O-Level Additional Math
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Hi can someone help me with this question, thanks!
Given that the line y+2nx-3m=0 is a tangent to the curve y=mx^2+nx+7m, where m and n are positive integers. Find the ratio of m:n and thus find the x-coordinate of the point where the line is tangent to the curve. -
S-H:
y+2nx-3m=0 => y= -2nx+3mHi can someone help me with this question, thanks!
Given that the line y+2nx-3m=0 is a tangent to the curve y=mx^2+nx+7m, where m and n are positive integers. Find the ratio of m:n and thus find the x-coordinate of the point where the line is tangent to the curve.
mx^2+nx+7m = = -2nx+3m
mx^2+3nx+4m = 0
Since the line is the tangent to the curve ,
D = b^2 - 4ac = 0
(3n)^2 - 4(m)(4m) = 0
9n^2 - 16m^2 = 0
(m/n)^2 = 9/16
m/n = 3/4 or m/n = - 3/4 (rej as m,n are positive integers)
m/n = 3/4 => 3n = 4m
mx^2+3nx+4m = 0
mx^2+4mx+4m = 0
m(x^2+4x+4) = 0
x^2+4x+4 = 0
(x+2)^2 = 0
x = -2 -
jieheng:
Thank you very much jieheng!
y+2nx-3m=0 => y= -2nx+3mS-H:
Hi can someone help me with this question, thanks!
Given that the line y+2nx-3m=0 is a tangent to the curve y=mx^2+nx+7m, where m and n are positive integers. Find the ratio of m:n and thus find the x-coordinate of the point where the line is tangent to the curve.
mx^2+nx+7m = = -2nx+3m
mx^2+3nx+4m = 0
Since the line is the tangent to the curve ,
D = b^2 - 4ac = 0
(3n)^2 - 4(m)(4m) = 0
9n^2 - 16m^2 = 0
(m/n)^2 = 9/16
m/n = 3/4 or m/n = - 3/4 (rej as m,n are positive integers)
m/n = 3/4 => 3n = 4m
mx^2+3nx+4m = 0
mx^2+4mx+4m = 0
m(x^2+4x+4) = 0
x^2+4x+4 = 0
(x+2)^2 = 0
x = -2 -
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. -
Vivian22:
Hi Vivian,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.
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. -
alwaysLovely:
Thanks for your reply, but isn't (2x^2)^n the 0th term?
Hi Vivian,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.
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.
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? -
Vivian22:
Thanks for your reply, but isn't (2x^2)^n the 0th term?
Hi Vivian,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.
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.
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 -
Can someone help me solve this Algebra question.
Make (y) the subject of the formulae
(1) x =y/y+1
Make (t) the subjects of the formulae
(2) a = 1-t/1+1
Thanks in advance -
shurley197323:
y=x/(1-x) for the first formula.Can someone help me solve this Algebra question.
Make (y) the subject of the formulae
(1) x =y/y+1
Make (t) the subjects of the formulae
(2) a = 1-t/1+1
Thanks in advance
Full solution can be found at http://mathtuition88.com/2013/06/03/make-y-the-subject-of-the-formulae-o-level-maths-tuition/
Regarding the second formula, I am not sure if it is typed correctly. Putting brackets would help to make the expression clearer.
Sincerely hope it helps!
Best regards, -
mathtuition88.Thanks for you help, But my girl did another way, can u help me to check it’s correct too.
(1) x = y/y+1
x(y+1) = y
xy + x = y
y-xy = x
y(1-x) = x
y = x/1-x
(2) a= 1-t/1+t
a(1+t)=1-t
a+at=1-t
1-a-at = t
t-at=1+a
t(1-a)=1+a
t=1-a/1+a
Thanks in advance