0rchid123:

Please help with the following question

When f(x) is divided by x-1 and x+2 the remainders are 4 and -2 respectively. Hence find the remainder when f(x) is divided by X^2+X-2

This is classified as a challenging question on Factor Remainder Theorem for O-Level Additional Maths.

[Basics]
Given:
By remainder theorem
f(1) = 4
f(-2) = -2

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

[Advance]
Do you know?
1) f(x) = Q(x) * Divisor + Remainder
2) The degree of the remainder is always less than that of the divisor. Example, a quadratic divisor (degree = 2) is given in this question, the remainder would have a degree of 1 or 0 so in general, we let remainder be Ax+B

f(x): Polynomial
Q(x): Quotient

f(x) = Q(x) *(x + 2)(x - 1) + (Ax + B)
When x = 1, Thinking process: How do we know what value of x to sub?
f(1) = A + B Notice Q(x) 'disappears'
4 = A + B --- (1)

When x = -2, Thinking process: How do we know what value of x to sub?
f(-2) = -2A + B Notice Q(x) 'disappears'
-2 = -2A + B --- (2)

Values of f(1) and f(-2) are obtained from the first few lines of the solution.

A = 2
B = 2

Hence remainder = 2x + 2 = 2(x + 1)

kiera:

can anyone tell me the solution to this problem sum? :please: http://i61.tinypic.com/wh26g.jpg\">

Isn't this a PSLE Math question?

Thx Jieheng

jieheng:

alwaysLovely:

Hi Jieheng,

Thank you for the clear explanation.

May I ask how do you know the closest point on the line passes through the centre? Is there any property you're using in this case?

http://i44.tinypic.com/16a9ycg.jpg\">

For the point A to be closest to the circle , the distance AB should be the shortest and this happen only when BD is the diameter.

jieheng:

Vivian22:

Hi, can someone please help me with this question?


http://i41.tinypic.com/amt9bc.png\">

a)

Circle C1 , x²+y²-12x-4y+15=0

2g = -12
g = -6

2f = -4
f = -2

the coordinates of the center = (-g,-f) = (6,2)

radius = √(g² +f² -c) = √[(-6)²+(-2)²-15] = √25 = 5 units

b)

Let A(x,y) be the point on the line L which is closest to the circle C1.

To be closest to the circle C1 , the point A has to lie on the Line L2 which is perpendicular to the line L and pass through the centre of the circle C1.

the line L , 2y=3x+12 => y=(3/2)x+6
Gradient of the line L = 3/2

Gradient of the line L2 = -(2/3)

Equation of the line L2 is
(y-2) / (x-6) = -(2/3)
3y-6 = -2x+12
3y= -2x+18

Solving the equations,
2y = 3x+12
3y = -2x+18

x= 0 , y=6

the coordinates of the point A is (0,6)

c)

the circle C2 , centre (a,b) , radius = 3
Given a>b and a<0

x=1 is a tangent to the circle C2 ,

1-a = 3
=> a = -2

the circle C2 touches the x-axis

0-b = 3
=> b = -3

Equation of the circle C2 is

[x-(-2)]² + [y-(-3)]² = 3²
(x+2)² + (y+3)² = 3²

d)

Circle C2 cut the y-axis => x=0

(x+2)² + (y+3)² = 3²
2² + (y+3)² = 3²
(y+3)² = 5
y+3 = ± √5
y = -3 ± √5

P (0, -3+√5) and Q (0, -3-√5)

the exact length PQ = -3+√5 - (-3-√5) = 2√5 units

Hi Jieheng,

Thank you for the clear explanation.

May I ask how do you know the closest point on the line passes through the centre? Is there any property you're using in this case?

S-H:

Can someone help me solve this question? Thanks.


The equation of a curve y=3sin2x
(a) Find the range of values of k such that 3 sin2x=k have 2 solutions for the given domain (- pi/2<equal x <equal pi/2)

You must first sketch the graph y = 3 sin2x

Amplitude = 3
Period = PI
No of cycles (from 0<x<PI/2) = 1/2 (from -PI/2<x<0)=1/2

From the graph,

-3<k<0 or 0<k<3

Explanation:
3 sin2x = k means draw a line y = k on the 3 sin2x graph so that the number of solutions is 2.
y = k is a horizontal line.

http://i44.tinypic.com/21a01ns.png\">

Vivian22:

alwaysLovely:

[quote=\"Vivian22\"]Hi, can someone please help me with these 2 questions?


1. In the expansion of (2x^2 - x^-4)^n, in descending powers of x, if the coefficient of the third term is seven times that of the first term. Find the value of n.

2. A graph, y=5x(e^-x), x>0, has a stationary point at S. FInd the x-coordinate of S and the value of d^2y/dx^2 (second derivative) at S.

Thanks in advance.

Hi Vivian,

1. Binomial expansion & Indices
Coeff refers to the constant (i.e numbers)

1st term: (2x^2)^n
=> coeff: 2^n
3rd term implies r = 2
i.e nC2 (2x^2)^(n-2) (x^-4)^2
=> coeff: nC2*2^(n-2)

Question: what's nC2? There's a formula in the GCE O-Level formula sheet when applied and simplified will give us:
nC2 = n(n-1)/2

2. Differentiation - Max & Min
Stationary point occurs when dy/dx = 0

Step 1: Differentiate y=5x(e^-x), apply product rule or quotient rule if y=5x(e^-x) is simplified to y=5x/(e^x)
Step 2: Set dy/dx = 0
Step 3: Find x. ( I believe it should involve solving of exponential eqn)

To find second derivative, Differentiate (dy/dx). Substitute the value from Step 3 into second derivative.

Thanks for your reply, but isn't (2x^2)^n the 0th term?
For the second question, I got -5x(e^-x) + 5(e^-x) = 0 (is this correct?), but I don't know how to solve this exponential eqn.
Can someone pls help?[/quote](2x^2)^n the 1st term. I don't think there's a term called '0th' term.
-5x(e^-x) + 5(e^-x) = 0: Factorise the common term e^-x to get
(e^-x)(-5x+ 5) = 0
Since e^-x can't be 0, therefore -5x+5 = 0 => x= 1

Vivian22:

Hi, can someone please help me with these 2 questions?


1. In the expansion of (2x^2 - x^-4)^n, in descending powers of x, if the coefficient of the third term is seven times that of the first term. Find the value of n.

2. A graph, y=5x(e^-x), x>0, has a stationary point at S. FInd the x-coordinate of S and the value of d^2y/dx^2 (second derivative) at S.

Thanks in advance.

Hi Vivian,

1. Binomial expansion & Indices
Coeff refers to the constant (i.e numbers)

1st term: (2x^2)^n
=> coeff: 2^n
3rd term implies r = 2
i.e nC2 (2x^2)^(n-2) (x^-4)^2
=> coeff: nC2*2^(n-2)

Question: what's nC2? There's a formula in the GCE O-Level formula sheet when applied and simplified will give us:
nC2 = n(n-1)/2

2. Differentiation - Max & Min
Stationary point occurs when dy/dx = 0

Step 1: Differentiate y=5x(e^-x), apply product rule or quotient rule if y=5x(e^-x) is simplified to y=5x/(e^x)
Step 2: Set dy/dx = 0
Step 3: Find x. ( I believe it should involve solving of exponential eqn)

To find second derivative, Differentiate (dy/dx). Substitute the value from Step 3 into second derivative.

Solving cubic equations is taught in Secondary 3 Additional Maths.


There are three methods:
1) Synthetic division
2) Comparing coefficients
3) Long division

I've two videos on demonstrating how to solve a cubic equation using the first two methods.

I believe this post will be useful for you.
http://www.singaporeolevelmaths.com/2013/04/21/amaths-solve-cubic-equation-by-synthetic-division-or-comparing-coefficients-video/

Cheers!

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