ssc111 series 1 final 2003

8/8/2019 SSC111 Series 1 Final 2003

1/17

SSC111 Homework Series 1

1:

1 + 3 = 3

x-2 x+1

[(x+1)+ (x-2)(3x)= (x+1)(x-2)=3 (x+1)(x-2)

3x2-5x+1=3x2-3x-6

-2x= -7

X= 3,5

2:

Sin2x-cosx=0

2sinx * cosx=cosx

2sinx=1

sin x=

x=

3:

sin2x-sinx=0

sin2x=sinx

sinx =1

x=

4:

sin2(x)- cos (x)-1=0sin2(x)- cos (x) =11-sin2(x)=-cos(x)

-cos2(x)=-cos(x)

-1=cos(x)X= v x 3


2/17

5:

Asin(x)+Bcos(x)=Csin(x+)Csin(x) * cos()+cos(x)sin()C cos()sin(x)+Csin()cos(x)

A=Ccos()B=Csin()

Asin(x)+Bcos(x)=0Asin(x)= Bcos(x)

Asin(x) = 1Bcos(x)

A * sin(x) = 1B cos(x)

B = sin(x)= tan(x)A cos(x)

tan(x)= BA

X= arctan BA

=-arctan BA

Asin(-)+Bcos(-)=C=D

6:

2e2x+2=3e-3x-3

eln(2)e2x+2= eln(3)e-3x-3

e2x+2+ln(2)=3e-3x-3+ln(3)

5x=-5+ln( )

X= -1


3/17

7:

P=2X

2x+ =2

P+ =2

p2+1=2p- > p2-2p+1=0

(p-1)2=0-> p=1

2x=1 -> x=0

8:

xe2x-3+4e2x+3=0

xe2x* e-3+4e2x*e3=0e2x(xe3x+ 4e3)=0e2x(xe3x+ 4e3)=0

e2x=0 V xe3x+ 4e3=0

xe3x= -4e3

x= -4e6

9:

=

ex-e-x=2(ex-e-x) -> 2 ex-2e-x

ex =3 e-x

x= ln3-x

x=


4/17

10:

ln(x+1)+ln(x-1)=3

lnx2-x+x-1=3

lnx2-1=3

elnx2-1=e3

x2-1=e3

x2=e3+1

x=

12:

Plot[{f1[x]+f2[x]},{x,-5,5}]

4 2 2 4

5

5

1

1

Even Function

4 2 2 4

6

4

2

2

4

Even Function


5/17

Plot[{g1[x]+g2[x]},{x,-5,5}]

4 2 2 4

1

5

5

1

Odd Function

4 2 2 4

1

2

3

4

5

6

7

Even Function

4 2 2 4

6

4

2

2

4

6

Odd Function


6/17

13:

Even part:

Odd parts:

4 2 2 4

4 0

2 0

2 0

4 0

6 0

8 0

Even part:

Odd parts:

4 2 2 4

1

1

2

3

4


7/17

14:

Standard function: f[x]=2x

Transformation

4 2 2 4

1

5

5

1

(negative/positive) increase or decrease of

the slope

Transformation

4 2 2 4

1

5

5

1

(negative/positive) increase or decrease ofthe slope

Transformation

4 2 2 4

1

5

5

1

shift (increase or decrease) of y at x=0


8/17

Transformation

4 2 2 4

1

5

5

1

Also a shift (increase or decrease) of y at x=0, although the difference with theformer is that a doesnt get multiplies by any multiplier infront of x (in this case 2)anymore.

15:

Although the outcome remains the same based on rules I learned at high school Istill would argue for scaling followed by shifting scaling:

If you have the function (ax+b)*c -> you first compute c*ax and after that add upb*c.

16:

1 . 0 . 0 . 1 .

1 .

0 .

0 .

1 .


9/17

x[t_]:=Cos[5t];

y[t_]:=Sin[7t];

ParametricPlot[{x[t],y[t]},{t,0,2}]

1 . 0 . 0 . 1 .

1 .

0 .

0 .

1 .

This doesnt look boring to me

x[t_]:=1/2 Cos[t]-1/4 Cos[2t];

y[t_]:=1/2 Sin[t]-1/4 Sin[2t];

ParametricPlot[{x[t],y[t]},{t,0,2}]

0 . 0 . 0 . 0 . 0 .

0 .

0 .

0 .

0 .

0 .

0 .


10/17

17:

In this case you could use differentiation:

F(x)= x3+3x2+kx

F(x) = 3x2+6x+k=0

F(x)= b2-4ac (discriminant) =0

62-4*3k=0

K= =3

Therefore k must be higher or equal to 3

18:

f(x)= =y

x+2=xy-2y

x-xy= -2y-2

x = -> -> -> -> = x

f(x)= -> -> = y

19:

F(x)= = y

ex-1 = y(ex+1)

ex-1 = yex+y

e

x

-ye

x

=1+y -> e

x

(1-y)=1+y

ex= -> x = ln ( )

Domain is -1,1Range is negative infinity till infinity20:


11/17

4 2 2 4

1 .

1 .

1 .

1 .

4 2 2 4

0 .

0 .

0 .

0 .

1 .

4 2 2 4

0 .

1 .

1 .

2 .


12/17

4 2 2 4

2

1

1

5

4 2 2 4

1 .

0 .

0 .

1 .

4 2 2 4

1 .

0 .

0 .

1 .


13/17

21:

3 2 1 1

1 .

0 .

0 .

Based on using the tool get coordinates the value of x0 is

{0., 0.},

{-0.9, 0.310345},

{-0.99, 0.331104},

{-0.999, 0.333111},

{-0.9999, 0.333311},

{-0.99999, 0.333331},

{-0.999999, 0.333333},

{-1., 0.333333},

{-1., 0.333333},

{-1., 0.333333},

{-1., 0.333333}

{-2., 0.5},

{-1.1, 0.354839},

{-1.01, 0.335548},

{-1.001, 0.333555},

{-1.0001, 0.333356},

{-1.00001, 0.333336},

{-1., 0.333334},

{-1., 0.333333},

{-1., 0.333333},

{-1., 0.333333},

{-1., 0.333333}

Based on these numbers we can further conclude that the value of x0 is


14/17

22:

2 3 4 5 6

1 .

1 .

2 .

2 .

Slope is is 1.77245

23:


15/17

24:

->

-> = -1

26:

->

-> * -

->

-> -> ->

27:

- -> -> -> ->

28:

=t -> x=t6

-> -> -> ->

29:

1stzero [2,3] = (2.21431,0)

2nd zero [-1,0] = (-0.539185,0)

3rd zero [-2,-1] = (-1.6748,0)


16/17

30:

= 1

-> =

-> = 1

x=t

-> = 1 * =

31:

2 1 1 2

0 .

1 .

1 .

2 .

32:

c(3)+1=c(3)2-1 -> 3c+1=9c-1 ->-6c=-2 -> c =

33:

This concept can be defined as a = a + 1

The function f(x) = x - x + 1 is

f(-1) = -1 < 0 f(2) = 7 > 0

Therefore, there should be a value such that f(a) = 0

If we then apply differentiation:

(x)=3x2-1=0


17/17

x1= x2= -

f( -min.

f(- -max.

f(-)=-

So the solution is close to x = -1.324717957497, but a number I dont have.

34:

x2cos20x=0

f(x)=x2

g(x)=x2

h(x)= x2cos(20x)

or

g(x)h(x)f(x)-x2 0 x2

-x2=0

-x2=0

Therefore =0

35:

3x=3

x3=1

1lim f(x)3

ssc111 series 1 final 2003

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