Exercise No 3_answer

Set No 3 1. If ( 2 x + 3 yi) 2 = 2 x + 3 yi where x and y are real, find the possible values of x and y. Answer: 4 x 2 + 12 xyi − 9 y 2 = 2 x + 3 yi (4 x 2 − 9 y 2 ) + 12 xyi = 2 x + 3 yi 4 x 2 − 9 y 2 = 2 x, 12 xy = 3 y 12 xy = 3 y →12 xy − 3 y = 0 → 3 y (4 x − 1) = 0 → y = 0, x =

1 4

y = 0 → 4 x 2 = 2 x → 4 x 2 − 2 x = 0 → 2 x(2 x − 1) = 0 → x = 0, x =

1 2

1 1 1 1 → 4( ) − 9 y 2 = 2( ) →1 − 36 y 2 = 2 → y 2 = − → no solution 4 16 4 36 ans : y = 0, x = 0 1 y = 0, x = 2 x=

2. Solve the equation log3 x + log 3 x − log 9 27 = 0 . Leave your answer in the surd form. Answer: log3 x 2 + log3 x − log9 27 = 0 2

log3 x 2 ( x) −

log3 27 =0 log3 9 1

log3 x 3 − log3 27 2 = 0 log3

x3 27

x3 1 2

1 2

=0

( )

1 3 2

= 1→ x = 3 3

3

1  12    → x =  3  → x = 32 = 3   3

27 3. Show that 1 is the only real root of x 3 + 3 x 2 + x − 5 = 0 . Answer: x3 + 3x 2 + x − 5 = 0

let p( x) = x 3 + 3 x 2 + x − 5 p (1) = 13 + 3(1) 2 +1 − 5 = 0

By using factor theorem, x=1 is the root of the equation x3 + 3x 2 + x − 5 = 0 p ( x) = x 3 + 3 x 2 + x − 5 = ( x −1)( x 2 + 4 x + 5) consider x 2 + 4 x + 5 b 2 − 4ac = 16 − 4(1)(5) = −4 < 0

no real roots, hence x = 1 is the only real root for the equation x3 + 3x 2 + x − 5 = 0

4. Show that x +1 is a factor of the polynomial x 3 − 12 x 2 + 23x + 36 . (a) Find all the real roots of the equation x 6 − 12 x 4 + 23x 2 + 36 = 0 .

(b) Determine the set of values of x such that x 3 − 12 x 2 + 23x + 36 < −8 x − 8 . Answer: p ( x ) = x 3 −12 x 2 + 23 x + 36 p ( −1) = −1 −12 − 23 + 36 = 0

Hence, x+1 is a factor of x 3 − 12 x 2 + 23x + 36 x 3 −12 x 2 + 23 x + 36 = ( x +1)( x 2 −13 x + 36) = ( x +1)( x − 4)( x − 9)

(a) x 6 − 12 x 4 + 23 x 2 + 36 = 0

( x ) −12( x ) + 23( x ) + 36 = 0 ( x +1)( x − 4)( x − 9) = 0 2 3 2

2 2

2

2

2

x 2 = −1 no solution x 2 = 4 → x = ±2 x 2 = 9 → x = ±3

(b) x 3 − 12 x 2 + 23 x + 36 < −8 x − 8 x 3 −12 x 2 + 23 x + 36 < −8 x − 8 ( x +1)( x 2 −13 x + 36) < −8( x +1) ( x +1)( x 2 −13 x + 36) + 8( x +1) < 0 ( x +1)( x 2 −13 x + 36 + 8) < 0 ( x +1)( x 2 −13 x + 44) < 0

Consider x 2 − 13x + 44 b 2 − 4ac = (−13) 2 − 4(1)(44) = −7 < 0

This show that the expression x 2 − 13 x + 44 > 0 Therefore x 3 − 12 x 2 + 23 x + 36 < −8 x − 8 when {x: x < -1}. 5. Prove that x 2 + y 2 ≥ 2 xy . If a and b are real numbers, prove that a 2 + b 2 + c 2 − ab − bc − ca ≥ 0 . Answer: ( x − y)2 ≥ 0

Since x and y are real numbers, → x 2 − 2 xy + y 2 ≥ 0 → x 2 + y 2 ≥ 2 xy

Using the above relation, a 2 + b 2 ≥ 2ab, a 2 + c 2 ≥ 2ac, b 2 + c 2 ≥ 2bc

Combining the 3 inequalities a 2 + b 2 + a 2 + c 2 + b 2 + c 2 ≥ 2ab + 2ac + 2bc 2(a 2 + b 2 + c 2 ) ≥ 2(ab + ac + bc) a 2 + b 2 + c 2 − ab − bc − ac ≥ 0

6. Show that the roots of the equation mx 2 + nx + k = 0 are given by x=

− n ± n 2 − 4mk . 2m

Answer:

mx 2 + nx + k = 0 n k x2 + x + = 0 m m 2  n 2  n   k ) −  + =0 ( x + 2m  2m   m  2

n  n2 k  − x +  = 2 2m  4m m  2

n  n 2 − 4mk  x +  = 2m  4m 2  x+

n n 2 − 4mk =± 2m 4m 2

x=− x=

n ± 2m

n 2 − 4mk 2m

− n ± n 2 − 4mk 2m

Deduce that if p+qi, p and q are real numbers is a root of the quadratic equation, then the other root is p-qi. − n ± n 2 − 4mk Answer:Since x = p + qi is the root, compare with x = , 2m

n n − 4mk , = qi , therefore the other root is x = p − qi 2m 2m (a) Show that 1+i is a zero of the polynomial f ( x) = 2 x 3 − 3 x 2 + 2 x + 2

p =−

2

and find the other zero. Answer: If 1+i is the root, then 1-i is also a root of the polynomial f ( x) = 2 x 3 − 3 x 2 + 2 x + 2

Then must show that [x-(1+i)][x-(1-i)]=x2-2x+2 is the factor of f(x) By using long division we can show that f ( x) = 2 x 3 − 3 x 2 + 2 x + 2 = ( x 2 − 2 x + 2)( 2 x + 1)

Hence the other zero is – ½ . (b) Find the polynomial g(x) such that f(x)-xg(x) = 2- x. By expressing g(x) in the form of a ( x −b) 2 +c , where a, b and c are real numbers, find the maximum value of Answer: f(x)-x g(x) = 2- x

1 . g ( x)

2 x 3 − 3 x 2 + 2 x + 2 − xg ( x ) = 2 − x xg ( x ) = 2 x 3 − 3 x 2 + 3 x g ( x) = 2 x 2 − 3x + 3 3  3 9   g ( x ) = 2 x 2 − x  + 3 = 2 ( x − ) 2 −  + 3 2  4 16    3 15 = 2( x − ) + 4 8 15 Since the minimum value of g(x) is , the maximum value of 8 1 8 is g ( x) 15 2

7. Find the set of values of x so that the series 1 +

x −1  x − 1  +  +  converges. x  x 

Find the sum to infinity of this series when x = 3. Answer: 2

1+

x − 1  x −1  +  +  is a geometric series. x  x 

In order for the series to converge, the absolute value of common ratio must be less than 1. x −1 x −1 x −1 x −1 , x < 1 → −1 < x < 1 → x > −1 and x x −1 + x x −1 − x > 0 and <0 x x 2 x −1 −1 > 0 and and <0 x x 1 x < 0, x > and x > 0 2 1   x : x >  2   2 1 r = , S∞ = =3 2 x=3, 3 1− 3

r=

x −1 <1 x