This test was once used to monitor the broad learning of university chemists at the end of the 1st year and is intended to check, somewhat lightly, a range of skills in only 50 minutes. It contains a mixture of what are perceived to be both easy and difficult questions so as to give the marker a good idea of the student's algebra skills and even whether they can do the infamous integration by parts .
(1) Solve the following equation for
x
{\displaystyle x}
x
2
+
2
x
−
15
=
0
{\displaystyle x^{2}+2x-15=0}
It factorises with 3 and 5 so :
(
x
+
5
)
(
x
−
3
)
=
0
{\displaystyle (x+5)(x-3)=0}
roots are -5 and +3, not 5 and -3!
(2) Solve the following equation for
x
{\displaystyle x}
2
x
2
−
6
x
−
20
=
0
{\displaystyle 2x^{2}-6x-20=0}
Divide by 2 and get
x
2
−
3
x
−
10
=
0
{\displaystyle x^{2}-3x-10=0}
This factorises with 2 and 5 so :
(
x
−
5
)
(
x
+
2
)
=
0
{\displaystyle (x-5)(x+2)=0}
roots are 5 and -2.
(3) Simplify
ln
w
6
−
4
ln
w
{\displaystyle \ln w^{6}-4\ln w}
Firstly
6
ln
w
−
4
ln
w
{\displaystyle 6\ln w-4\ln w}
2
ln
w
{\displaystyle 2\ln w}
(4) What is
log
2
1
64
{\displaystyle \log _{2}{\frac {1}{64}}}
64 = 8 x 8 so it also equals
2
3
{\displaystyle 2^{3}}
2
3
{\displaystyle 2^{3}}
i.e.
1
64
{\displaystyle {\frac {1}{64}}}
2
−
6
{\displaystyle 2^{-6}}
(5) Multiply the two complex numbers
3
+
5
i
a
n
d
3
−
5
i
{\displaystyle 3+5i~~~~{\rm {and}}~~~~3-5i}
These are complex conjugates so they are
3
2
{\displaystyle 3^{2}}
i
2
{\displaystyle i^{2}}
5
2
{\displaystyle 5^{2}}
i.e. plus 25 so the total is 34.
(6) Multiply the two complex numbers
(
5
,
−
2
)
a
n
d
(
−
5
,
−
2
)
{\displaystyle (5,-2)~~~~{\mathrm {a} nd}~~~~(-5,-2)}
The real part is -25 plus the
4
i
2
{\displaystyle 4i^{2}}
−
10
i
{\displaystyle -10i}
+
10
i
{\displaystyle +10i}
(7) Differentiate with respect to
x
{\displaystyle x}
1
3
x
2
−
3
x
2
{\displaystyle {\frac {1}{3x^{2}}}-3x^{2}}
Answer:
−
2
3
x
3
−
6
x
{\displaystyle ~~~~~~~-{\frac {2}{3x^{3}}}-6x}
(8)
6
x
4
+
3
x
3
{\displaystyle {\frac {6}{x^{4}}}+3x^{3}}
Answer:
9
x
2
−
24
x
5
{\displaystyle ~~~~~~~9{x^{2}}-{\frac {24}{x^{5}}}}
(9)
2
x
+
2
x
{\displaystyle {\frac {2}{\sqrt {x}}}+2{\sqrt {x}}}
Answer:
1
x
−
1
x
3
{\displaystyle ~~~~~~{\frac {1}{\sqrt {x}}}-{\frac {1}{\sqrt {x^{3}}}}}
(10)
x
3
(
x
−
(
2
x
+
3
)
(
2
x
−
3
)
)
{\displaystyle x^{3}(x-(2x+3)(2x-3))}
Expand out the difference of 2 squares first.....collect and multiply....then just differentiate term by term giving:
20
x
4
−
4
x
3
+
27
x
2
{\displaystyle ~~~~20x^{4}-4x^{3}+27x^{2}}
(11)
3
x
3
cos
3
x
{\displaystyle 3x^{3}\cos 3x}
This needs the product rule.... Factor out the
9
x
2
{\displaystyle 9x^{2}}
9
x
2
(
cos
3
x
−
x
sin
3
x
)
{\displaystyle 9x^{2}(\cos 3x-x\sin 3x)}
(12)
ln
(
1
−
x
)
2
{\displaystyle \ln(1-x)^{2}}
This could be a chain rule problem.......
1
(
1
−
x
)
2
.2
.
(
−
1
)
.
(
1
−
x
)
{\displaystyle {\frac {1}{(1-x)^{2}}}.2.(-1).(1-x)}
or you could take the power 2 out of the log and go straight to the same answer with a shorter version of the chain rule to:
−
2
(
1
−
x
)
{\displaystyle -{\frac {2}{(1-x)}}}
(13) Perform the following integrations:
∫
(
2
cos
2
θ
+
2
θ
)
d
θ
{\displaystyle \int \left(2\cos ^{2}\theta +2\theta \right){\rm {d}}\theta }
cos
2
{\displaystyle \cos ^{2}}
cos
θ
sin
θ
+
θ
+
θ
2
{\displaystyle \cos \theta \sin \theta +\theta +\theta ^{2}}
(14)
∫
(
8
x
−
3
−
4
x
+
8
x
3
)
d
x
{\displaystyle \int \left(8x^{-3}-{\frac {4}{x}}+{\frac {8}{x^{3}}}\right){\rm {d}}x}
Apart from
−
4
x
{\displaystyle -{\frac {4}{x}}}
ln
{\displaystyle \ln }
16
x
3
{\displaystyle {\frac {16}{x^{3}}}}
−
(
8
x
2
+
4
ln
x
)
{\displaystyle -({\frac {8}{x^{2}}}+4\ln x)}
(15) What is the equation corresponding to the determinant:
|
b
1
2
0
1
2
b
1
0
1
b
|
=
0
{\displaystyle {\begin{vmatrix}b&{\frac {1}{\sqrt {2}}}&0\\{\frac {1}{\sqrt {2}}}&b&1\\0&1&b\\\end{vmatrix}}=0}
The first term is
b
(
b
2
−
1
)
{\displaystyle b(b^{2}-1)}
−
1
2
(
b
2
−
0
)
{\displaystyle -{\frac {1}{\sqrt {2}}}({\frac {b}{\sqrt {2}}}-0)}
b
3
−
3
/
2
b
{\displaystyle b^{3}-3/2b}
(16) What is the general solution of the following differential equation:
d
ϕ
d
r
=
A
r
{\displaystyle {\frac {{\rm {d}}\phi }{{\rm {d}}r}}={\frac {\rm {A}}{r}}}
where A is a constant..
θ
=
A
ln
r
+
k
{\displaystyle \theta =A\ln r+k}
(17) Integrate by parts:
∫
x
sin
x
d
x
{\displaystyle \int x\sin x{\rm {d}}x}
Make
x
{\displaystyle x}
sin
x
−
x
cos
x
{\displaystyle \sin x-x\cos x}
(18)The Maclaurin series for which function begins with these terms?
1
+
x
+
x
2
/
2
!
+
x
3
/
3
!
+
x
4
/
4
!
+
…
{\displaystyle 1+x+x^{2}/2!+x^{3}/3!+x^{4}/4!+\dots }
It is
e
x
{\displaystyle e^{x}}
(19)Express
x
−
2
(
x
−
3
)
(
x
+
4
)
{\displaystyle {\frac {x-2}{(x-3)(x+4)}}}
It is .....
1
7
(
x
−
3
)
+
6
7
(
x
+
4
)
{\displaystyle ~~~~~{\frac {1}{7(x-3)}}+{\frac {6}{7(x+4)}}}
(20) What is
2
e
i
4
ϕ
−
cos
4
ϕ
{\displaystyle 2e^{i4\phi }-\cos 4\phi }
This is just Euler's equation.....
2
e
i
4
ϕ
=
2
cos
4
ϕ
−
2
i
sin
4
ϕ
{\displaystyle 2e^{i4\phi }=2\cos 4\phi -2i\sin 4\phi }
so one
cos
4
ϕ
{\displaystyle \cos 4\phi }
cos
4
ϕ
−
2
i
sin
4
ϕ
{\displaystyle \cos 4\phi -2i\sin 4\phi }
50 Minute Test II
edit
(1) Simplify
2
ln
(
1
/
x
3
)
+
5
ln
x
{\displaystyle 2\ln(1/x^{3})+5\ln x}
(2)What is
log
10
1
10
000
{\displaystyle \log _{10}{\frac {1}{10~000}}}
(3) Solve the following equation for
t
{\displaystyle t}
t
2
−
3
t
−
4
=
0
{\displaystyle t^{2}-3t-4=0}
(4) Solve the following equation for
w
{\displaystyle w}
w
2
+
4
w
−
12
=
0
{\displaystyle w^{2}+4w-12=0}
(5) Multiply the two complex numbers
(
−
4
,
3
)
a
n
d
(
−
5
,
2
)
{\displaystyle (-4,3)~~~~{\rm {and}}~~~~~(-5,2)}
(6) Multiply the two complex numbers
3
+
2
i
a
n
d
3
−
2
i
{\displaystyle 3+2i~~~~{\rm {and}}~~~~~~3-2i}
(7) The Maclaurin series for which function begins with these terms?
x
−
x
3
/
6
+
x
5
/
120
+
…
{\displaystyle x-x^{3}/6+x^{5}/120+\dots }
(8) Differentiate with respect to
x
{\displaystyle x}
x
3
(
2
−
3
x
)
2
{\displaystyle x^{3}(2-3x)^{2}}
(9)
x
2
−
3
2
x
{\displaystyle {\frac {\sqrt {x}}{2}}-{\frac {\sqrt {3}}{2{\sqrt {x}}}}}
(10)
x
4
−
3
x
2
+
k
{\displaystyle x^{4}-3x^{2}+{\rm {k}}}
where k is a constant.
(11)
2
3
x
4
−
A
x
4
{\displaystyle {\frac {2}{3x^{4}}}-{\rm {A}}x^{4}}
where A is a constant.
(12)
3
x
3
e
3
x
{\displaystyle 3x^{3}e^{3x}}
(13)
ln
(
2
−
x
)
3
{\displaystyle \ln(2-x)^{3}}
(14) Perform the following integrations:
∫
(
3
w
4
−
2
w
2
+
6
5
w
2
)
d
w
{\displaystyle \int \left(3w^{4}-2w^{2}+{\frac {6}{5w^{2}}}\right){\rm {d}}w}
(15)
∫
(
3
cos
θ
+
θ
)
d
θ
{\displaystyle \int \left(3\cos \theta +\theta \right){\rm {d}}\theta }
(16) What is the equation belonging to the determinant
\begin{vmatrix}
x & 0 & 0\\
0 & x & i \\
0 & i & x \\
\end{vmatrix}
= 0</math>
(17)
What is the general solution of the following differential equation:
d
y
d
x
=
k
y
{\displaystyle {\frac {{\rm {d}}y}{{\rm {d}}x}}=ky}
(18) Integrate by any appropriate method:
∫
(
ln
x
+
4
x
)
d
x
{\displaystyle \int \left(\ln x+{\frac {4}{x}}\right){\rm {d}}x}
(19) Express
x
+
1
(
x
−
2
)
(
x
+
2
)
{\displaystyle {\frac {x+1}{(x-2)(x+2)}}}
as partial fractions.
(20)
What is
2
e
i
2
ϕ
+
2
i
sin
2
ϕ
{\displaystyle 2e^{i2\phi }+2i\sin 2\phi }
(1) Solve the following equation for
t
{\displaystyle t}
t
2
−
4
t
−
12
=
0
{\displaystyle t^{2}-4t-12=0}
(2) What is
log
4
1
16
{\displaystyle \log _{4}{\frac {1}{16}}}
(3) The Maclaurin series for which function begins with these terms?
1
−
x
2
/
2
+
x
4
/
24
+
…
{\displaystyle 1-x^{2}/2+x^{4}/24+\dots }
x
{\displaystyle x}
5
x
2
−
8
x
4
{\displaystyle {\frac {5}{x^{2}}}-8{x^{4}}}
(5)
4
x
−
2
x
{\displaystyle {\frac {4}{\sqrt {x}}}-{{\sqrt {2}}x}}
(6)
5
x
+
6
x
3
{\displaystyle 5{\sqrt {x}}+{\frac {6}{x^{3}}}}
(7)
5
x
3
−
5
x
3
{\displaystyle {\frac {5}{x^{3}}}-5x^{3}}
(8)
x
2
(
2
x
2
−
(
5
+
2
x
)
(
5
−
2
x
)
)
{\displaystyle x^{2}(2x^{2}-(5+2x)(5-2x))}
(9)
2
x
2
sin
x
{\displaystyle 2x^{2}\sin x}
(10) Multiply the two complex numbers
(
2
,
3
)
a
n
d
(
2
,
−
3
)
{\displaystyle (2,3)~~~~~~{\rm {and}}~~~~~~(2,-3)}
(11) Multiply the two complex numbers
3
−
i
a
n
d
−
3
+
i
{\displaystyle 3-i~~~~~{\rm {and}}~~~~-3+i}
(12) Perform the following integrations:
∫
(
1
3
x
+
1
3
x
2
−
5
x
−
6
)
d
x
{\displaystyle \int \left({\frac {1}{3x}}+{\frac {1}{3x^{2}}}-5x^{-6}\right){\rm {d}}x}
(13)
∫
(
6
x
−
2
+
2
x
−
8
x
2
)
d
x
{\displaystyle \int \left(6x^{-2}+{\frac {2}{x}}-{\frac {8}{x^{2}}}\right){\rm {d}}x}
(14)
∫
(
cos
2
θ
+
θ
)
d
θ
{\displaystyle \int \left(\cos ^{2}\theta +\theta \right){\rm {d}}\theta }
(15)
∫
(
sin
3
θ
cos
θ
+
2
θ
)
d
θ
{\displaystyle \int \left(\sin ^{3}\theta \cos \theta +2\theta \right){\rm {d}}\theta }
(16) Integrate by parts:
∫
2
x
cos
x
d
x
{\displaystyle \int 2x\cos x{\rm {d}}x}
(17) What is the equation corresponding to the determinant:
|
x
−
1
0
−
1
x
0
0
0
x
|
=
0
{\displaystyle {\begin{vmatrix}x&-1&0\\-1&x&0\\0&0&x\\\end{vmatrix}}=0}
(18) Express
x
−
1
(
x
+
3
)
(
x
−
4
)
{\displaystyle {\frac {x-1}{(x+3)(x-4)}}}
(19)What is the general solution of the following differential equation:
d
θ
d
r
=
r
A
{\displaystyle {\frac {{\rm {d}}\theta }{{\rm {d}}r}}={\frac {r}{A}}}
(20) What is
e
i
2
ϕ
−
2
i
sin
2
ϕ
{\displaystyle e^{i2\phi }-2i\sin 2\phi }
