+0

# I dont rly need help - some challenging questions

+2
487
19
+7002

1) [Logarithms]

Simplify $$\log_3(81\sqrt[3]{\dfrac{\sqrt{3+\ln1}}{3}})$$

2. [Complex numbers]

Find the value for $$\ln(\sqrt[n]{-1})$$ in terms of n.

3. [Trigonometry]

(a) Express $$4\sin x + 3\cos x$$ in the form of $$a\sin(x+\alpha)$$

(b) Solve the equation $$4\sin x + 3 \cos x = 2$$ for the range $$0^{\circ} \leq x <2\pi$$

(x and alpha are in radians)

4. [Algebra]

(a) Factorize $$x^4 + x^2 + 1$$

(b) If $$y = x\sqrt x$$, express $$x^2+x+1$$ in terms of y(That means no 'x's in the final answer.)

5. [Integral Calculus]

Compute $$\displaystyle \int\sqrt{\tan x}dx$$

(Do not show steps. This will make the whole forum lag.)

6. [Algebra]

(a) State the exact value of the golden ratio.

(b) Find a quadratic equation whose one of the roots is the golden ratio. The equation must be in the simplest form.

7. [Algebra]

If 3xy = a + 2, find the value of $$(27^x)^y$$, in the form of $$R\cdot Q^a$$, where Q and R are positive integers.

(So easy this one.......)

8. [Differential Calculus]

Differentiate $$u(x)$$, where $$u(x)$$ is the unit-step function.

9. [Complex numbers]

Compute $$\sqrt{1+i}$$

10. [Trigonometry]

Simplify $$\cos \arctan x$$.

MaxWong  Mar 24, 2017
edited by MaxWong  Mar 24, 2017
edited by MaxWong  Mar 24, 2017
edited by MaxWong  Mar 24, 2017
edited by MaxWong  Mar 24, 2017
edited by MaxWong  Mar 24, 2017
edited by MaxWong  Mar 26, 2017
#1
+5250
+3

x4+x2+1

(x2-x+1)(x2+x+1)

I don't believe that there're many steps to this one, but I took a shot at it. :/

rarinstraw1195  Mar 24, 2017
#2
+7002
+3

You're correct.

But how you do it in one step...... I took 2 steps at it:

$$x^4 + x^2 + 1\\ =(x^2 + 1)^2 - x^2\\ =(x^2 -x+ 1 )(x^2+x+1)$$

Step 2 I used a^2 - b^2 = (a+b)(a-b)

And pls do challenge the b part of the question I just posted.

MaxWong  Mar 24, 2017
#3
+2

1)   4/3^(1/6)

2)   ((i π)/2)^n

4a)   (x^2 - x + 1) (x^2 + x + 1)

5)  integral sqrt(tan(x)) dx = (-2 tan^(-1)(1 - sqrt(2) sqrt(tan(x))) + 2 tan^(-1)(sqrt(2) sqrt(tan(x)) + 1) + log(tan(x) - sqrt(2) sqrt(tan(x)) + 1) - log(tan(x) + sqrt(2) sqrt(tan(x)) + 1))/(2 sqrt(2)) + constant

6a     1/2 (1 + sqrt(5))

Guest Mar 24, 2017
#4
+7002
+3

be careful, always.

MaxWong  Mar 24, 2017
#5
+87604
+4

6.  The Golden Ratio is irrational.....so it won"t have an exact value

It equals  ≈   1.618   = "Phi"  =  [ 1 + sqrt(5) ] / 2

We can find an equation by using similar triangles in a regular pentagon

We have  that

x^2  - x  - 1  = 0    solving this with the quadratic formula gives

[ 1 ± sqrt (5) ] / 2

BTW :  Interestingly, as the Fibonacci Series grows larger and larger, the ratio of a successive term to a preceding term approaches Phi

CPhill  Mar 24, 2017
#8
+7002
+1

I say exact value but I mean the value 1 + sqrt(5)/2......

That is a mistake.

MaxWong  Mar 25, 2017
#6
+2

9)    2^(1/4) e^((i π)/8)

10)   1/sqrt(x^2 + 1)

Guest Mar 24, 2017
#10
+7002
+1

Both correct.

MaxWong  Mar 25, 2017
#7
+87604
+2

1 )

log3  ( 81 * (1/3)1/3 * (3 + ln 1)1/6 )  =

log3  ( 81 * (1/3)1/3 * (3 + 0 )1/6 ) =

log3  ( 81 * (1/3)1/3 * (3 )1/6 ) =

log3  ( 34 * (3)-1/3 * (3 )1/6 )  =

log3 (34 -1/3 + 1/6 ) =

log3 323/6   =

(23/6) log3 3  =

(23/6) * 1  =

23/6

CPhill  Mar 24, 2017
#9
+7002
+1

Correct.

MaxWong  Mar 25, 2017
#11
+7002
+1

Unsolved Problems:

3, 4(b), 7(why no one answer the easiest question), 8.

MaxWong  Mar 25, 2017
#12
+87604
+3

4b

y =  x*sqrt (x)   = x^(3/2)

Therefore....

y^(2/3)    =  x

So

x^2  +   x  +  1 =

[y^(2/3)]^2 + y^(2/3)  + 1 =

y^(4/3)  + y^(2/3)  +  1

CPhill  Mar 25, 2017
#13
+87604
+3

7

3xy  = a + 2

xy  =  [a + 2 ] / 3

So

27xy =

27 [a + 2] / 3  =

[ 271/3 a + 2  =

[3 ] a + 2  =

3a  *  32   =

9 * 3a

CPhill  Mar 25, 2017
#14
+7002
+1

Both correct :)

MaxWong  Mar 26, 2017
#15
+7002
+2

Unsolved question: 3, 8

MaxWong  Mar 26, 2017
#16
+7002
+1

8 is straight-forward if you do some research......

MaxWong  Mar 26, 2017
#17
+7155
+4

Man, these are hard! I can't figure any of them out...

hectictar  Mar 26, 2017
#18
+87604
+3

3

(a)   I do not know the proof, but there is a trig identity that says that

a sin x   +  b cos x  = c sin (x + α)      where c =  sqrt (a^2 + b^2)   and

α = atan2(b,a)

So

c =  sqrt (4^2 + 3^2)  = sqrt (25)   = 5    and

atan2 (b , a)   =  atan2( 3 , 4)  ≈  .6435     (in rads)

4 sin x  + 3 cos x   =   5 sin ( x + .6435 )

(b)  4sin x + 3 cos x  = 2

4sin x    =   2 - 3 cos x        square both sides

16sin^2 x   =  4 - 12cosx  +   9cos^2 x

16(1 - cos^2 x)   =  4 -  12cos x   + 9cos^2 x

16 - 16 cos^2 x   = 4  - 12cos x  + 9cos^2 x

25 cos^2x - 12 cos x - 12  = 0        let cos x  = a

25a^2    -  12 a     -   12   = 0

a  = [ 6  ±  4 sqrt (21)] / 25

So

a  = cos x  = [ 6  ±  4 sqrt (21)] / 25

Wnen    cos x  = [ 6  + 4 sqrt (21)] / 25

arccos [ [ 6  + 4 sqrt (21)] / 25 ]  = x  ≈   .232 rads   and  ≈ 6.051 rads

When  cos x  = arccos [ [ 6  - 4 sqrt (21)] / 25 ]

arccos [ [ 6  - 4 sqrt (21)] / 25 ]   = x ≈ 2.0866 rads    and  ≈ 5.228 rads

However.......232 rads  and 5.228 rads are extraneous  [ because of squaring ]

As the graph here shows.....the two solutions are   x ≈ 2.0866 rads  and  x ≈ 6.051  rads

https://www.desmos.com/calculator/pjvewx15zw

CPhill  Mar 26, 2017
#19
+7002
+1

Chris: Both are correct.

But in (b) part you are supposed to use the answer from (a)...... XD

MaxWong  Mar 27, 2017