Title Logarithm & Modulus Function(Nu).pdf ✅

MATHEMATICS Q.1 Find the number of integral values of '' parameter for which the inequality 1 + log2 (2x2 + 2x + 7/2)  log2 (x 2 + ) has at least one root. [0008] Q.2 Sum of integer solutions of the inequality (log3 x)3 – 4(log3 x)2 + 5 (log3 x) – 6 < 0 is........ [0351] Q.3 If x = = 2000 n 1 n , then the value of the expression, log x 1 ..... log x 1 log x 1 1 2 3 2000 + + + is ........... [0001] Q.4 The expression log (0.1) 20 (0.05) is a perfect square of the natural number .............. [0009] Q.5 Simplify : 3 10 log 5 1 log (0.1) 1 5 7 − + [0002] Q.6 If b – c log x = c – a log y = a – b log z Then answer the following (i) xyz = ............ (ii) xay b z c = ............ Sol. (i) 1 (ii) 1 Q.7 The value of 2 (0.5) (log 4) is .................. Sol. 2 Q.8 If x + log10(2x +1) = log106 + x log105 then the value of x is................... Sol. 1 Q.9 The value of b satisfying the equation, loge2. logb625 = log1016.loge10 is ____. [0005] Q.10 The expression log 8 2 0.5 has the value equal to____. [0003] Q.11 The number of integral value(s) of x satisfying the equation |x4 .3|x–2 | . 5x–1 | = –x 4 .3|x–2 | . 5x–1 is...... Sol. 0001 Q.12 Number of ordered pair (x,y) satisfying the system of equations 2 log (x2 + y2 ) – log 5 = log {2(x2 + y2 ) + 75} and       3 x log + log (5y) = 1 + log 2, is..... Sol. 0002 Q.13 The number N = 1og 5 1og 10 1og 5 log 250 1250 5 50 5 − when simplified reduces to a natural number N. find N Sol. [2] N = log 5 log 10 log 5 log 250 1250 5 50 5 − = (3 + log52) (2 +log52) – (1 + log52) (4 + log52) = (log52)2 + 5 log52 + 6 – [(log52)2 + 5 log52 + 4] = 2 Q.14 The number N = 6log102 + log1031 lies between two successive integers whose sum equals ⎯⎯ Sol. [7]  N = log1064 + log1031 = log101984  3 < N < 4 Q.15 If a, b, c are distinct positive numbers, each different from 1,such that (logba logca – logaa) + (logab logcb – logbb) + (logac logbc – logcc) = 0, then abc = Sol. 0001 Q.16 If the value of a = log0.75 log2 2 0.125 − the value of log0.01 1000 + log0.1 0.0001 is b then the value of (125)a + 620b must be - Sol. 1675

Q.27 Find solution of ( x 4x 3 1 2 − + + ) log5 5 x + x 1 ( 8x 2x 6 1 2 − − + )  0. Q.28 Consider the inequation x2 + |x + a| – 9 < 0. Find the maximum integral value of the real parameter 'a' so that the given inequation has at least one negative solution. Q.29 Find all positive values of 'a' for which the equation log (ax) = 2 log (x + 1) has the unique root. Sol. [0004] log ax = 2 log (x + 1)  ax = (x + 1)2  x 2 + (2 –a) x + 1 = 0 Let x1 and x2 be roots x1 = 2 a 2 a 4a 2 − + − and x2 = 2 a 2 a 4a 2 − − − For solution exists a2 –4a  0 a  4, a  0 For positive values a  4 For a > 4, x1 and x2 are different For a = 4, x1 and x2 are same Hence a = 4. Questions Add (23-10-09) Q.30 If no. of zeroes after decimal in (0.15)20 is ab. Find b – a. (assume log 2 = 0.3010, log 3 = 0.4771) Sol.[5] log x = 20 (log 0.15) = 20 [1 – log 2 + log 3 – 2] = 20 [–1 + 0.176] = – 20 + 352 = – 17 + 0.52 c = – 17 no. of zeroes = | – 17 + 1 | = 16 ab = 16  b – a = 5 Q.31 If x, y, z be positive real numbers such that log2x z = 3, log5y z = 6 and logxy z = 2/3 then the value of z is in the form of m/n in lowest form then find value of n – m. Sol.[9] z = 8x3 , z = 5 6y 6 , z = x2/3y 2/3 x = 2 z 1/ 3 , y = 5 z 1/ 6 , z = 2/3 2 1 z 2/9 . 2 / 3 2 /18 5 z z 2/3 = 2 / 3 (10) 1  z = 10 1  n – m = 9 Q.32 (679854321)! contains ‘10’ a times while ‘15’ b times then find the value of a – b. Sol.[0] 10  5 also 15  5 so a = b hence a – b = 0 Q.33 Let P = log5 (log5 3). If P C 5 3 − + = 405 then C equals Sol.[4] 5 – P = log log (3) 5 5 5 −  5 –P = log3 5  C log3 5 3 + = 405 (3C ) (5) = 405  3 C = 81  C = 4 Q.34 Number of solution for | 3x2 – 2 | = [–2] is ( [] denotes greatest integer) Sol. [0] [– 2] = – 7  L.H.S. is positive R.H.S. is negative  no solution Q.35 The number of positive integers satisfying the equation x + log10(2x + 1) = xlog105 + log106 is Sol. [0001] x[1 – log105] + log10 (2x + 1) = log106 x[log1010 – log105] + log10 (2x + 1) = log106 x log102 + log10 (2x + 1) = log106 log102 x (2x + 1) = log106 (2x ) 2 + 2x – 6 = 0 2 x = 2 2 x = – 3 x = 1, which is not possible +ve integer Q.36 If a = log12 18 & b = log2454 then find the value of ab + 5 (a – b) Sol. [1] Q.37 If x, y > 0,logyx + logxy = 3 10 and xy = 144, then 2 x + y = N where N is a natural number, find the value of N. Sol. [507]