Solutions to list of exercises MQ 1

1

Solutions to List of Exercises MQ 1

1. This question simply involves plugging the appropriate values of x into the formulae.

(a) x=0, f(0) = 3(02₎ – 40 + 2 = 2.

x=2, f(2) = 3(22_{) –} 42 + 2 = 6.

x=–1, f(–1) = 3(–1)2– 4(–_{1) + 2 = 9.}

(b) x=0, f(0) =4(02) + (20) – _{3 =} –_3.

x=3, f(3) =4(32) + (23) – _{3 = 39.}

x=a, f(a) =4(a2) + (2a) – _{3 = 4a}2_+2a–_3.

x=(3+a), f(3+a) =4((3+a)2) + (2(3+a)) – 3 = 4(9+6a+a2_)+6+2a–_{3 =} 4a2_+26a+39.

(c) In general, the answer is no. We can see from the final example compared with the sum of the previous two in part (b) above that the rule does not follow when f is a quadratic function – in other words, f(3) + f(a)  f(3+a).

2. To answer this question, we simply need to use the rules for manipulating powers so that we group like terms together.

(a) 4x5  6x3 _{= 24x}8_.

(b) 3x2  4y2  8x4  –_2y4 ₌ –_192x6_y6_. (c) (4p2_q3₎3 ₌₄3_p23_q33 _{= 64p}6_q9_.

(d) 6x5_÷3x2 _{= (6/3)x}5–2 _{= 2x}3_. (e) 7y2_÷2y5 _{= (7/2)y}2–5 _{= 7/(2y}3_).

(f) _{5 2} 2 4

4 3 7

3 2 2

4 4 3 3

3 2

4 3

9 9

9 )

( 2

) ( 6 ) ( 3

yz x y

x z y x x

y x

z x y x x

xy xz xy

 

 

.

2 (h) (xy)3– _x3_y3 _{= x}3_y3– _x3_y3 _{= 0.}

3. Again, here we use the rules for manipulating powers, but now the powers are not integers and thus effectively involve taking roots.

(a) 1251/3 ₌ 3 ₁₂₅_5.

(b) 641/3 ₌ 3 ₆₄ _4.

(c) 161/4 ₌ 4₁₆ _2.

(d) 93/2 ₌ 2 ₉3 2 ₇₂₉_27.

(e) 92/3 ₌ 3 ₉2 3 ₈₁_{4.33(to 2 decimal places).}

(f) 811/2 _{+ 64}1/2 _{+ 64}1/3 _{= 81} ₆₄3 ₆₄₉₈₄_21.

4. This question involves doing the reverse of the steps in the previous one. (a) 9 = 32_.

(b) 625 = 54_. (c) 125–1 _{= 5}–3_.

5. Answering this question involves simply rearranging the expression so that the terms in x are gathered together on one side of the equation and all of the number terms are on the other side.

(a) 3x–6 = 6x –12, 3x = 6, x = 2.

(b) 2x–304x + 8 = x + 9–3x+4, –302x = –2x+5, 300x = –5, x = –5/300 = –0.017 (to 3 decimal places).

(c)

3 6 2 2

3 



 x

x

, 3(x+3) = 2(2x–6), 3x+9 = 4x–12, x–21 = 0, x = 21.

6. This question involves using the rules of sigma and pi notation to list and simplify all of the terms in the expressions.

3 (a)



1

j j = 1+2+3 = 6.

(b) 5



 

 

 

 



2              



(c)



i 1x(with n = 4 and x = 3) = 3 3 3 3 3 12 4

1     



(d)



j x(with x = 2) = 2 2 2 2 8

3

1    



(e) 3 1 2 3 6 1    



7. Recall that the equation for a straight line is y = a + bx where a is the intercept and b is the gradient. In the first three cases, the answers arise naturally by plugging the appropriate values directly into this formula, but the others require more thought.

(a) y = –1 + 3x. (b) y = 4 – 2x. (c) y = 3 + ½x.

(d) Care is needed here. If the line crosses the x–axis at 3, this is not the intercept. To get the intercept, we need to solve y = a + (1/2)x for a with y = 0 and x = 3, which gives 0 = a + (1/2)3; so a = –3/2 and thus the line is y = – 3/2 + (1/2)x.

(e) We first need to find the gradient by solving y = a + bx for 2, x = 3 and y = 1; so 1 = 2 + 3b, b = –1/3 and thus the line is y = 2 –(1/3)x.

(f) We first need to find the intercept by solving y = a + bx for a with b = 4, x = – 2, y = –2; so –2 = a +4–2, a = 6 and thus the line is y = 6 + 4x.

4 8. (a) y = 6x, 6

dx dy

, ₂ 0 2  dx y d .

(b) y = 3x2 _{+ 2, x}

dx dy

6

 , ₂ 6 2  dx y d .

(c) y = 4x3 _{+ 10,} 2 12x dx dy

 , x

dx y d 24 2 2  .

(d) 1 1, 2 ₂1

x x dx dy x x

y       , ₂ 3 ₃

2 2 2 x x dx y d    .

(e) y = x, 1

dx dy

, ₂ 0 2  dx y d .

(f) y = 7, 0

dx dy

, ₂ 0 2  dx y d .

(g) 6 3 6₃ 12 3, 36 4 36₄

x x dx dy x x x

y         , ₂ 5 ₅

2 144 144 x x dx y d    .

(h) y = 3ln x,

x dx dy _ 3

, ₂ 2 ₂

2 3 3 x x dx y

d _  _ 

.

(i) y = ln(3x2_),

x x x x f x f dx dy 2 3 6 ) ( ) ( ' 2  

 , ₂ 2 ₂

2 2 2 x x dx y

d _  _

.

(j) _{3 2 3} 1 2 3

2 4 4 6 3 4 1 6 3 4 6

3    _{ } _ _ _{ }  _  _ 

 x x x x

x x x x x x x x y , 4 3 2 4 3

2 6 2 12

3 12 2 6 3 x x x x x x dx

dy _{ }  _  _  _{    ,}

5 4 3 5 4 3 2 2 48 6 12 48 6 12 x x x x x x dx y

d _  _  _  _ _ _

.

9. All we do here is to apply the standard rule for differentiation for x (treating all the terms in y as if they were constants) and then separately for y (treating all the terms in x as if they were constants).

(a) z = 10x3 _{+ 6y}2– _7y, 2 30x x z   

; 12 7   y y z .

(b) z = 10xy2– _6, 2 10y x z _

 

; xy

5 (c) z = 6x, 6

 

x z

; 0  

y z

.

(d) z = 4, 0  

x z

; 0  

y z

.

10. (a) x2_{– 7x – 8 = (x – 8)(x+1).} (b) 5x – 2x2 _{= x(5} – _2x).

(c) 2x2– _x – _{3 = (2x} – _{3)(x + 1).} (d) 6 + 5x – 4x2 _{= (2} – _{x)(3 + 4x).} (e) 54 – 15x – 25x2 _{= (9 + 5x)(6} – _5x). 11. (a) 53 _{= 125, log5 125 = 3.}

(b) 112 _{= 121, log11 121 = 2.} (c) 64 _{= 1296, log6 1296 = 4.} 12. (a) log10 10000 = 4.

(b) log2 16 = 4. (c) log10 0.01 = –2. (d) log5 125 = 3. (e) loge e2 = 2.

13. (a) log5 3125 = 5, 55 _{= 3125.} (b) log49 7 = ½, 491/2 _{= 7.} (c) log0.5 8 = –3, 0.5–3 _{= 8.}

14. (a) log 60 = log(435) = log(2235) = 2log 2 + log 3 + log 5.

6 15. (a) log 27 – log 9 + log 81 = log 33– _{log 3}2 _{+ log 3}4 _{= 3log 3} – _{2log 3 + 4log3 =}

5log 3.

(b) log 8 – log 4 + log 32 = log 23– _{log 2}2 _{+ log 2}5 _{= 3log 2} – _{2log 2 + 5log 2 =} 6log 2.

16. (a) log x4 – _{log x}3 _{= log 5x} – _{log 2x, }            

x x x

x

2 5 log log ₃

4

, 

           

x x x

x

2 5 3 4

, x =

5/2.

(b) log (x–1) + log (x+1) = 2log (x+2), log((x1)(x1))log(x2)2, (x+1)(x– 1) = (x+2)2_{, x}2– _{1 = x}2 _{+ 4x + 4, 4x =} –_{5, x =} –_5/4.

(c) log10 x = 4, x = 104 _{= 10000.}

17. (a) log 16 = log (82) = log 8 + log 2 = log 8 + log 81/3 = 4/3 log 8 = 4/3  2.1 = 2.9 (to one decimal place)

(b) log 64 = log (88) = log 8 + log 8 = 2log 8 = 4.2 (to one decimal place). (c) log 4 = log (8/2) = log 8 – log 2 = log 8 – log 81/3 _{= log 8 – 1/3 log 8 = 2/3}

log 8 = 2/3  2.1 = 1.5 (to one decimal place).

18. (a) 4x _{= 6, x = log4 6 = 1.3 (to one decimal place).}

(b) 42x _{= 3, 2x = log4 3 = 0.79, x = 0.79/2 = 0.4 (to one decimal place).} (c) 32x–1 _{= 8, (2x}–_{1) = log3 8 = 1.9, x = 1.4 (to one decimal place).}

19. To find the minimum of a function, we need to differentiate the function and set this first derivative to zero. We would then find the second derivative and ensure that this is above zero to verify that the turning point is indeed a minimum. To find the value of the function at the minimum, we need to plug the value of x obtained back into the original formula for y.

(a)y = 6x2– _10x – _8, _12x₁₀₀

dx dy

, x = 5/6. ₂ 12 0 2

 

dx y d

. If x = 5/6, y = –12.2 (to one decimal place).

7 and so on) or we could use the rule for differentiating a power of a function, which is ) ( ' ) ( )

(x nf x 1f x f

dx

d n  n

. So for y = f(x) = (6x2_{– 8)}2_,

0 192 144 12 ) 8 6 (

2 2   3 

 x x x x

dx dy

.

If we solve this, we get three solutions: x=0, x=4/3 and x=–4/3. The second derivative is 432 2 192

2 2   x dx y d

. We need to plug each of the roots x=0, x=4/3 and x=–4/3 into this equation sequentially to identify which corresponds to a minimum. When x=0, 192 0

2 2    dx y d

; when x=4/3, 576 0 2 2   dx y d

; and when

x=–4/3, 576 0 2 2   dx y d

. So the function has two minima when x=4/3 and when x=–4/3 and one maximum when x=0. When x=4/3, y=64/9 and when x=–4/3, y=64/9.

20. Suppose we picked:

               7 2 1 4 , 6 1 0 3 B A .

The first thing to note is that the matrices are conformable for multiplication as A is 22 and B is 22 so AB will also be 22.

                     41 8 3 12 7 2 1 4 6 1 0 3

AB , _

                 12 8 3 41 468 1 41 8 3 12 1 1 AB .

We also have _

         4 2 1 7 26 1 1

B , ,

3 1 0 6 18 1 1        

A so that

                          12 8 3 41 468 1 3 1 0 6 4 2 1 7 18 1 26 1 1 1 A B .

Hence this example has illustrated that (AB)–1 _=B–1_A–1_.

8 To multiply matrices by a scalar (including scalars less than one), we simply multiply each element of that matrix by the scalar. So

, 8 4 12 2 4 2 6 1 2 2 _                A _                12 18 24 9 4 6 8 3 3

3B ,

                          0 5 . 1 5 . 0 0 1 3 0 3 1 0 2 6 2 1 2 1 D .

(c) Recall that the trace is the sum of the elements on the leading diagonal. So Tr(A) = 1 + 4 = 5, Tr(B) = –3+4 = 1.

                         8 4 2 2 4 6 8 3 4 2 6 1 B

A and Tr(A+B) = –2 + 8 = 6.

Tr(A) + Tr(B) = 5 + 1 = 6 = Tr(A+B).

(d) The rank is the number of linearly independent rows or columns in a square matrix. We can see in the case of the matrix A that all the columns and rows are linearly independent of one another and hence the matrix is of full rank, 2.

(e) To find the eigenvalues, , of the matrix A+B, we need to solve 0

)

(AB I  , where I is a 22 identity matrix. From part (c) above,          8 4 2 2 B A . 0 1 0 0 1 8 4 2 2               

 , so 0

8 4 2 2             

and thus (–2–)(8–) – (4–2) = 0. So we have –16 – 8 + 2 + 2 _{+ 8 = 0,} 2 – 6 – _{8 = 0. This} equation does not factorise and thus the quadratic formula can be used, giving the eigenvalues as  = –1.12 and  = 7.12 to two decimal places.

(f) An identity matrix of order 12 will be a square 1212 matrix having one as each element of the leading diagonal and zero elsewhere. Since the trace is the sum of

the elements on this leading diagonal, it will be 1+1+…+1 = 12.

22. (a) _

9

(b) _

  

  

 

   

 

  



        

 

      

  



8 14

1 5 4

4 7 7

0 1 3 2 4 7

0 3 4

7 1 2

.

(c)The inverse of _   

  

 2 4

1 3

is _

  

      

 

5 . 1 2

5 . 0 1 3 4

1 2 2 1

.

(d)No, since the determinant is 32 – 32 = 0, the matrix is not of full rank and therefore its inverse does not exist.

23. (a) E(ax + by) = aE(x) + bE(y). (b) E(axy) = aE(x)E(y).