 ## Question

junior college 2 | H2 Maths

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##### Mathnerd

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Not sure how to do pls help thanks

Date Posted: 11 months ago
Views: 106
Eric Nicholas K
11 months ago
Sent
J
11 months ago
3 - cosy = x² - xy

Differentiate both sides with respect to x,

siny(dy/dx) = 2x - y - x(dy/dx)

(x + siny)(dy/dx) = 2x - y

dy/dx = (2x - y)/(x + siny)

2 ways to show :

① For the x-axis, y = 0 so dy/dx = 0

For dy/dx = 0, (2x - y)/(x + siny) = 0

Since denominator ≠ 0,

2x - y = 0
y = 2x

Sub y = 2x,

3 - cos2x = -x²

x² + 3 = cos2x

x² + 3 ≥ 3 for all real x since x² ≥ 0 for all real x.

But -1 ≤ cos2x ≤ 1 for all real x.
So x² + 3 ≠ cos2x for all real x.

The two functions will never meet for all real values of x, i.e there are no real values of x that satisfies the equation for dy/dx = 0. So dy/dx ≠ 0

x = 0 on the y - axis.

the gradient of the y-axis is infinity/undefined.

If the tangent to C is parallel to the y-axis, dy/dx is also undefined.

So this means the denominator is 0 since dividing by 0 gives an undefined result.

Then x + siny = 0
siny = -x

sin²y + ysiny + cosy

= (-x)² + y(-x) + cosy
= x² - xy + cosy
= 3 - cosy + cosy (refer to equation of C)
= 3

So this y-coordinate satisfies sin²y + ysiny + cosy = 3
Eric Nicholas K
11 months ago
The idea here is to show that in the case dy/dx = 0, we are unable to solve for x. We use reasoning while trying to solve

3 - cos 2x = -x^2

for values of x.
J
11 months ago
It's not that we are unable to solve for x. Rather , it's that there are no real values of x that satisfies the equation. (There are actually two complex roots) 