Mathematics Marathon HSC 09 (1 Viewer)

[star of the sea]

so whats the next question?

 

[boxhunter91]

Ahh these answers take too long to type up
(secx)^2 = sec^2x
.:V=pi Fpi/3-0 (sec^2x) dx
.: V=pi(tanx)pi/3-o
.:V=pi(root3-0)
.:V=piroot3 units cubed.

Differentiate cosx/x

 

[iMAN2]

y=cosx/x
y'=(-xsinx-cosx)/x^2
=-(xsinx+cosx)/x^2

 

[star of the sea]

-xsinx-cosx/x^2

 

[00iCon]

2 ppl answer and no next question???

 

[star of the sea]

the graph of y=f(x) passes through the point (1,3) and f'(x)=3x^2-2 find f(x)

 

[addikaye03]

f'(x)=3x^2-2

f(x)=x^3-2x+C

Sub in point (1,3)

3=1-2+C therefore C=4

f(x)=x^3-2x+4



Question: Factorise x^4-1

 

[star of the sea]

(x-1)(x+1)(x^2+1)

differentiate (1+sinx)^5

 

[addikaye03]

(x-1)(x+1)(x^2+1)

differentiate (1+sinx)^5

d/dx[(1+sinx)^5]=5(1+sinx)^4 x(cosx)

=5cosx(1+sinx)^4

Question(Harder 2u):

Using the fact that: (a-b)^2>=0 where a>b

show that a^2+b^2+c^2>= ab+ac+cb

 

[moshizzle]

Find the values of k if has equal roots.

 

[untouchablecuz]

d/dx[(1+sinx)^5]=5(1+sinx)^4 x(cosx)

=5cosx(1+sinx)^4

Question(Harder 2u):

Using the fact that: (a-b)^2>=0 where a>b

show that a^2+b^2+c^2>= ab+ac+cb

if

(a-b)²≥0 (1)

then similarly

(b-c)²≥0 (2)

(a-c)²≥0 (3)

=> (1)+(2)+(3)

(a-b)²+(b-c)²+(a-c)²≥0

expanding

a²-2ab+b²+b²-2bc+c²+a²-2ac+c²
= 2(a²+b²+c²)-2(ab+bc+ac)
≥ 0

.'. a²+b²+c²≥ab+bc+ac

 

[hermand]

you gotta askkk a question again.

 

[00iCon]

f'(x)=3x^2-2

f(x)=x^3-2x+C

Sub in point (1,3)

3=1-2+C therefore C=4

f(x)=x^3-2x+4

Are u sure?

y=3x^2-2
inverse:
x=3y^2-2

 

[addikaye03]

Find the values of k if has equal roots.

equal roots means a double root so discriminant is=0

∆=b^2-4ac=0

(k^2-2k+1)+4(2k+1)=0

k^2+6k+5=0

Factors: 5,1

(k+5)(k+1)=0

Hence k=-5 or k=-1

 

[untouchablecuz]

Find the values of k if has equal roots.

Δ=0

(k-1)²+4(1)(2k+1)
= k²-2k+1+8k+4
= k²+6k+5
= (k+5)(k+1)

.'. k = -5 or -1

 

[addikaye03]

Are u sure?

y=3x^2-2
inverse:
x=3y^2-2

The question is when you have the gradient of the tangent to the curve and a point, hence to find an expression for f(x) you just integrate and find C.

Should be right

 

[00iCon]

f'(x)=3x^2-2

f(x)=x^3-2x+C

Sub in point (1,3)

3=1-2+C therefore C=4

f(x)=x^3-2x+4

Are u sure?

y=3x^2-2
inverse:
x=3y^2-2
x+2=3y^2
3y=squareroot(x+2)
y=squareroot(x+2)/3
??????

OH FUCK! derivative not inverse!!!!

 

[untouchablecuz]

if your cos(C)

and im the reciprocal of you

what am i

 

[untouchablecuz]

2-6+8-8+32-10+...

find the sum of the first 20 terms of this series

 

[00iCon]

if your cos(C)

and im the reciprocal of you

what am i

sec(C)?

ahahaha i c what u did there!

integrate: [1-sqroot(x)]/[1-x]