help with questions from practise paper (1 Viewer)

jjuunnee · Sep 11, 2016

Solve 2sin²x = sinx for 0 degrees <= x <= 360 degrees

Given equation x² - (k + 2)x + (2k + 4) = 0, determine the values of k such that one root is twice the other

Is the curve y = x² - 4 / x³ + 8 differentiable for all x? Give a reason for your answer.

Sorry if this is hard to read

jjuunnee · Sep 11, 2016

sorry there's more

The line l is tangent to the curve y = 3x² at the point (3, 27). What is the gradient of line l?

Simplify |x + 2| / x² - 4 for x does not equal to + or - 2

InteGrand · Sep 11, 2016

jjuunnee said:
Solve 2sin²x = sinx for 0 degrees <= x <= 360 degrees

Given equation x² - (k + 2)x + (2k + 4) = 0, determine the values of k such that one root is twice the other

Is the curve y = x² - 4 / x³ + 8 differentiable for all x? Give a reason for your answer.

Sorry if this is hard to read

$\noindent First one: we have $2\sin ^2 x -\sin x = 0$, so $\sin x = 0$ or $\sin x = \frac{1}{2}$. Just solve these two now.$

$\noindent Second one: Let the roots be $a,2a$. From sum of roots: $k+2 = 3a$. From product of roots: $2k + 4 = 2a^2$. Solve these two equations in two unknowns to find the values of $k$.$

$\noindent Third one: Yes, it's differentiable \emph{everywhere that it's defined} (which is what just saying `differentiable' means). Incidentally, that expression is undefined at $x=-2$ but this is a removable discontinuity. The denominator is $\left(x+2\right)\left(x^2 -2x + 4\right)$ and the numerator is $(x+2)(x-2)$, so for all $x \neq -2$, we have $ y = \frac{x-2}{x^2 -2x + 4}$, and the denominator is non-zero for all real $x$. Therefore, whilst the given function hasn't been defined at $x=-2$, it is possible to give it a value here that will make the resulting function differentiable everywhere.$

InteGrand · Sep 11, 2016

jjuunnee said:
sorry there's more

The line l is tangent to the curve y = 3x² at the point (3, 27). What is the gradient of line l?

Simplify |x + 2| / x² - 4 for x does not equal to + or - 2

$\noindent First one here: just find the value of the derivative at $x=3$, i.e. answer is $6\times 3 = 18$.$

$\noindent Second one: note that $x^2 -4 = (x+2)(x-2)$. So for $x \neq \pm 2$, the given expression equals $\mathrm{sgn}\left(x+2\right)\cdot \frac{1}{x-2}$, where sgn is the \textsl{sign function}. You can write this in case form too: it equals $\frac{1}{x-2}$ if $x>-2, x \neq 2$, and it equals $-\frac{1}{x-2}$ if $x < -2$.$

Sign function: https://en.wikipedia.org/wiki/Sign_function .

kashkow · Sep 11, 2016

For the first question:

> Simplify down to: sinx(2sinx-1) = 0

> then solve: sinx = 0 and 2sinx-1 = 0 for x

For the second question:

> let (alpha) be one root and (beta) be another root such that (beta) = 2(alpha)

> Thus roots are (alpha) and 2(alpha)

> now (2k+4) = product of roots = (alpha)*2(alpha)= 2(alpha)^2

> Also (k+2) = sum of roots = (alpha) + 2(alpha) = 3(alpha)

> Solve simultaneous equations to solve for k (substitute alpha).

For the third question:

> Recall that differentiation is finding the tangent to the curve

> As f'(x) = 2x +12/x^4 is not continuous you cannot find tangent to the curve for ALL x (ie. x=0).

> :. answer is No; reason because the original function is a non-continuous function

For the fourth question:

> Recall that finding gradient of the tangent at a point is the same as finding the gradient of the curve of which the tangent belongs at that point, (3, 27)

> thus simply differentiate the curve y=3x^2

> Then use the gradient function and substitute for f'(3)

For your fifth question:

|x + 2| can be + or -

ie x+2 > 0
or x+2 < 0

Thus x>-2 or x<-2

When x>-2
x+2 / (x+2)(x-2)

1/(x-2)

When x<-2
(-x-2)/(x+2)(x-2)

-(x+2)/(x+2)(x-2)
-1/(x-2)

:. expression becomes +- 1/(x-2)

I think the fifth one is right but I'm not too sure; I'm not the best with absolute values... sorry :S Also not 100% on the third question... I'm not too sure what "differentiable for all x" means exactly. You can differentiate it and find the gradient function; however you cannot use all x within the gradient function to find the tangent. Thus that is why I think it's not "differentiable" for all x.

Rats; i was too slow

help with questions from practise paper (1 Viewer)

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