# 數條MATHS~~~

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(13) Just simply multiply the function by 2 and we get y = 2x3 - 2

(14) First translation: y = 2x - 6 (Negate the y for reflection about the x-axis)

Second translation: y = 4x - 12 (Multiply by 2)

(15) First translation: Replace x by 2x to get y = x2/4 - x/2 + 1

Second translation: Replace x by x + 1 to get:

y = (x + 1)2/4 - (x + 1)/2 + 1

= x2/4 + x/2 + 1/4 - x/2 - 1/2 + 1

= x2/4 + 3/4

(16) First translation: Shft to the right by 2 units

Second translation: Reflect about the x-axis.

(17) It should be first reflected about the x-axis to botain -f(x), followed by shifting down by 1 unit to get -1 - f(x)

Alternatively, we can shft it up by 1 unit to get f(x) + 1, followed by reflecting about the x-axis to get -1 - f(x).

So ans = D

(18) First translation: Shifting to the left by 3 units.

Second translation: Reducing to 1/2 of original along the y-axis.

(18 Long Q) (a) (i) Original volume of water in cone = π(9)2 x 24/3 = 648π cm3.

After transfer, depth of water in cone = (h + 5) cm

By similar triangle, base radius of water portion in the cone = 3(h + 5)/8 and so the volume is:

[3(h + 5)/8]2 x (h + 5)π/3 = 3(h + 5)3π/64 cm3

Hence volume of water in cylinder = 36πh cm3

Thus:

3(h + 5)3π/64 + 36πh = 648π

(h + 5)3/64 + 12h = 216

h3 + 15h2 + 75h + 125 + 768h = 13824

h3 + 15h2 + 843h - 13699 = 0

(ii) The line y = 13699 should be added and the solution of h is 11.8 cm (corr. to 0.1 cm)

(b) Volume of frustum = 648π - 648π(53/243) = 41097π/64 cm3.

Suppose that the depth of the water is h cm in both cylinder and frustum:

Volume of water in frustum = 3(h + 5)3π/64 - 648π(53/243) cm3

Volume of water in cylinder = 36πh cm3

So,

3(h + 5)3π/64 - 648π(53/243) + 36πh = 41097π/64

3(h + 5)3 - 375 + 2304h = 41097

(h + 5)3 - 125 + 768h = 13699

h3 + 15h2 + 843h - 13699 = 0

By (a), we have h = 11.8 cm

Source(s): My Maths knowledge
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