Basic Maths & Logarithm Exercise # 3

BASIC MATHS & LOGARITHM Exercise # 3 [Matrix Match Type Questions] Part # I Following question contains statements given in two columns, which have to be matched. The statements in Column-I are labelled as A, B, C and D while the statements in Column-II are labelled as p, q, r and s. Any given statement in Column-I can have correct matching with one or more statement(s) in Column-II. Column –  1. (A) (B) (C) (D) Column –  Set of all values of x satisfying the inequation 5x  1 < 1 is (x  1) 2 Set of all values of x satisfying the inequation |x| + |x – 3| > 3 is (p) (–, 0) (0, 2)  (2, ) Set of all values of x satisfying the inequation 1 1  is | x | 3 2 Set of all values of x satisfying the inequation (r) (–, –1)(–1, 0)(3, ) (q) (–, –5)(–3, 3) (5, ) (s) (0, 3)  (4, ) 4 x > 0 is (x  2) 2 (t) (–, 0)  (3, ) 2. Match the column for values of x which satisfy the equation in Column-I Column-I (A) (B) log10 ( x  3) = log10 ( x 2  21) Column-II 1 2 xlogx +4 = 32, where base of logarithm is 2 log x log x 1 log x 1 (C) 5 3 3 of logarithm is 10 (D) 91+logx – 31+logx – 210 = 0 ; 5 log x 1 where the base (p) 5 (q) 100 (r) 2 (s) 1 32 where base of log is 3 3. Column- I (A) Column-II Interval containing all the solutions of the    (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. [Comprehension Type Questions] Part # II Comprehension # 1 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. 3.  1  (C)  2 3 ,     (B) [–1, 0]  [2, ) (D) none of these The complete solution set of the equation |x2 – x| + |x + 3| = |x2 – 2x – 3| is (A) [1, ) (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 (B) f(x)  0, g(x)  0 (C) f(x) . g(x)  0 (D) f(x) . g(x) = 0 Comprehension # 2 In comparison of two numbers, logarithm of smaller number is smaller, if base of the logarithm is greater than one. Logarithm of smaller number is larger, if base of logarithm is in between zero and one. For example log 2 4 is smaller than log 2 8 and log 1 4 is larger than log 1 8 . 2 2 On the Basis of Above Information, Answer the Following Questions 1. 2. Identify the correct order :(A) log 2 6 < log 3 8 < log 3 6 < log 4 6 (C) log 3 8 > log 2 6 > log 3 6 > log 4 6 log 1 40 20 is- (A) greater than one (C) greater than zero and smaller than one 3. log 2 3 (B) log 2 6 > log 3 8 > log 3 6 > log 4 6 (D) log 3 8 > log 4 6 > log 3 6 > log 2 6 (B) smaller than one (D) none of these 5 6 is- (A) less than zero (C) greater than one (B) greater than zero and less than one (D) none of these BASIC MATHS & LOGARITHM Comprehension # 3 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 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 EXERCISE - 3 : PART # I 1. 3. A  r, A  prs, B  t, C  q, D  p B  rs, C  t, D  prs 2. A  q, B  r s, C  q, PART - II Comprehension # 1 : Comprehension # 2 : Comprehension # 3 : 1. 1. 1. A B C 2. 2. 2. D B A 3. 3. 3. C B C D p