Find the constant term in the binomial expansion $\left( {1\over 2x^3}+x\right)^{20}$

BOSTES Formula Reference Sheet 2U/3U/4U

http://www.boardofstudies.nsw.edu.au/syllabus_hsc/pdf_doc/maths-ref-sheet.pdf

GROVE HSC 3U EX2.8 - REGARDING Q15

JUst regarding finding the second derivative of this (good luck to you!)

WOLFRAM ALPHA SAYS

AND THIS CAN NEVER BE ZERO, SO THERE ARE NO INFLEXIONS.

(yes x here should be h)

GROVE HSC 3U - EX9.2 - Q5

CAMB Y12 - 3U - EX6E - Q4ab 8a

QUESTIONS: 4abc, 8a

GROVE HSC 3U - EX2.8 - SIMILAR QUESTION

Question: Find any non horizontal inflexions.

(by finding second derivative)

[this is not Q15, but looks similar.

It's good practice for derivatives]

GROVE HSC 3U TEST YOURSELF 10 - Q2

GROVE HSC 3U TEST YOURSELF 10 - Q13

GROVE HSC 3U. TEST YOURSELF 10 - Q19

GROVE HSC 3U - TEST YOURSELF 10 - Q13

Q13

GROVE HSC 3U - Q15 test yourself 10

GROVE HSC 3U - TEST YOURSELF 10 - Q5

(a)  Put $x=0$  on LHS and RHS!

(b)  Put $x=-1$  on LHS and RHS!

Keep in mind $(-1)^2=1$, $(-1)^3=-1$, $(-1)^4=1$, ...

GROVE PRELIM 2U/3U TEST YOURSELF 8 - Q18

$y={1\over 3}x^{-1}$
$y'=-{1\over 3}x^{-2}$
$y'=-{1\over 3x^2}$

$y'=-{1\over 3\left({1\over 6}\right)^2}=-{36\over 3}=-12$

When $x=1/6$, $y=1/(3(1/6))=2$

$y-2=-12(x-1/6)$
$y-2=-12x+2$

$12x+y-4=0$

JH :-)

GROVE PRELIM 2U/3U TEST YOURSELF 8 - Q20

$y'={vu'-uv' \over v^2}$

$y'={(2x+1)4-(4x-3)2 \over (2x+1)^2}$

Put $x=1$

$y'={(2(1)+1)4-(4(1)-3)2 \over (2(1)+1)^2}$

$y'=\displaystyle{10\over 9}$

When $x=1$, $y=(4-3)/(2+1)=1/3$.

$(1, 1/3)$.

$y-{1\over 3}={10\over 9}(x-1)$

$9y-3=10(x-1)$

$9y-3=10x-10$

$10x-9y-7=0$

When $y=0$,

$10x-7=0$

$x=\displaystyle {7\over 10}$

GROVE PRELIM 2U/3U TEST YOURSELF 8 - Q15

Find $y'$ then sub $x=2$

Use product rule: If $y=u(x)\times v(x)$ where $u,v$ are functions of $x$, then
$$y'=uv'+u'v$$
Here $$u(x)=(3x-1)^3, v(x)=(2x-1)^2$$ $y'=(3x-1)^3\times 2(2x-1)\times 2\\+3(3x-1)^2\times 3(2x-1)^2$

$y'=(3(2)-1)^3\times 2(2(2)-1)\times 2\\+3(3(2)-1)^2\times 3(2(2)-1)^2$

$y'=3525$

GROVE PRELIM 2U/3U EX 8.9 - Q8

$y'=2(x+3)^2+2x\times 2(x+3)$

$y'=2(x^2+6x+9)+4x^2+12x$

$y'=2x^2+12x+18+4x^2+12x$

$y'=6x^2+24x+18$

$\therefore 6x^2+24x+18=14$

$6x^2+24x+4=0$

$3x^2+12x+2=0$

$x=\displaystyle { -12\pm \sqrt{12^2-4(3)(2)} \over 6}$

$x=\displaystyle { -12\pm \sqrt{120} \over 6}$

$x=\displaystyle { -12\pm 2\sqrt{30} \over 6}$

$x=\displaystyle { -6\pm \sqrt{30} \over 3}$

GROVE PRELIM 2U/3U TEST YOURSELF 8 - Q15

QUESTION:

Find the equation of the tangent to the curve $y=x^2+2x-5$ that is parallel to the line $y=4x-1$.

SOLUTION:

We need to find the $x$ value on the curve at which the tangent has gradient $m=4$ 
(since the line $y=4x-1$ has gradient $4$)

Find $y'$.
 (since $y'$ is the gradient of the tangent at $x$)

$y'=2x+2$

Solve $y'=4$.

$2x+2=4$

$2x=2$

$x=1$

When $x=1$, $y=(1)^2+2(1)-5=-2$

So the point is $(1,-2)$.

Tangent has equation $y-y_1=m(x-x_1)$

$$y--2=4(x-1)$$

$$y+2=4x-4$$

$$4x-y-6=0$$

GROVE PRELIM 2U/3U BOOK - TEST YOURSELF 8 - Q13

Regarding $\pi$ as a constant and differentiating with respect to $x$.

$${dS\over dr}=2\times 4\pi r^{2-1}$$
$${dS\over dr}=8\pi r^{1}$$

$${dS\over dr}=8\pi r$$

C'est fini.

JH 

Find the Values of x at which a Curve has a Given Gradient [Solution to Grove Prelim 2u Ex8.10 Q4]

Firstly, the question should read, "find a quadratic equation...." as there are many....

(a) 

$(x-4)(x- -7)=0$

$x^2-4x+7x-28=0$

$x^2+3x-28=0$

(b) 

$(x-5-\sqrt{7})(x-5+\sqrt{7})=0$
$((x-5)-\sqrt{7})((x-5)+\sqrt{7})=0$

Thin difference of two squares.
$(x-5)^2-\sqrt{7}^2=0\\$
$(x-5)^2-7=0\\$
$x^2-10x+25 -7=0\\$
$x^2-10x+18=0$

GROVE HSC 3U - TEST YOURSELF 9 (APPROX OF ROOTS) - Q6

It's up to you to find to values of $x$ such that the function value changes sign. 
It will be easy trial and error. (unlikely in an actual test)

Let's randomly (with a bit of intuition) try $x=0$ and $x=2$! 

$f(x)=x^3+3x-7$

$f(0)=-7<0$
$f(2)=2^3+3(2)-7=8+6-7=7>0$

Since $f(0)<0$ and $f(2)>0$, then there is at least one root  between $x=0$ and $x=2$.  

Now proceed to half the interval 3 times.....

GROVE PRELIM 2U TEST YOURSELF 9 / GROVE PRELIM 3U TEST YOURSELF 10 - Q13

Multiply both sides by $x$.

$\displaystyle 2x=5+{3\over x}$

$\displaystyle 2x\times x=5\times x+{3\over x}\times x$

$$2x^2=5x+3$$

$$2x^2-5x-3=0$$

$$2x^2-6x+1x-3=0$$

$$2x(x-3) + 1(x-3)=0$$

$$(2x+1)(x-3)=0$$

$$\therefore x=-0.5, 3$$

JH2015 :-)

GROVE PRELIM 2U TEST YOURSELF 9 / GROVE PRELIM 3U TEST YOURSELF 10 - Q12

Here you use sum and product of roots formulas:

$\alpha+\beta = -b/a$ and 

$\alpha\beta=c/a$

GROVE PRELIM 2U/3U TEST YOURSEWLF 8 - Q10

Grove HSC 2u Chapter 2 Test Yourself 2 Q15

GROVE Prelim 8.8 Q1(t)(u)&(v)

GROVE HSC 3U EX 2.1 Q14/15

GROVE HSC 3U BOOK (QUADRATIC FUNCTION) - TEST YOURSELF 10 - Q3

GROVE HSC 3U BOOK - EX 8.7 - Q2h

GROVE HSC 3U (APPROX OF ROOTS) EX 9.1 - Q10 -

Question (Quadratic identity): Find $A,B,C$ so $4x^2-9x-2=Ax(x+1)+B(x-1)+C$

Question (Quadratic Functions): For what values of $k$ will $3x^2+kx+12$ always be positive definite.

QUESTION: solve $4^x-12(2^x)+32=0$. - Equations reducible to quadratic equations

GROVE HSC CHALLENGE EX2 - Q2 - Curve sketching, stat points, inflexions

Thanx Bec Rae.

GROVE HSC 2U/3U PRACTICE ASSESSMENT TASK 1 - Q43 Curve Sketching

GROVE HSC EX 10.6 Q6 , Reducible to Quadratics

GROVE HSC EX 10.6 Q3d , Reducible to Quadratics

Quadratic Identity question

QUESTION:
Write $3x^2+7$ in the form $a(x-2)^2+b(x+3)+c$.


ANSWER:
$3x^2+7=a(x-2)^2+b(x+3)+c$ holds for all $x$.

Best way is to just expand the RHS and then equate coefficients of the powers of $x$ on LHS and RHS.

$3x^2+7=a(x^2-4x+4)+b(x+3)+c$
$3x^2+7=a x^2-4ax+4a+bx+3b+c$
$3x^2+7=a x^2+(b-4a)x+(4a+3b+c)$

Therefore 

1. $a=3$  (comparing coefficient of $x^2$ on LHS and RHS)
2. $b-4a=0$ since coefft of $x$ on LHS is 0. (there is no $x$ term)
3. $4a+3b+c=7$ (Constant terms on LHS and RHS)

So, $b-4(3)=0$.

$b=12$.

Finally, $4(3)+3(12)+c=7$.

So, $c=7-12-36=-41$.

Final answer: $a=3, b=12, c=-41$. 

EXAMPLES - EQUATIONS REDUCIBLE TO QUADRATIC EQUATIONS

Grove HSC 3U EX 10.4 - Q3, 9, 11c, 15, 16, 18a, 20ac, 21abe -: Binomial Theorem

QUESTIONS: Q3, 9, 11c, 15, 16, 18a, 20ac, 21abe,  

Q3 =====================

Fanks Tatiana :-)

Q9 =====================

Thx Becster :-)

Q11c =====================

Q15 =====================

Q16 =====================

Q18a =====================

Thanks Juliette :-)

Q20a =====================


Q20c =====================
This is a second version of this solution.

Thanks Tat & Ellie

Q20c - version 2-=====================

Q21a=====================

Q21b =====================

Q21e =====================