P1 mistakes + theory [IAL]

pic shows discriminant way, here is easier way:

f(x) > 2, so min x value > 2

dy/dx = 2x - 6 = 0 (for minimum)
x = 3
plug in x =3 to f(x), it will be:
-9 + c > 2
c > 11

(a) The cubic curve touches the x-axis at $(1, 0)$, which means $(x-1)^2$ is a factor. It also passes through $(4, 0)$, so $(x-4)$ is another factor. Therefore, the equation is of the form $f(x) = A(x-1)^2(x-4)$. We can find $A$ using the point $(0, 3)$

$3 = A(0-1)^2(0-4)$
$3 = A(1)(-4)$
$A = -\frac{3}{4}$

Thus, the equation is: $f(x) = -\frac{3}{4}(x-1)^2(x-4)$ (ans)

(b)Transforming $f(x)$ into $k(f(x+h))$

• (0, 3) → (-h, 3k)

• (1, 0) → (1 - h, 0) (X-1)² so still a “bounce”

• (4, 0) → (4 - h, 0)$</span></p></li></ul><p><span style="font-size: inherit;">$ 3$$

  
  (warnning —> answer for (c )and (d) is incomplete btw)

(b) the transformation is:
translated by π/5 units to the right followed by stretching with scale factor of ½ parallel to x-axis

Find f(x)

DON’T forget C, find C by plugging in coords, i.e (4,-1)

no C means -2 marks

y = cos(θ)

[cos theta curve)

(i) 3cos(270/n) = 0
270/n = cos^-1(0) = 90
n = 3

(ii) period = 1 full cylce

DO NOT DO dy/dx at (−1,28), thats gonna give grad of curve, q. wants grad of l.

Do this:
grad of l = y2 - y1 / x2-x1 (first find the coords of p, not always present)

(a) (5/2π, 12)

c. The shortest distance from $A$ to the diagonal $BD$ is the perpendicular distance. Let's call this distance $h$. The area of triangle $ABD$: $\frac{1}{2} \times AB \times AD \times \sin(\angle DAB)$
OR Since area of parallelogram = 40 cm$^2$, the area of triangle $ABD$ is half that, which is 20 $\operatorname{cm}^2$

Area of triangle $ABD = \frac{1}{2} \times BD \times h = 20$
$20 = \frac{1}{2} \times 15.0 \times h$
$40 = 15.0 \times h$
$h = \frac{40}{15.0} \approx 2.67$ cm

Ans: shortest length = 2.67 cm

given AB longest side
∴ angle opposite to AB(longest side) = largest angle

so if we took our x = 43, the other angle = 180-43 -25 = 112 (N/P) as angle opposite as largest angle shud be opposite to AB

∴ angle opposite to AB = x = 180 - 43.05 = 136.95

NOTE:

sin(180-θ) = sin(θ)

sin(180+θ) = - sin(θ)

sin (θ-180) = - sin(θ)


given sin(θ) = p, so sin (θ-180) = - sin(θ) = -p

for ( c ), every sin graph has 2 solutions as yk → θ and 180 - θ
(so using this and 2x = a(alpha), we can find in terms of a)

we multiply by x² as for any real number, x² is always positive

(iii) from -2 to 2 pi there was 5 pi, 2×2 + 1 from origin = 5

therefore for -100 to 100pi, it is 100×2 + 1 from origin = 201

x >3/2 as curve will keep rising so it will keep being f(x)>0

y = 9-x does not cross or touch C, therefore no intersections, therefore b²-4ac < 0

(btw atleast one real solution : b²-4ac >= 0
2 real solutions → b²-4ac > 0)

max point is the line of symmetry, so if its 2 units away from origin so other also 2 units away
hence→ (4,0)


read Q. carefully, passes origin
therefore at (0,0) …….. a = -5