Mr Johari's Interesting question...
(a) Solve the equation 2x^3 - 7x^2 - 7x + 30 = 0
2x^3 - 7x^2 - 7x + 30 = 0
(x + 2)(2x - 5)(x - 3) = 0
x = -2 or 3 or 2.5
(b) When the expression x^2 + bx + c is divided by x-2, the remainder is R. When the expression is divided by x+1, the remainder is also R
(i) Find the value of b
let f(x) = x^2 + bx + c
f(2) = f(-1)
4 + 2b + c = 1 - b + c
b = -1
(ii) When the expression is divided by x-4, the remainder is 2R. Find the value of c and R
f(4) = 2f(2)
16 + 4b + c = 2(2 + 2b + c)
sub b
16 - 4 + c = 2(4 - 2 + c)
c = 8
R = f(2) = (4 - 2 + 8) = 10
(iii) When the equation is divided by x-t, the remainder is 5R. Find the two possible values of t
f(t) = 5 x 10
t^2 + tb + c = 50
t^2 - t + 8 = 50
t = 7 or -6
(c)
the sketch shows part of the graph y = x^3+px^2+qx+r where p, q, and r are constants.
The points A, B, and C have co-ordinates (-2,0), (2,0) and (4,0) respectively.
Find p, q, r
let f(x) = x^3 + px^2 + qx + r
f(-2) = 0
f(2) = 0
f(4) = 0
Therefore: f(x) = (x + 2)(x - 2)(x - 4)
f(x) = x^3 - 4x^2 - 4x + 16
comparing coef.
-4x^2 = px^2
p = -4
-4x = qx
q = -4
r = 16
(a) Solve the equation 2x^3 - 7x^2 - 7x + 30 = 0
2x^3 - 7x^2 - 7x + 30 = 0
(x + 2)(2x - 5)(x - 3) = 0
x = -2 or 3 or 2.5
(b) When the expression x^2 + bx + c is divided by x-2, the remainder is R. When the expression is divided by x+1, the remainder is also R
(i) Find the value of b
let f(x) = x^2 + bx + c
f(2) = f(-1)
4 + 2b + c = 1 - b + c
b = -1
(ii) When the expression is divided by x-4, the remainder is 2R. Find the value of c and R
f(4) = 2f(2)
16 + 4b + c = 2(2 + 2b + c)
sub b
16 - 4 + c = 2(4 - 2 + c)
c = 8
R = f(2) = (4 - 2 + 8) = 10
(iii) When the equation is divided by x-t, the remainder is 5R. Find the two possible values of t
f(t) = 5 x 10
t^2 + tb + c = 50
t^2 - t + 8 = 50
t = 7 or -6
(c)
the sketch shows part of the graph y = x^3+px^2+qx+r where p, q, and r are constants.
The points A, B, and C have co-ordinates (-2,0), (2,0) and (4,0) respectively.
Find p, q, r
let f(x) = x^3 + px^2 + qx + r
f(-2) = 0
f(2) = 0
f(4) = 0
Therefore: f(x) = (x + 2)(x - 2)(x - 4)
f(x) = x^3 - 4x^2 - 4x + 16
comparing coef.
-4x^2 = px^2
p = -4
-4x = qx
q = -4
r = 16
No comments:
Post a Comment