 

[rohanz]

Nice 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

Can't assume x to be sinx
Sin x comes to be (3/2)^1/2 which can never be true. (greater than 1)