Basic Maths & Logarithm Mock Test

MOCK TEST SECTION - I : STRAIGHT OBJECTIVE TYPE 1. Greatest integer less than or equal to the number log2 15. log1/6 2. log3 1/6 is (A) 4 2. (B) 3 (C) 2 (D) 1 The solution set of the inequation 1 + log 1 (x2 + x + 1) > 0 is 3 (A) (– , –2)  (1, ) 3. (D) (– ,  ) (B) (– , –2)  (3, ) (C) (– , 3) (D) [–3, –2]  [2, 3] (C) [7, ) (D) (–, –2] Solution set of |x2 – 5x + 7| + |x2 – 5x – 14| = 21 is (B) (–, –2]  [7, ) (A) [–2, 7] 5. (C) (–2, 1) If |x2 – 9| + |x2 – 4| = 5, then the set of values of x is (A) (– , –3)  (3, ) 4. (B) [–1, 2] Consider the following statements : log11 7 and y = S1 : x = log 7 11 , then ey n7 – x n11 is equal to 1. S2 : logx 3 > logx 2 is true for all values of x  (0, 1)  (1, ) S3 : |x – 2| = [–], then x is 6, –2 S4 : log25 (2 + tan2 ) = 0.5, then  may be 2 4 or 3 3 State, in order, whether S1, S2, S3, S4 are true or false (A) T F F T 6. (B) F F T T (B) – 4 (B) 4 (B) 2 (C) 3 (D) Infinite 1 is 2 (D) 4 Solution set of the inequality loge2 [2x] – loge [2x]  0 is (A) [1, 3) (B) (0, 3) 1 log5 9 10. (D) none of these (C) 3 Number of values of x satisfying the equations 5{x} = x + [x] and [x] – {x} = (A) 1 9. (C) 0 The number of solutions of |[x] – 2x| = 4 is (where [.] denotes greatest integer function) (A) 2 8. (D) T T F T The product of all the solutions of the equation (x –2)2 – 3 |x – 2| + 2 = 0 is (A) 2 7. (C) F T F T N= 81 (A) 0 3 409 3 log 6 3 (C) {1, 2} 1 3  (D)  ,  2 2  2   log 25 7 – (125)log25 6  , then log N has the value  ( 7)   2   (B) 1 (C) –1 (D) None of these BASIC MATHS & LOGARITHM SECTION - II : MULTIPLE CORRECT ANSWER TYPE 11. If n (x + z) + n (x  2y + z) = 2 n (x  z), then (A) y = 12. 13. 14. 2 xz xz (B) y2 = xz (D) xy x = yz z log10 5 . log10 20 + (log10 2) 2 when simplified reduces to (A) an odd prime number (B) an even prime (C) a rational number (D) an integer If x2 x < 2, then the set of values of x is x (A) (– , 1)  (2, ) (B) (– , 0)  (1, ) (C) (– , –1) (0, ) (D) none of these Which of the following when simplified reduces to unity ? (A) log3 log27 log4 64 (C) log2 15. (C) 2y = x + z (B) 2 log18 ( 2  8 )  2  10 + log2    5 (D) – log 2 1 ( 2 + 1) If x, y, z  R+ and z R then the system x + y + z = 2, 2xy – z2 = 4 (A) is satisfied for x = 2, y = 2, z = – 2 (B) has only one real solution (C) has only two real solution (D) has infinite solutions. SECTION - III : ASSERTION AND REASON TYPE 16. Statement-I : If 2, 3 & 6 are the sides of a triangle then it is an obtuse angled triangle. Statement-II : If b2 > a2 + c2, where b is the greatest side, then triangle must be obtuse angled. (A) Statement-I is True, Statement-II is True; Statement-II is a correct explanation for Statement-I (B) Statement-I is True, Statement-II is True; Statement-II is NOT a correct explanation for Statement-I (C) Statement-I is True, Statement-II is False (D) Statement-I is False, Statement-II is True 17. Statement-I : Minimum value of |x – 2| + |x – 5| + |x + 3| is 8. Statement-II : If a < b < c, then the minimum value of |x – a| + |x – b| + |x – c| is |b – a| + |b – c|. (A) Statement-I is True, Statement-II is True; Statement-II is a correct explanation for Statement-I (B) Statement-I is True, Statement-II is True; Statement-II is NOT a correct explanation for Statement-I (C) Statement-I is True, Statement-II is False (D) Statement-I is False, Statement-II is True 18. Statement-I : logex > logey  x > y, x, y > 0 Statement-II : If logax > logay, then x > y (x, y > 0 & a > 0, a  1) (A) Statement-I is True, Statement-II is True; Statement-II is a correct explanation for Statement-I (B) Statement-I is True, Statement-II is True; Statement-II is NOT a correct explanation for Statement-I (C) Statement-I is True, Statement-II is False (D) Statement-I is False, Statement-II is True 19. Statement-I : [x] + [– x] = x2 – 5x + 6 has only two real solution.  1 , x  I Statement-II : [x] + [–x] =   0 , xI (A) Statement-I is True, Statement-II is True; Statement-II is a correct explanation for Statement-I (B) Statement-I is True, Statement-II is True; Statement-II is NOT a correct explanation for Statement-I (C) Statement-I is True, Statement-II is False (D) Statement-I is False, Statement-II is True 20. Statement-I : If ax2 +x +  = 0 , where ,   R, has roots a, b and logab < 0, then  +  < 0. Statement-II : If ax2 +x +  = 0 , where ,   R, has roots a, b and logab < 0, then a +  + must be negative. (A) Statement-I is True, Statement-II is True; Statement-II is a correct explanation for Statement-I (B) Statement-I is True, Statement-II is True; Statement-II is NOT a correct explanation for Statement-I (C) Statement-I is True, Statement-II is False (D) Statement-I is False, Statement-II is True SECTION - IV : MATRIX - MATCH TYPE 21. Match the following Column –  Column –  (A) If logsinx (log3 (log0.2 x)) < 0, then (p) x  [–1, 1] (B) If (q) x  [–3, 6) (C) If |2 – | [x] – 1| |  2 , then (r) 1   x   0,   125  (s) x  (1, ) (t) 3  x  (–, –1)   ,   2  (D) 22. (e x  1)(2x  3)(x 2  x  2)  0, then (sin x  2) x(x  1) (where [.] represents greatest integer function).  If |sin–1 (3x – 4x3)|  , then 2 Match the column Column- I (A) Interval containing all the solutions of the Column-II    (p)  – ,   2 2 inequality 3 – x > 3 1– x 2 is (B) Interval containing all the solutions of the 1 inequality    3 (C) (D) x 2 < 3 –x is Interval containing all the solutions of the 1 1 inequality log5(x – 3) + log53 < log5 (2x2 – 6x + 7) is 2 2 Interval containing all the solutions of the 1 equation 7x + 2 – . 7x + 1 – 14.7x – 1 + 2.7x = 48 is 7 (q) (, 2) (r) (–, ) (s) (– e, e) (t) ([], – [–2]), where [.] is G.I.F. BASIC MATHS & LOGARITHM SECTION - V : COMPREHENSION TYPE 23. Read the following comprehension carefully and answer the questions. The general procedure for solving equation containing modulus function is to split the domain into subintervals and solve the various cases. But there are certain structures of equations which can be solved by a different approach. For example, for solving the equation |f(x)| + |g(x)| = f(x) – g(x) one can follow this method. First find the permissible set of values of x for the equation. Since LHS  0  f(x) – g(x)  0. Now squaring both sides, we get f2 + g2 + 2|f.g| = f2 + g2 – 2fg  |fg| = – fg. The equation can hold if f.g  0 and f  g. This can be simplified to f  0, g  0. Answer the following questions on the basis of this method 1. The complete solution of the equation |x3 – x| + |2 – x| = x3 – 2 is (A) [2, ) 2. (D) none of these (B) [–3, 0]  [1, ) (C) (–, –3] (D) (–, –3]  [0, 1] All the condition(s) for which |f(x) – g(x)| = |f(x)| + |g(x)| is true, is (A) f(x)  0, g(x)  0 24.  1  (C)  2 3 ,     The complete solution set of the equation |x2 – x| + |x + 3| = |x2 – 2x – 3| is (A) [1, ) 3. (B) [–1, 0]  [2, ) (B) f(x)  0, g(x)  0 (C) f(x) . g(x)  0 (D) f(x) . g(x) = 0 Read the following comprehension carefully and answer the questions. An equation of the form 2m loga f(x) = loga g(x), a > 0, a 1, m  N f (x)  0 is equivalent to the system  2m f (x)  g(x) 1. The number of solutions of 2 loge 2x = loge(7x – 2 – 2x2) is (A) 1 2. (B) 2 (B) 1 (C) 2 (D) Infinite The number of solutions of log(3x2 + x – 2) = 3 log(3x – 2) is (A) 1 25. (D) Infinite The number of solutions of ln 2x = 2 ln(4x – 15) is (A) 0 3. (C) 3 (B) 2 (C) 3 Read the following comprehension carefully and answer the questions. Let a1 < a2 < a3 < ........< an , n is an odd natural number and m, k  N Consider the equation |x –a1| + |x – a2| + |x – a3| +....+ | x – an| = kx + d. Case-1 When 2m – n = k for some m < n, then the equation has (i) no solution if (am +1 + am + 2 + ....+ an) – (a1 + a2 + ....+ am) > d (ii) infinite solutions if (am +1 + am + 2 + ....+ an) – (a1 + a2 + ....+ am) = d (iii) two solutions if (am +1 + am + 2 + ....+ an) – (a1 + a2 + ....+ am) < d (D) 0 Case-2 Let when 2m – n  k for any m < n Two cases arise 1. (A) If |k| > n, then there is one solution. (B) If |k| < n, then there is m such that 2m – (n – 1) = k. (i) If (am +2 + am + 3 + ....+ an) – (a1 + a2 + ....+ am) > d no solution (ii) If (am +2 + am + 3 + ....+ an) – (a1 + a2 + ....+ am) = d one solution (iii) If (am +2 + am + 3 + ....+ an) – (a1 + a2 + ....+ am) < d two solutions Number of solutions of |x – 1| + |x – 3| + |x – 4| + | x – 7| + |x – 9| = 3x + 5 is (A) 0 2. (B) 1 (D) infinite Number of solutions of |x – 1| + |x – 3| + |x – 4| + | x – 7| + |x – 10| = 2x + 1 is (A) 0 3. (C) 2 (B) 1 (C) 2 (D) None of these Number of solutions of |x – 1| + |x – 2| + |x – 4| + | x – 6| + |x – 7| = 2x + 10 is (A) 0 (B) 1 (C) 2 (D) None of these SECTION - VI : INTEGER TYPE 26. The inequality log x 2 |x – 1| > 0 is not defined for some integral values of x, find the sum of their magnitudes. 27. If set of all real values of x satisfying |x2 – 3x – 1| < |3x2 + 2x + 1| + |2x2 + 5x + 2|, x2 – 3x – 1  0 is (– , – a)  (– b, ), then find the value of a + log ab. 28. If value of logplogp p p p p ..... p p is –, then find the value of . 1 44 2 4 43 2008 times 29. , 2  x  1  1  x  2 , 1  x  1 Let f(x) =  4  x , 1  x  2  Find number of solutions of {f(x)} = 30. 1 (where {.} denotes fractional part function) 2 Find the number of integral solution of the equation x   x    x   2     {x}   3   = 3.       (where [ ] denotes greatest integer function) MOCK TEST 1. C 2. C 3. D 4. A 5. 11. AD 12. CD 19. A 20. A 21. A  r, B  t, C  pqr, D  pq 23. 1 A 2 26. 4 27. 2 D 3 13. B A 10. A 6. 14. BD C 24. 1 B 2 28. 2008 29. 3 30. 1 B C 15. AB 7. B 8. 16. D A 9. 17. A 22. A  prs, B  rs, C  t, D  prs 3 D 25. 1 C 2 A D 18. C 3 C