Exercise No 4_answer

Exercise No 4(Online)-With answers 1. The complex numbers z1 and z2 satisfy the equation z 2 = 1 − 2 2i . (a) Express z1 and z2 in the form of a+bi, where a and b are real numbers. (b) Represent z1 and z2 in the argand diagram. (c) For each z1 and z2, find the modulus and argument (in radian). Answer: (a) Let z = a + bi (a + bi ) 2 = 1 − 2 2i a 2 − b 2 + 2abi = 1 − 2 2i Equate the real parts and the imaginary parts: 2 a 2 − b 2 = 1, 2ab = −2 2 → b = − a 2 a 2 − 2 = 1 → a 4 − a 2 − 2 = 0 → (a 2 − 2)(a 2 + 1) = 0 → a = ± 2 , [a 2 + 1 = 0 reject ] a a = 2 , b = −1, a = − 2 , b = 1 z1 = 2 − i

z2 = − 2 + i

(b) y

(-√2,1) z2 0

z1

x (√2,-1)

(c) z1 = z2 = 2 + 1 = 3 −1 Argz1 = tan −1 ( ) = −0.515 look at argand diagram. 2 1 Argz2 = tan −1 ( ) = π − 0.615 = 2.526 − 2 2. Using the method of completing the square, or otherwise, solve the equation z 2 + 4 z = 4 − 6i . Hence determine z and arg z. Answer: z 2 + 4 z = 4 − 6i ( z + 2) 2 − 4 = 4 − 6i ( z + 2) 2 = 8 − 6i

Let z = a + bi ( z + 2) 2 = ( a + 2 + bi ) 2 → (a + 2)2 − b 2 + 2(a + 2)bi = 8 − 6i

(a + 2) 2 − b 2 = 8, 2(a + 2)b = −6 → b = − (a + 2) 2 −

[(a + 2)

2

3 a+2

9 = 8 → (a + 2) 4 − 8(a + 2) − 9 = 0 2 (a + 2)

][

]

− 9 (a + 2) 2 + 1 = 0

[

a + 2 = ±3 (a + 2) 2 + 1 = 0 has no solution a = −5,1 a = −5, b = 1 a = 1, b = −1 z = −5 + i, 1 − i π z = 26 , 2 , and Argz = 2.94,− 4

]

3. Using the algebraic laws of sets, show that ( A ∩ B )′ − ( A′ ∩ B ) = B′ . Answer: LHS = ( A ∩ B)′ − ( A′ ∩ B ) = ( A ∩ B )′ ∩ ( A′ ∩ B )′ = ( A′ ∪ B′) ∩ ( A ∪ B′) = ( B′ ∪ A′) ∩ ( B′ ∪ A) = B′ ∪ ( A′ ∩ A) = B′ ∪ φ = B′ = RHS 4. Given x = log a b, y = logb c and z = log c a , show that xzy=1. Answer: xyz = (log a b)(logb c)(logc a ) =

lg b lg c lg a × × =1 lg a lg b lg c

5. Find the least integral value of n such that log10 (2n + 1) − log10 2n < log10 1.0025 . Answer: 2n + 1 log10 < log10 1.0025 2n 2n + 1 < 1.0025 2n 1 1 1+ <1+ 2n 400 1 1 < 2n 400 1 1 < n 200 n > 200 Least integral value of n is 201. 6. One of the roots of the equation 21x 3 − 50 x 2 − 37 x − 6 = 0 is a positive integer. Find this root and hence, solve the equation completely.

Answer: Let f ( x ) = 21x 3 − 50 x 2 − 37 x − 6 By trial and error, f (3) = 21(3)3 − 50(3) 2 − 37(3) − 6 = 0 Hence, x=3 is the positive integer root. f ( x) = ( x − 3)(21x 2 + 13 x + 2) = ( x − 3)(7 x + 2)(3x + 1) = 0 2 1 x = 3,− , − 7 3 7. Find the set of values of x such that − 4 < x 3 − 2 x 2 + 2 x − 4 < 0 . Answer: − 4 < x3 − 2 x 2 + 2 x − 4 < 0 − 4 < x 3 − 2 x 2 + 2 x − 4 and x 3 − 2 x 2 + 2 x − 4 < 0 x 3 − 2 x 2 + 2 x > 0 and x 2 ( x − 2) + 2( x − 2) < 0 x( x 2 − 2 x + 2) > 0 and ( x 2 + 2)( x − 2) < 0 Consider x 2 − 2 x + 2 Since b 2 − 4ac = 4 − 16 < 0 x 2 − 2 x + 2 > 0 For x( x 2 − 2 x + 2) > 0 , x>0→I Consider ( x 2 + 2)( x − 2) < 0 , since x 2 + 2 > 0 , for ( x 2 + 2)( x − 2) < 0 , x<2→II Therefore the answer is the intersection between I and II, {x: 0<x < 2} 8. The function f is defined as f ( x) = 2 x 2 + 4 x + 5, x ∈ ℜ . (a) Find the set of values of x such that f(x) <3x2. Answer: 2 x 2 + 4 x + 5 < 3 x 2 → x 2 − 4 x − 5 > 0 → ( x − 5)( x + 1) > 0 → { x : x < −1, x > 5} (b) Find the set of values of k so that the equation f(x) =kx has no real roots. Answer: 2 x 2 + 4 x + 5 = kx 2 x 2 + (4 − k ) x + 5 = 0 No real roots, b 2 − 4ac < 0 (4 − k ) 2 − 4(2)(5) < 0 16 − 8k + k 2 − 80 < 0 k 2 − 8k − 64 < 0 Consider k 2 − 8k − 64 = 0 k =4±4 5

{

Ans : k : 4 − 4 5 < k < 4 + 4 5

9. Express

}

x3 in partial fractions. ( x + 1)( x + 2)

Answer: By doing long division, x3 7x + 6 = ( x + 3) + ( x + 1)( x + 2) ( x + 1)( x + 2) 7x + 6 A B = + → A = −1, B = 8 ( x + 1)( x + 2) x + 1 x + 2 x3 1 8 = ( x + 3) − + ( x + 1)( x + 2) x +1 x + 2 10. The 2nd, 4th and 8th terms of an AP are 2x+3, 5x+1, and x2-23 respectively. Find the value of x and hence the sum of the first 12 terms of the series. Answer: a + d = 2 x + 3 → (i ) a + 3d = 5 x + 1 → (ii ) a + 7 d = x 2 − 23 → (iii ) (ii ) − (i ) → 2d = 3 x − 2 → (iv) (iii ) − (ii ) → 4d = x 2 − 5 x − 24 → (v ) (v ) − 2(iv) → 0 = x 2 − 5 x − 24 − (6 x − 4) x 2 − 11x − 20 = 0 11 ± 201 2 11 + 201 27 + 201 29 + 3 201 x= ,a = ,d = , 2 4 4 Taking 1119 + 105 201 S12 = 2 [Note: I think something must be wrong with this question since the answer is quite ridiculous, but what important is the method of getting the values for x, then finding a, d and sum of the first 12 terms.] x=