Exercise # 5 1. Part # I [Previous Year Questions] If z and  are two non-zero complex numbers such that |z| = 1, and arg (z) – arg() = (A) 1 (B) – 1 (C) i  , then z  is equal to : 2 (D) – i x 2.  1 i  If   = 1, then 1– i  (A) x = 4 n, where n is any positive integer (B) x = 2 n, where n is any positive integer (C) x = 4 n + 1, where n is any positive integer (D) x = 2n + 1, where n is any positive integer 3. Let z,w be complex numbers such that z  iw  0 and arg zw = . Then arg z equals : (A) 4.  4 (B)  2 (C) The conjugate of a complex number is (A) – 1 i –1 (B) 1 i 1 3 4 1 . Then, that complex number isi –1 1 (C) – i 1 (D) 5 4 (D) 1 i –1 5. Let , be real and z be a complex number. If z2 + z +  = 0 has two distinct roots on the line Re z = 1, then it is necessary that : (A)   (0, 1) (B)  (–1, 0) (C) || = 1 (D)   (1, ) 6. The sum of all real values of x satisfying the equation (x2 – 5x + 5)x (A) 3 (B) – 4 (C) 6 2 4x60 Part # II (D) 5 [Previous Year Questions] 2 1. Find sum of all the real roots of the equation x  2  x  2  2  0 2. Find the set of all solutions of the equation 2 |y| – | 2 y–1 – 1| = 2 y–1 + 1. 3. The equation x  1 – (A) no solution (C) two solutions 4. = 1 is : x 1 = 4x  1 has Find all real numbers x which satisfy the equation 2 log2 log2 x + log1/2 log2 (2 2 x) = 1. (B) one solution (D) more than two solutions BASIC MATHS & LOGARITHM 1 1 1 (1) If z 1, z 2, z 3 are complex numbers such that z1 = z2 = z3 = z  z  z 1 2 3 5. (A) equal to 1 (B) less than 1 (C) greater than 3 = 1, thenz1 + z 2 + z3 is : (D) equal to 3 (2) If arg (z) < 0, then arg ( z)  arg (z) = (A)  (B)  (C)   2 (D) 6. Solve the equation log3/4 log8 (x2 + 7) + log1/2 log1/4 (x2 + 7) 1 =  2. 7. Number of solutions of log4(x–1) = log2(x – 3) is (A) 3 (B) 1 (C) 2 (D) 0 8. The set of all real numbers x for which x2 – |x + 2| + x > 0 is (A) (– , – 2 )  (2, ) (B) (– , – 2 )  ( 2 , ) (C) (– , – 1)  (1, ) (D) ( 2 , ) 9. If z1 and z2 are two complex numbers such that |z 1| < 1 < |z 2| then prove that 10. Let f(x) = 11. 1  z1 z2 z1  z 2 < 1. x 2  6x  5 x 2  5x  6 Column –  If – 1 < x < 1, then f(x) satisfies If 1 < x < 2, then f(x) satisfies If 3 < x < 5, then f(x) satisfies If x > 5, then f(x) satisfies (A) (B) (C) (D)  2 Column –  0 < f(x) < 1 f(x) < 0 f(x) > 0 f(x) < 1 (p) (q) (r) (s) Let (x0, y0) be the solution of the following equations 2x ln 2 lnx 3 =2 Then x0 is (A)  (3 y ) ] ln 3 lny 1 6 (B) 1 3 (C) 1 2 (D) 6 12.   1 1 The value of 6  log 3  1 4  1 4 4 ......  is   3 2 3 2 3 2 2 3 2   13. If 3x = 4x–1, then x = 2 (B) 2  log 3 2 2log3 2 (A) 2log 2  1 3 1 (C) 1  log 3 4 2log 2 3 (D) 2log 3  1 2 EXERCISE - 5 : PART # I 1. D 2. A 3. C 4. C 2 A 7. B 5. D 6. A PART - II 3. A 13. ABC 5. 1 A 8. B 10. A  prs, B  qs, C  qs, D prs 12. 4

Basic Maths & Logarithm Exercise # 5

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