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

(Also you can ask your mathematics related questions or queries
we will try to solve them)
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Click to expand...

If I =each annual installment
P = total money borrowed
r = annual compound interest
n=no. of years

Then formula of annual installment is

[I/(1+r/100)]+[I/(1+r/100) ² ]+[I/(1+r/100)³]+-----+[I/(1+r/100)ⁿ ]= P

Here I =882
r = 5
n=2
Using in above formula

[I/(1+r/100)]+[I/(1+r/100) ² ]= P
[882/(1+5/100)]+[882/(1+5/100) ² ]= P

solving we get
P=1640
 
OP
rst

rst

Youngling
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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
we will try to solve them)
----------------------------------------------------------------------------------------------------------------------
 
Niilesh

Niilesh

Padawan
rst said:
It is not necessary
for example
if a=1,b=1,c=1
then a[sup]3[/sup] + b[sup]3[/sup] + c[sup]3[/sup] = 3abc
but a + b + c = 3
Click to expand...

Can you prove that (a[SUP]2[/SUP] + b[SUP]2[/SUP] + c[SUP]2[/SUP] − ab − bc − ca) is non zero? where a,b,c = f(x),x,-1?
i.e. f(x)[SUP]2[/SUP] + x[SUP]2[/SUP] + 1 + f(x) + x -f(x)x ≠ 0
 
OP
rst

rst

Youngling
Niilesh said:
Can you prove that (a[SUP]2[/SUP] + b[SUP]2[/SUP] + c[SUP]2[/SUP] − ab − bc − ca) is non zero? where a,b,c = f(x),x,-1?
i.e. f(x)[SUP]2[/SUP] + x[SUP]2[/SUP] + 1 + f(x) + x -f(x)x ≠ 0
Click to expand...

Yeah
I can prove above thing
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if a[sup]3[/sup] + b[sup]3[/sup] + c[sup]3[/sup] = 3abc and (a[SUP]2[/SUP] + b[SUP]2[/SUP] + c[SUP]2[/SUP] − ab − bc − ca) ≠ 0
then a + b + c = 0

In such case you will get your answer
 
OP
rst

rst

Youngling
Suppose f(x)[sup]2[/sup] + x[sup]2[/sup] + 1 + f(x) + x -f(x)x =0
let f(x)=y
then
y[sup]2[/sup] + x[sup]2[/sup] + 1 + y + x -yx =0
y[sup]2[/sup] + y -yx+ x[sup]2[/sup] + 1+ x =0
y[sup]2[/sup] + y(1 -x)+ x[sup]2[/sup] + 1+ x =0

Here a=1 , b=(1 -x) and c=x[sup]2[/sup] + 1+ x

D=b[sup]2[/sup] -4ac
=(1 -x)[sup]2[/sup] -4(x[sup]2[/sup] + 1+ x )
= -(3 x[sup]2[/sup]+6x+3)
= -3(x[sup]2[/sup]+2x+1)
= -3 (x+1)[sup]2[/sup]
= - ve [as (x+1)[sup]2[/sup] is always positive]

So there is no real value of y [or f(x)] which satisfy this equation

Hence f(x)[sup]2[/sup] + x[sup]2[/sup] + 1 + f(x) + x -f(x)x =0 is not possible for any real fuction f(x)
 
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OP
rst

rst

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

*img694.imageshack.us/img694/6659/principala.png

Plz explain your answer

(Also you can ask your mathematics related questions or queries
we will try to solve them)
----------------------------------------------------------------------------------------------------------------------
Click to expand...

S.I for 2 years =6800 -6080
=720

So S.I for 3 years =(720/2) x 3
=1080

Amount for 3 years =6080

Hence Principal= 6080 - 1080
=5000
 
Niilesh

Niilesh

Padawan
rst said:
Suppose f(x)[sup]2[/sup] + x[sup]2[/sup] + 1 + f(x) + x -f(x)x =0
let f(x)=y
then
y[sup]2[/sup] + x[sup]2[/sup] + 1 + y + x -yx =0
y[sup]2[/sup] + y -yx+ x[sup]2[/sup] + 1+ x =0
y[sup]2[/sup] + y(1 -x)+ x[sup]2[/sup] + 1+ x =0

Here a=1 , b=(1 -x) and c=x[sup]2[/sup] + 1+ x

D=b[sup]2[/sup] -4ac
=(1 -x)[sup]2[/sup] -4(x[sup]2[/sup] + 1+ x )
= -(3 x[sup]2[/sup]+6x+3)
= -3(x[sup]2[/sup]+2x+1)
= -3 (x+1)[sup]2[/sup]
= - ve [as (x+1)[sup]2[/sup] is always positive]

So there is no real value of y [or f(x)] which satisfy this equation

Hence f(x)[sup]2[/sup] + x[sup]2[/sup] + 1 + f(x) + x -f(x)x =0 is not possible for any real fuction f(x)
Click to expand...
:) Thanks
 
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)
----------------------------------------------------------------------------------------------------------------------
 
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)
----------------------------------------------------------------------------------------------------------------------
 
darkv0id

darkv0id

Journeyman
My answer is coming out to be [2*(a^2)*(b^2)]/[sqrt(a^2 + b^2)].

First we find the equation of the radical axis, i.e. S-S'= 0

=>ax -by = 0

Then we find distance of the above line from any one centre, and then use Pyth. Theorem to find the length of the chord.

Alternatively, we can simply find the points of intersection of the common chord with any one circle, then use distance formula to calculate the length.

Please confirm if my answer is correct.
 
OP
rst

rst

Youngling
darkv0id said:
My answer is coming out to be [2*(a^2)*(b^2)]/[sqrt(a^2 + b^2)].

First we find the equation of the radical axis, i.e. S-S'= 0

=>ax -by = 0

Then we find distance of the above line from any one centre, and then use Pyth. Theorem to find the length of the chord.

Alternatively, we can simply find the points of intersection of the common chord with any one circle, then use distance formula to calculate the length.

Please confirm if my answer is correct.
Click to expand...

Yeah
your method is correct
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equation of common chord,S-S'= 0
ax -by = 0

Distance of centre (a,0) from common chord,
d= a[sup]2[/sup] / √ (a[sup]2[/sup]+b[sup]2[/sup])
Radius of circle,r =a

By Pythagoras th,
d[sup]2[/sup] +p[sup]2[/sup] =r[sup]2[/sup]
[a[sup]4[/sup] / (a[sup]2[/sup]+b[sup]2[/sup])]+p[sup]2[/sup] =a[sup]2[/sup]
solving we get
p= ab/√ (a[sup]2[/sup]+b[sup]2[/sup]

Length of chord =2p
=2ab/√ (a[sup]2[/sup]+b[sup]2[/sup])
 
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OP
rst

rst

Youngling
This thread is only for maths questions

For P&C questions ,we have to start new thread
 
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