Maths new edition

["1.32 Chapter 1 \t69.\t ( 6 15 \u2212 2 56 )( 3 7 + 2 2 ) \u2234 1 = 7+ 4 3 3 15 \u2212 2 56 3 7 + 2 2 P = 3 8 \u2212 7 3 7 + 8 = 3 8 \u2212 7 =1. P + 1 = P2 +1 = 7 \u22124 3 +7+4 3 =14 P P 7\t 0.\t P = 7 \u2212 4 3 \u21d2 P2 +1 = 14 = 2. \t\t 7P 7 1 = 7 1 3 = 7+4 3 P \u22124 49 \u2212 48 Hints and Explanation","122CChhaapptteerr LKoignaemriathtimcss REMEMBER Before beginning this chapter, you should be able to: \u2022 Explain laws of indices \u2022 Solve exponential equations KEY IDEAS After completing this chapter, you should be able to: \u2022 Understand the system of logarithms \u2022 Learn properties and laws of logarithms \u2022 Understand variation of logax with x and learn the signs of logax for different values of x and a \u2022 Find the log of a number using log tables obtain the antilog from a given problem Figure 1.1","2.2 Chapter 2 INTRODUCTION Please recall that you have earlier learnt about indices. One of the results we learnt is that if 2x = 23, x = 3 and if 4x = 4y, then x = y, i.e., if two powers of the same base are equal and the base is not equal to -1, 0 or 1, then the indices are equal. But when 3x = 52, just by using the knowledge of indices, we cannot find the numerical value of x. The necessity of the concept of logarithms arises here. Logarithms are useful in long calculations involving multiplication and division. Definition The logarithm of any positive number to a given base (a positive number not equal to 1) is the index of the power of the base which is equal to that number. If N and a (\u22601) are any two positive real numbers and for some real number x, ax = N, then x is said to be logarithm of N to the base a. It is written as logaN = x, i.e., if ax = N, then x = logaN. If in a particular relation, all the log expressions are to the same base, we normally do not specify the base. Examples: 1.\t 23 = 8 \u21d2 3 = log28 2.\t 54 = 625 \u21d2 log5625 = 4 From the definition of logs, we get the following results: When a > 0, b > 0 and b \u2260 1, 1.\t logaan = n, e.g., log443 = 3 2.\t alogab = b, e.g., 2log216 = 16 . SYSTEM OF LOGARITHMS Though we can talk of the log of a number to any positive base not equal to 1, there are two systems of logarithms, natural logarithms and common logarithms, which are used most often. 1.\t \u0007Natural logarithms: These were discovered by Napier. They are calculated to the base e which is approximately equal to 2.71. These are used in higher mathematics. 2.\t \u0007Common logarithms: Logarithms to the base 10 are known as common logarithms. This system was introduced by Briggs, a contemporary of Napier. For the rest of this chapter, we shall use the short form log rather than logarithm. Properties 1.\t Logs are defined only for positive real numbers. 2.\t Logs are defined only for positive bases different from 1. 3.\t In logba, neither a nor b is negative but the value of logba can be negative. \t Example: As 10-2 = 0.01, log100.01 = -2","Logarithms 2.3 4.\t \u0007Logs of different numbers to the same base are different, i.e., if a \u2260 b, then logma \u2260 logmb. In other words, if logma = logmb, then a = b. Example: log102 \u2260 log103 \tlog102 = log10y \u21d2 y = 2 5.\t L\u0007 ogs of the same number to different bases have different values, i.e., if m \u2260 n, then logma \u2260 logna. In other words, if logma = logna, then m = n. Example: log216 \u2260 log416 \tlog216 = logn16 \u21d2 n = 2 6.\t Log of 1 to any base is 0. Example: log21 = 0 (\u2234 20 = 1) 7.\t Log of a number to the same base is 1 Example: log44 = 1. 8.\t Log of 0 is not defined. Laws 1.\t logm(ab) = logma + logmb Example: log\u200915 = log(5 \u00d7 3) = log\u20095 + log\u20093 2.\t logm\uf8eb\uf8ec\uf8ed a \uf8f6 = logma \u2212 logmb b \uf8f8\uf8f7 Example: log \uf8eb 15 \uf8f6 = log\u200915 - log\u200920 \uf8ec\uf8ed 20 \uf8f8\uf8f7 3.\t log\u2009am = mlog\u2009a Example: log\u200936 = log\u200962 = 2log\u20096 4.\t logba logcb = logca (chain rule) Example: log24 log416 = log216 5.\t logba = logca logcb Example: log416 = log216 (change of base rule) log2 4 In this relation, if we take a = c, we get the following result: logba = 1 b . loga Variation of logax with x For 1 < a and 0 < p < q, logap < logaq For 0 < a < 1 and 0 < p < q, logap > logaq Example: log102 < log103 and log0.12 > log0.13","2.4 Chapter 2 Bases which are greater than 1 are called strong bases and bases which are less than 1 are called weak bases. Therefore, for strong bases log increases with number and for weak bases log decreases with number. Sign of logax for Different Values of x and a Strong bases (a > 1) 1.\t If x > 1, logax is positive. Example: log210, log525 are positive. 2.\t If 0 < x < 1, then logax is negative. Example: log30.2, log100.25 are negative. Consider log30.2 = log 0.2 = log 2 \u2212 log 10 log 3 log 3 log\u20092 < log\u200910 and 0 < log\u20093 for strong bases. As log 2 \u2212 log 10 < 0, log30.2 < 0 log 3 Weak bases (0 < a <1) 3.\t If x > 1 and, then logax is negative. Example: Consider log0.42 log 2 = log 0.4 = log log 2 10 4 \u2212 log log\u20094 < log\u200910 (for any base) log\u20094 - log\u200910 < 0 log log 2 < 0. (for strong bases) 4 \u2212 log 10 4.\t If 0 < x < 1, then logax is positive. For example, log0.10.2, log0.40.3 are positive. To summarize, logs of big numbers (>1) to strong bases and small numbers (<1) to weak bases are positive. Example 2.1 If x2 + y2 = 3xy, then choose the correct answer of 2log(x - y) from the following options: (a)\t log\u2009x - log\u2009y\t (b)\u2002log\u2009x + log\u2009y\t\t (c)\u2002log(xy)\t\t (d)\u2002 Both (b) and (c) Hint Find (x \u2212 y)2. Example 2.2 Choose the correct answer from the following option for: log2[log4{log3(log327)}] = (a)\t 0\u2003\u2003\u2003\u2003(b)\u20021\u2003\u2003\u2003\u2003(c)\u2002log23\u2003\u2003\u2003\u2003(d)\u2002undefined","Logarithms 2.5 Solution log2\u2009[log4\u2009{log3 (log327)}] = log2\u2009[log4\u2009{log3 (log333)}] = log2\u2009[log4\u2009{log33(log33)}] = log2\u2009[log4\u2009{log33(1)}] = log2 [log4 {log33}] = log2 [log41] = log2 0, which is not defined. Example 2.3 If log\u20092 = 0.301, then find the number of digits in 21024 from the following options: (a)\t 307\t (b)\u2002308\t (c)\u2002309\t (d)\u2002310 Solution Let x = 21024 \u21d2 log\u2009x = log\u200921024 = 1024 log\u20092 = 1024(0.301) \u21d2 log\u2009x = 308.22 \u21d2 The characteristic is 308 \\\\ The number of digits in 21024 is 309. To Find the log of a Number to Base 10 Consider the following numbers: 2, 20, 200, 0.2 and 0.02. We see that 20 = 10(2) and 200 = 100(2) \u2234 log\u200920 = 1 + log\u20092 and log\u2009200 = 2 + log\u20092 Similarly, log\u20090.2 = -1 + log\u20092 and log\u20090.02 = -2 + log\u20092 From the tables, we see that log\u20092 = 0.3010. (Using the tables, this is explained in more detail in later examples.) \u2234 log\u200920 = 1.3010, log\u2009200 = 2.3010, log\u20090.2 = -1 + 0.3010 and log\u20090.02 = -2 + 0.3010. We note two points: 1.\t Multiplying or dividing by a power of 10 changes only the integral part of the log, not the fractional part. 2.\t For numbers less than 1, (for example 0.2) it is more convenient to leave the log value as -1 + 0.3010 instead of changing it to -0.6090. We refer to the first form (in which the fraction is positive) as the standard form and the second form as the normal form. Both the forms represent the same number. \t \t For numbers less than 1, it is more convenient to express the log in the standard form. As the negative sign refers only to the integral part, it is written above the integral part, rather than in front, i.e., log 0.2 = 1.3010 and not -1.3010. \t \t The convenience of the standard form will be clear when we learn how to take the antilog, which is explained in more detail later. \t \t antilog (-0.6090) = antilog (-1 + 0.3010) = antilog 1.3010 = 0.2.","2.6 Chapter 2 \t \t When the logs of numbers are expressed in the standard form, (for numbers greater than 1, the standard form of the log is the same as the normal form), the integral part is called the characteristic and the fractional part (which is always positive) is called the mantissa. Example 2.4 Express -0.5229 in the standard form and locate it on the number line. Solution 0 -0.5229 = -1 + 1 - 0.5229 = 1.4771 0.4771 \u2022 \u20131 \u20130.5229 The Rule to Obtain the Characteristic of log\u2009x 1.\t If x > 1 and there are n digits in x, the characteristic is n - 1. 2.\t If x < 1 and there are m zeroes between the decimal point and the first non-zero digit of x, the characteristic is -m, more commonly written as m. \u2002Note\u2002\u2002 \u22124 = 4 but \u2212 4.01 \u2260 4.01 To Find the log of a Number from the log Tables Example 2.5 Find the value of log\u200925, log\u2009250 and log\u20090.025. Solution In the log table, we find the number 25 in the left-hand column. In this row, in the next column (under 0), we find 0.3979. (The decimal point is dropped in other columns) For the log of all numbers whose significant digits are 25 and this number 0.3979, is the mantissa. Prefixing the characteristics we have, \t log\u200925 = 1.3979 \t log\u2009250 = 2.3979 \t log\u20090.025 = 2.3979. Example 2.6 Find the value of log\u20092.54, log\u20090.254 and log\u200925400. Solution In the log table, we locate 25 in the first column. In this row, in the column under 4 we find 0.4048. As done in the earlier example, the same line as before gives the mantissa of logarithms of all numbers which begin with 25. From this line we pick out the mantissa which is located in the column numbered 4. This gives 4048 as the mantissa for all numbers whose significant digits are 254. \t log\u20092.54 = 0.4048 \t log\u20090.254 = 1.4048 \t log\u200925400 = 4.4048.","Logarithms 2.7 Example 2.7 Find the value of log\u20092.546 and log\u200925460. Solution As found in the above example, we can find the mantissa for the sequence of digits 4048. Since there are four significant digits in 2546, in the same row where we have found 4048, under column 6 in the mean difference column, we find the number 10, the mantissa of the logarithm of 2546 will be 4048 + 10 = 4058. Thus,\t log\u20092.546 = 0.4058 \t log\u20090.2546 = 1.4058 \t log\u200925460 = 4.4058. ANTILOG As log28 = 3, 8 is the antilogarithm of 3 to the base 2. Antilog b to base m is mb. In example 3 above, we saw that, log\u20092.546 = 0.4058. Therefore, Antilog 0.4058 = 2.546 To Find the Antilog Example 2.8 To find the antilog of 1.301. Solution Step 1: In the antilog table, we find the number 30 in the left hand column. In that row in the column under 1, we find 2000. Step 2: As the characteristic is 1, we place the decimal after two digits from the left. That is, antilog 1.301 = 20.00. If the characteristic was 2, we would place the decimal after three digits from the left. That is, antilog 2.301 = 200.0. If the characteristic was 3, we would place the decimal after four digits from the left. That is, antilog 3.301 = 2000. Example 2.9 To find the antilog of 2.3246. Solution We have to locate 0.32 in the left hand column and slide along the horizontal line and pick out the number in the vertical column headed by 4. We see that the number is 2109. The mean difference for 6 in the same line is 3. The significant digits in the required number are = 2109 + 3 = 2112. As the characteristic is 2, the required antilog is 211.2.","2.8 Chapter 2 Example 2.10 Find the value of 5.431 \u00d7 0.061 to four significant digits. 12.38 \u00d7 0.041 Solution log of a fraction = (log of numerator) - (log of denominator) log of numerator = log\u20095.431 + log\u20090.061 = 0.7349 + 2.7853 = 1.5202 log of denominator = log\u200912.38 + log\u20090.041 = 1.0927 + 2.6128 = 1.7055 log of given expression = 1.5202 - 1.7055 = 1.8147 antilog of 8147 is 6516 + 11 = 6527 Since the characteristic is 1, the decimal should be kept before the four digits. \u2234 The answer is 0.6527. Example 2.11 If log103 = 0.4771, and log102 = 0.3010, Find the value of log1048. Solution log1048 = log1016 + log103 = log1024 + log103 = 4log102 + log103 = 4(0.3010) + 0.4771 = 1.6811.","Logarithms 2.9 TEST YOUR CONCEPTS Very Short Answer Type Questions \t1.\t logxAn = _____. 1\t 8.\t If 2log\u2009x + 2log\u2009y = k and xy = 1, then k = _______. \t2.\t Expand log3 \uf8eb xy2 \uf8f6 . 1\t 9.\t If log\u2009198.9 = 2.2987, then the characteristic of \uf8ec z3 \uf8f7 log\u2009198.9 = and mantissa of log\u2009198.9 = _______. \uf8ed \uf8f8 \t3.\t Can we write logx a as logxa ? \t20.\t When 0 < a < 1 and m < n, then which is greater, b logxb logam (or) logan? \t4.\t Express 0.001 = (0.1)3 in the logarithmic form. \t21.\t Given log10x = y. If the characteristic of y is 10, then the number of digits to the left the decimal \t5.\t (5)2log5 2 = _______ . point in x is _____. \t6.\t log52 + log520 \u2212 log58 = _______. 2\t 2.\t Find the value of log 816. \t7.\t The value of 3 + log10(10)2 is _______. \t23.\t logy\u2009x \u00d7 logzy \u00d7 logxz = _____. log5 5 2\t 4.\t If the characteristics of the logarithm of two \t8.\t logxab = (logxa) \u00d7 (logxb). State True or False. numbers abcd\u2009\u22c5\u2009abef and a\u2009\u22c5\u2009bcdabef are x and y respec- tively, then x \u2212 y = _____. \t9.\t If log102 = 0.3010, then log10 2000 = _______. 1\t 0.\t Evaluate 3 \u2212 log10100. \t25.\t If log\u20092 = 0.3010, log\u20093 = 0.4771 and log\u20097 = 0.8451, then find the values of log\u2009210. \t11.\t Given 3 = log2x + 4log28. Then the value of x = _______. 2\t 6.\t Given, antilog(2.375) = x. Characteristic of log\u2009x is _____. \t12.\t If log102 = 0.3010, then log105 = _______. \t13.\t If x = log53 and y = log58, then log524 in terms of \t27.\t If log(21.73) = 1.3371, then find the values of log(2.173). x, y is equal to _______. \t14.\t If log1625 = k log25, then k = _______. 2\t 8.\t If antilog (0.2156) = 1.643, then find the values of antilog (1.2156). \t15.\t If 5log\u20093 + log\u2009x = 5log\u20096, then x = _______. PRACTICE QUESTIONS 1\t 6.\t If log\u2009x = \uf8eb loga y \uf8f6 k then k = _______. \t29.\t Without using the logarithm tables find the value \uf8ed\uf8ec loga x \uf8f7\uf8f8 of 3log327. , 1\t 7.\t If a > 1 and m > n, then which is greater, logam \t30.\t Find the value of log0.6 \uf8eb 9\uf8f6 . (or) logan? \uf8ed\uf8ec 25\uf8f7\uf8f8 Short Answer Type Questions \t31.\t Prove that log\u20095040 = 4log\u20092 + 2log\u20093 + log\u20095 3\t 6.\t Prove that log2\u2009[log4\u2009{log5\u2009(625)4}] = 1. + log\u20097. 3\t 7.\t If log102 = 0.3010, then find the number of digits 3\t 2.\t Find the value of log2\u22121 (0.0625). in (16)10. \t33.\t Express the following as a single logarithm. \t38.\t If log\u20092 = 0.3010, log\u20093 = 0.4771 and log\u20097 = 0.8451, 1 8 7 then find the value of log\u200975. 3 5 2 log x \u2212 log y + log z. 3\t 9.\t If x4 + y4 = 83x2y2, then prove that log \uf8eb x2 \u2212 y2 \uf8f6 \uf8ec\uf8ed 9 \uf8f7\uf8f8 \t34.\t If x2 + y2 = 25xy, then prove that 2log(x + y) = = log\u2009x + log\u2009y. 3log\u20093 + log\u2009x + log\u2009y. 4\t 0.\t Prove that \t35.\t If x2 + y2 = z2, then prove that logy(z + x) + 2 log 35 + 2 log 114 + log 48 + 2 log \uf8eb 13 \uf8f6 = log \uf8eb 75 \uf8f6 . logy(z \u2212 x) = 2. 192 91 \uf8ed\uf8ec 19 \uf8f8\uf8f7 \uf8ec\uf8ed 64 \uf8f7\uf8f8","2.10 Chapter 2 \t41.\t Solve for x: 4\t 3.\t Using the tables find the value of 4 (32)3 . \t\tlog\u2009x + log\u20095 = 2 + log\u200964 \t44.\t Using the tables find the value of (0.12)3 . \t42.\t If x6 - y6 = z6, then prove that \t45.\t Given log\u20093 = 0.4771, then the number of digits in \t\tlogz(x2 - y2) + logz(x2 + y2 \u2212 xy) + logz(x2 + y2 + 31000 is _____. xy) = 6. Essay Type Questions \t46.\t If logx+12x \u2212 1 + log2x\u22121x + 1 = 2, find x. 4\t 9.\t If logyx \u2212 logy3x2 = 9(logxy)2 and x = 9y, find y. \t47.\t I=f a b1=\/3 c=1\/5 d=1\/7 e1\/9 , find logaabcde. 5\t 0.\t log \uf8eb a2 \uf8f6 + log \uf8eb a4 \uf8f6 + log \uf8eb a6 \uf8f6 + + log \uf8eb a2n \uf8f6 = ? 4\t 8.\t Arrange the following in ascending order. \uf8ed\uf8ec b \uf8f7\uf8f8 \uf8ec\uf8ed b3 \uf8f8\uf8f7 \uf8ec\uf8ed b3 \uf8f8\uf8f7 \uf8ed\uf8ec bn \uf8f7\uf8f8 \t\tA = log96561 \t\tB = log 5 625 \t\tC = log 3 243 \t\tD = log 2 256 CONCEPT APPLICATION Level 1 \t1.\t logya \u00d7 logxy = ______. \t5.\t logz2 x2y2 = ? \t\t(a) logay\t\t (b) logxa \t\t(a)\t2(log\u2009x + log\u2009y - log\u2009z) \t \t (c) logya\t\t (d) logax \t\t(b) log\u2009x + log\u2009y - log\u2009z \t2.\t If log\u2009x = 123.242, then the characteristic of log\u2009x \t\t(c) log x2 + log y is log z PRACTICE QUESTIONS \t\t(a) 0.242\t\t (b) 122 \t\t(d) log x + log y log z \t\t(c) 123\t\t (d) 124 \t6.\t If x3 - y3 = 3xy(x - y), then log(x - y)3 = ______. \t3.\t Pick up the false statement. \t\t(a) 0\t\t (b) 1 \t\t(A)\tLogarithms are defined only for positive real \t\t(c) Undefined\t (d) None of these numbers. \t7.\t log(a3 + b3) - log(a + b) - log(a2 - ab + b2) = ______. \t\t(B)\tlogaN is always unique. \t\t(a) a3 - b3 \t\t(C)\tThe log form of 23 = 8 is 3 = log82. \t\t(b) 0 \t\t(D)\tlog\u20091 = 0 \t\t(c) log\u20091 \t\t(d) Both (b) and (c) \t\t(a) B\t\t (b) C \t\t(c) D\t\t (d) A \t8.\t Which is greatest among the following: \t4.\t log \uf8eb 169\uf8f6 - 2log\u200913 + 2log\u20093 = ? \t\t(a) log220\t\t (b) log735 \uf8ec\uf8ed 9 \uf8f8\uf8f7 \t\t(c) log570\t\t (d) log368 \t\t(a) 1\t\t (b) 0 \t9.\t log(a + b) + log(a - b) - log(a2 - b2) = ______. \t\t(c) log \uf8eb 13 \uf8f6 \t(d) log \uf8eb x \uf8f6 \t\t(a) 0\t\t (b) 1 \uf8ec\uf8ed 3 \uf8f7\uf8f8 \uf8ec yz \uf8f7 \uf8ed \uf8f8 \t\t(c) (a2 - b2)\t\t (d) a2 + b2","Logarithms 2.11 \t10.\t If x3 + y3 = 4xy(x + y). Then log(x + y)3 = ______. \t\t(a) 11\t\t (b) 121 \t\t(a) log\u2009x + log\u2009y + log(x + y) - log\u20097 \t\t(c) 0\t\t (d) 1 \t\t(b) log(x) - log\u2009y + log(x + y) + log\u20097 1\t 9.\t log3729 + log6216 = ______ . 4 + log216 \u2212 2 log464 \t\t(c) log\u2009x + log\u2009y + log(x - y) + log\u20097 \t\t(d) log\u2009x + log\u2009y + log(x + y) + log\u20097 \t\t(a) 9\t\t (b) 4 1\t 1.\t If log x = log 49 , then the relation between x and y. \t\t(c) 9 \t\t (d) 1 log y log 7 2 2 \t\t(a) x = y \t20.\t If logxn ym \u2009= klogxy, then the value of k is \t\t(b) x = y3 m n \t\t(c) y = x2 \t\t(a) \t\t (b) mn \t\t(d) x = y2 \t\t(c) mn\t\t (d) nm \t12.\t log(x) - log(2x - 3) = 1, then x =? \t21.\t If logyx = 2, then a loga(logxy) = _____. \t\t(a) 30 \t\t (b) 20 \t\t(a) \u22122\t\t (b) 4 19 19 \t\t(c) 1 \t\t (d) \u22121 19 2 4 \t\t(c) 30 \t\t (d) 19 \t22.\t If p = log6216 and q = log325 then pq = ______. 20 \t\t(a) 3\t\t (b) 25 \t13.\t If 2log(x + 4) = log\u200916, then x =? \t\t(a) 0, -8\t\t (b) -8 \t\t(c) 15\t\t (d) Cannot be determined \t\t(c) -2\t\t (d) 0 2\t 3.\t 2(16\u2212log2 1024) = _______. \t14.\t The value of x when logx 343 = 3, is \t\t(a) 16\t\t (b) 32 \t\t(a) 7\t\t (b) 8 \t\t(d) 64\t\t (d) 8 \t\t(c) 3 \t\t (d) 27 \t24.\t 23 log2 2 + 32 log3 2 = _______. PRACTICE QUESTIONS \t15.\t log163 \u22c5 log174 \u22c5 log917 = _______. \t\t(a) 8\t\t (b) 4 1 1 \t\t(c) 9\t\t (d) 2 2 4 \t\t(a) \t\t (b) \t25.\t loga+b(a3 + b3 ) \u2212 loga+b(a2 \u2212 ab + b2 ) = _______. \t\t(a) loga\u2009+\u2009b(a - b)\t (b) 2 \t\t(c) 1 \t\t (d) 2 \t\t(c) a + b\t\t (d) 1 8 3 \t26.\t 4log1625 = _______. \t16.\t log2 [log2\u2009{log2\u2009(log381)}] = \t\t(a) 25\t\t (b) 5 \t\t(a) 1 \t\t(b) 0 \t\t(c) 16\t\t (d) 4 \t\t(c) log\u20093 2\t 7.\t 1 x + 1 y = _______? logxy logxy \t\t(d) Undefined \t\t(a) 1\t\t (b) 2 1\t 7.\t log113 \u22c5 log31331 = _______. \t\t(c) 0\t\t (d) 1 \t\t(a) 3\t\t (b) 11 logxyx \u00d7 logxyy \t\t(c) 121\t\t (d) 9 \t28.\t If y = logx-3(x2 - 6x + 9), then find y. 1\t 8.\t log121 \uf8eb 14641 \uf8f6 = _______. \t\t(a) 4\t\t (b) 8 \uf8ec\uf8ed 121 \uf8f7\uf8f8 \t\t(c) 2\t\t (d) 32","2.12 Chapter 2 Level 2 \t29.\t If log102 = 0.3010, then the number of digits in 3\t 8.\t If aloga n = 3 then a2logb \u2212bloga = _______. 1612 is \t\t(a) 6\t\t (b) 9 \t\t(a) 14\t\t (b) 15 \t\t(c) 3\t\t (d) Cannot be determined \t\t(c) 13\t\t (d) 16 3\t 9.\t If log3(x - 5) + log3(x + 2) = log38, then x = _______. =3\t 0.\t If log64 p2 1=32 , then log2 1p6 ______. \t\t(a) -3\t\t (b) 6 \t\t(a) 16\t\t (b) 2 \t\t(c) 6, \u22123\t\t (d) 3, \u22126 \t\t(c) 32\t\t (d) 1 4\t 0.\t If log(x + y) = log\u2009x + log\u2009y, then x = _______. 3\t 1.\t log2\u2009log2\u2009log5125 = _______. \t\t(a) \u2212y \t\t (b) \u2212y 1+ y 1\u2212 y \t\t(a) 4\t\t (b) 8 \t\t(c) 1\t\t (d) y \t\t(c) \u22121\t\t (d) 1 1+ y \t32.\t If logx y = loga y , then the value of p is \t41.\t If 2log5 \u22c5 5log2 = 2logx, then log5 3 x2 = _______. p 4 3 \t\t(a) logy\u2009x\t\t (b) logx\u2009a \t\t(a) 3 \t\t (b) 4 \t\t(c) loga\u2009x\t\t (d) loga\u2009y \t\t(c) 1 \t\t (d) 3 3 \uf8ee 3 x2 \u00d7 y \uf8f9 \uf8ef 5 z2 \uf8fa 4\t 2.\t If 3log\u2009x + xlog\u20093 = 54, find log\u2009x. \t33.\t log \uf8f0\uf8ef \uf8fa\uf8fb = ______ \t\t(a) log\u2009x2\/3 - log\u2009z2\/5 + log\u2009y \t\t(a) 3\t\t (b) 2 \t\t(c) 4\t\t (d) Cannot be determined \t\t(b) log\u2009x3\/2 + log\u2009y - log\u2009z5\/2 \t43.\t If log10x - log10y = 1 and x + y = 11, then x = _______. PRACTICE QUESTIONS \t\t(c) log\u2009x2\/3 - log\u2009y + log\u2009z2\/5 \u2009 \t\t(d) None of these \t\t(a) 10\t\t (b) 1 \t\t(c) 11\t\t (d) 2 \t34.\t If log[4 \u2212 5log32(x + 3)] = 0, find x. \t44.\t If log493 \u00d7 log97 \u00d7 log28 = x, then find the value \t\t(a) 32\t\t (b) 8 4x 3 \t\t(c) 3\t\t (d) 5 of . \t35.\t If x = log3 log2 log2256, then 2log4 22x\u2009= _______. \t\t(a) 3\t\t (b) 7 \t\t(a) 4\t\t (b) 8 \t\t(c) 8\t\t (d) 1 \t\t(c) 2\t\t (d) 1 \t45.\t The value of loga\u2212b(a3 \u2212 b3 ) \u2212 loga\u2212b(a2 + ab + b2 ) is _____. (a > b) \t36.\t If loga \uf8eb 132 \uf8f6 = 2loga 13 - loga 5 - x then \t\t(a) 0\t\t (b) 1 \uf8ec\uf8ed 23 \u00d7 5\uf8f8\uf8f7 \t\t(3) 3\t\t (c) Undefined \t\t(a) ax = 23\/2\t\t (b) xa = 23\/2 \t46.\t If log3\u20092 = x, then the value of log10 72 is log10 24 \t\t(c) ax = 22\/3\t (d) xa = 22\/3 1+ x 2 + 3x \t37.\t If log\u200981 \u2212 log\u20093 = log\u2009a, then 49log a = _______. \t\t(a) 1\u2212x \t\t (b) 1\u2212 3x \t\t(1) 4\t\t (2) 16 \t\t(c) 2 \u2212 3x \t\t (d) 3x +1 2 + 3x 3x + 2 \t\t(3) 2\t\t (4) 8","Logarithms 2.13 \t47.\t log \uf8eb 1 \uf8f6 + log \uf8eb 2 \uf8f6 + log \uf8eb 3 \uf8f6 + + log \uf8eb 99 \uf8f6 \t\t(a) \u221e\t\t (b) 0 \uf8ec\uf8ed 2 \uf8f7\uf8f8 \uf8ec\uf8ed 3 \uf8f7\uf8f8 \uf8ec\uf8ed 4 \uf8f8\uf8f7 \uf8ec\uf8ed 100 \uf8f8\uf8f7 \t\t(c) 1\t\t (d) Cannot be determined = ______. \t51.\t If x2 + y2 - 3xy = 0 and x > y, then find the value of logxy(x \u2212 y). \t\t(a) -2\t\t (b) -1 \t\t(c) 0\t\t (d) 2 1 4 \t48.\t If x2 - y2 = 1, (x > y), then find the value of \t\t(a) \t\t (b) 4 log(x\u2009-\u2009y) (x + y). \t\t(a) -2\t\t (b) 2 \t\t(c) 1 \t\t (d) 2 2 \t\t(c) -1\t\t (d) 1 4\t 9.\t If 3log35 + 5logx3 = 8, then find the value of x. \t52.\t If log\u20093 = 0.4771, then find the number of digits in 3100. \t\t(a) 3\t\t (b) 5 \t\t(c) 4\t\t (d) 8 \t\t(a) 47\t\t (b) 48 \t50.\t log21 \u22c5 log32 \u22c5 log43 \u22c5 log54 \u22c5 log65 ... log200199 \t\t(c) 49\t\t (d) 50 =\u2009\u2009______. Level 3 \t53.\t If log5x - log5y = log54 + log52 and x \u2212 y = 7, then 5\t 8.\t The value of 1 + 1 + 1 b + 1 + 1 a x = _____. 1 + logab c logac logbc \t\t(a) 1\t\t (b) 8 equals \t\t(c) 7\t\t (d) 6 \t\t(a) 2\t\t (b) 0 \t54.\t If log2 \uf8ee\uf8f0\uf8ef\u22121 + x2 \u2212 14x + 49 \uf8f9 = 4, then x = _____. \t\t(c) 1\t\t (d) log\u2009abc \uf8fb \t59.\t If x2 + y2 = z2, then 1 + 1 = ____ . \t\t(a) 24\t\t (b) \u221210 log(z + x)y log(z \u2212x)y \t\t(c) 24, \u221210\t\t (d) 10 \t\t(a) 4\t\t (b) 3 PRACTICE QUESTIONS \t55.\t If lo=2g p l=o4g q=lo8g r k=and pqr 100,then k \t\t(c) 2\t\t (d) 1 = _____. 6\t 0.\t If log\u20092 = 0.301, then find the number of digits in \t\t(a) 14\t\t (b) 1 21024. 6 \t\t(a) 307\t\t (b) 308 \t\t(c) 1 \t\t (d) 2 \t\t(c) 309\t\t (d) 310 7 6\t 1.\t If x2 - y2 = 1, (x > y), then find the value of log(x\u2009\u2212\u2009y) 5\t 6.\t If log\u20092 = 0.3010, and log\u20093 = 0.4771 then log\u2009150 (x + y). = _____. \t\t(a) -2\t\t (b) 2 \t\t(a) 2.1761\t\t (b) 2.8751 \t\t(c) -1\t\t (d) 1 \t\t(c) 2.5762\t\t (d) 2.6126 \t62.\t If x2 + y2 - 3xy = 0 and x > y, then find the value of logxy(x \u2212 y). 5\t 7.\t If log102 = 0.3010 and log103 = 0.4771, then the 1 value of log10 \uf8eb 23 \u00d7 32 \uf8f6 is \t\t(a) 4 \t\t (b) 4 \uf8ed\uf8ec 52 \uf8f7\uf8f8 \t\t(a) 0.4592.\t\t (b) 0.5492. \t\t(c) 1 \t\t (d) 2 2 \t\t(c) 0.4529.\t\t (d) 0.5429.","2.14 Chapter 2 \t63.\t If 2log39 + 25log93 = 8logx9 , \u2009then x = _____. \t66.\t If x = log327 and y = log927, then 1 + 1 = ______ . x y \t\t(a) 9\t\t (b) 8 \t\t(c) 3\t\t (d) 2 \t\t(a) 1 \t\t (b) 1 3 9 6\t 4.\t If logax = m and logbx = n, then log\uf8eb a \uf8f6x = _____. \t\t(c) 3\t\t (d) 1 \uf8ed\uf8ec b \uf8f8\uf8f7 \t\t(a) m \t\t (b) mn \t67.\t If log6x + 2log36x + 3log216x = 9, then x m\u2212n m\u2212n = _____. \t\t(c) n \t\t (d) mn \t\t(a) 6\t\t (b) 36 \u2212 n\u2212m m n \t\t(c) 216\t\t (d) None of these \t65.\t log5 6 1 = log5 2 + \t\t(a) log26\t\t (b) log25 (d) log1030 \t\t(c) log106\t\t PRACTICE QUESTIONS","Logarithms 2.15 TEST YOUR CONCEPTS Very Short Answer Type Questions \t1.\t nlogxA 1\t 6.\t \u22121 \t2.\t log3x + 2log3y \u2212 3log3z \t3.\t No 1\t 7.\t logam 1\t 8.\t0 \t4.\t log0.1(0.001) = 3 1\t 9.\t 2; 0.2987 \t5.\t 4 \t20.\t logam \t6.\t 1 \t21.\t 11 \t7.\t 5 \t22.\t 8 3 \t8.\t False \t23.\t 1 \t9.\t 3.3010 2\t 4.\t 3 1\t 0.\t 1 2\t 5.\t 2.3222 1\t 1.\t 1 2\t 6.\t 2 512 \t12.\t 0.6990 2\t 7.\t 0.3371 1\t 3.\t x + y \t28.\t 16.43 \t14.\t 1\/2 2\t 9.\t 27 \t15.\t 32 \t30.\t 2 Short Answer Type Questions \t32.\t 4 \t38.\t 1.8751 \t41.\t x = 1280 3\t 3.\t log \uf8eb x1\/3 z 7\/2 \uf8f6 \t43.\t 13.45 ANSWER KEYS \uf8ec\uf8ed y8\/5 \uf8f8\uf8f7 \t44.\t 0.04158 4\t 5.\t 478 3\t 6.\t 1 \t37.\t 13 Essay Type Questions \t49.\t 3 \t46.\t 2 5\t 0.\t log an(n +1) 4\t 7.\t 25 4\t 8.\t ABCD n(n +1) b2","2.16 Chapter 2 CONCEPT APPLICATION Level 1 \t1.\u2002(b)\t 2.\u2002 (c)\t 3.\u2002 (b)\t 4.\u2002 (b)\t 5.\u2002 (d)\t 6.\u2002 (c)\t 7.\u2002 (d)\t 8.\u2002 (a)\t 9.\u2002 (a)\t 10.\u2002 (d) \t11.\u2002 (d)\t 12.\u2002 (a)\t 13.\u2002 (d)\t 14.\u2002 (a)\t 15.\u2002 (b)\t 16.\u2002 (b)\t 17.\u2002 (a)\t 18.\u2002 (c)\t 19.\u2002 (c)\t 20.\u2002 (a) \t21.\u2002 (c)\t 22.\u2002 (b)\t 23.\u2002 (c)\t 24.\u2002 (b)\t 25.\u2002 (d)\t 26.\u2002 (b)\t 27.\u2002 (d)\t 28.\u2002 (c) Level 2 \t29.\u2002 (b)\t 30.\u2002 (d)\t 31.\u2002 (c)\t 32.\u2002 (c)\t 33.\u2002 (a)\t 34.\u2002 (d)\t 35.\u2002 (c)\t 36.\u2002 (a)\t 37.\u2002 (d)\t 38.\u2002 (a) \t39.\u2002 (b)\t 40.\u2002 (b)\t 41.\u2002 (a)\t 42.\u2002 (a)\t 43.\u2002 (a)\t 44.\u2002 (d)\t 45.\u2002 (b)\t 46.\u2002 (b)\t 47.\u2002 (a)\t 48.\u2002 (c) \t49.\u2002 (b)\t 50.\u2002 (b)\t 51.\u2002 (c)\t 52.\u2002 (b) Level 3 55.\u2002 (c)\t 56.\u2002 (a)\t 57.\u2002 (a)\t 58.\u2002 (a)\t 59.\u2002 (c)\t 60.\u2002 (c)\t 61.\u2002 (c)\t 62.\u2002 (c) 65.\u2002 (c)\t 66.\u2002 (d)\t 67.\u2002 (c) \t53.\u2002 (b)\t 54.\u2002 (c)\t \t63.\u2002 (b)\t 64.\u2002 (d)\t ANSWER KEYS","Logarithms 2.17 CONCEPT APPLICATION Level 1 \t1.\t Apply laws of logarithm. \t13.\t (i)\tUse log\u2009m = log\u2009n \u21d2 m = n. \t2.\t Recall the definition of characteristic. \t\t(ii)\tRemove logarithms to set (x + 4)2 = 16, then \t3.\t Verify all the options. find x. \t4.\t Apply laws of logarithm. \t5.\t Apply laws of logarithm. \t14.\t (i)\tUse log\u2009m = log\u2009n \u21d2 m = n. \t\t(ii)\tUse, logba = N \u21d2 a = bN. \t6.\t (i)\tApply log on both the sides \t15.\t Apply laws of logarithms. \t\t(ii)\tx3 \u2212 y3 = 3xy (x \u2212 y) \u21d2 (x \u2212 y)3 = 0. 1\t 6.\t (i)\tlogbam = mlogba and loga a = 1 \t\t(ii)\tExpress 81 as 34 and then use log\u2009am = m loga \t7.\t (i)\tlog x \u2212 log y \u2212 log z = log \uf8eb x \uf8f6 \uf8ed\uf8ec yx \uf8f8\uf8f7 and proceed until all the brackets are removed. \t\t(ii)\tlog\u2009M \u2212 log\u2009N = log \uf8eb M \uf8f6 and log\u2009M + log\u2009N 1\t 7.\t Apply laws of logarithms. \uf8ed\uf8ec N \uf8f8\uf8f7 1\t 8. \tFind 14641. = log\u2009MN. 1\t 9.\t Simplify the expression by applying laws of \t8.\t (i)\tlog216 < log220 < log232 logarithms. 2\t 0.\t Apply laws of logarithm. \t\t(ii)\tGiven four options are to be compared. 2\t 1.\t Apply laws of logarithm. 2\t 2.\t Find p and q. \t\t(iii)\tFor example log216 < log220 < log232 \u21d2 4 < \t23.\t Write 1024 as a power of 2. Hints and Explanation log220 < 5. 2\t 4.\t Apply laws of logarithm. \t25.\t Apply laws of logarithm. \t\t(iv)\t7 < 35 < 72; 52 < 70 < 53; 33 < 68 < 34 apply logarithms to suitable bases and evaluate log715, log570 and log368. \t9.\t (i)\tUse the identity log\u2009m + log\u2009n - log\u2009p = log\u2009mn\/p \t\t(ii)\tUse log\u2009x + log\u2009y = log\u2009xy and log\u2009p - log\u2009q = log(p\/q). \t10.\t Find (x + y)3. \t26.\t (i)\tRecall the laws of logarithm. \t\t(ii)\tUse a loga N \u2009= N after simplifying the given 1\t 1.\t (i)\tUse the identity log a = logb a log b term. 2\t 7.\t (i)\tSimplify and then use laws of logarithm. \t\t(ii)\t log x = 2 \u21d2 log y x = 2 \t\t(ii)\tTake LCM and simplify. log y \t\t(iii)\tUse logax + logay = logaxy. \t29.\t (i)\tUse loga a = 1 to find y and then substitute y \t\t(iii)\tlogb a = n \u21d2 a = bn . value. 1\t 2.\t (i)\tUse log\u2009m - log\u2009n = log m \t\t(ii)\tWrite x2 \u2212 6x + 9 as (x \u2212 3)2 and use logaa = 1. n \t\t(ii)\tUse log\u2009m - log\u2009n = log m and simplify LHS. n \t\t(iii)\tWrite 1 as log1010 and proceed.","2.18 Chapter 1 Level 2 \t29.\t Find the value of 1612 by applying logarithm. 4\t 4.\t log493 \u00d7 log97 \u00d7 log28 = x 3\t 0.\t Find p. \t\t2lologg37 \u00d7 log 7 \u00d7 3 log 2 = x 2 log 3 log 2 \t31.\t Apply laws of logarithm. \t\t43 4x 4 3 \t32.\t (i)\tRecall the laws of logarithm. =x \u21d2 3 = 3 \u00d7 4 = 1. \t\t(ii)\t p = log ay . Use log\u2009pq = q log\u2009p. \t45.\t loga\u2212b(a3 \u2212 b3 ) \u2212 loga\u2212b a2 + ab + b2 log xy \uf8eb a3 \u2212 b3 \uf8f6 3\t 3.\t (i)\tUse the identities log\u2009mn = log\u2009m + log\u2009n, = loga\u2212b \uf8ec a2 + ab + b2 \uf8f7 log\u2009mn = log\u2009m - log\u2009n and log\u2009xn = n log\u2009x \uf8ed \uf8f8 = loga\u2212b (a \u2212 b)(a2 + ab + b2 ) (a2 + ab + b2 ) \uf8eb 3 x2 \u22c5 y \uf8f6 \uf8eb \uf8eb x2\/3 \u22c5 y \uf8f6 \uf8f6 \t\t(ii)\tlog \uf8ed\uf8ec\uf8ec 5 z2 \uf8f8\uf8f7\uf8f7 log \uf8ec \uf8ec z 2\/5 \uf8f7 \uf8f7 = loga\u2212b(a \u2212 b) = 1. \uf8ed \uf8ed \uf8f8 \uf8f8 \t46.\t Given, log32 = x a \t\t(iii)\tUse log \uf8eb b \uf8f6 = log\u2009a \u2212 log\u2009b and log\u2009ab = log\u2009a log10 72 = log24 72 = log 24 ( 24 \u00d7 3) \uf8ed\uf8ec \uf8f8\uf8f7 log10 24 + log\u2009b simplify. log 3 log 24 \t34.\t (i)\tlog\u2009x = 0 \u21d2 x = 1 = log24 24 + log24 3 = 1 + \t\t(ii)\tlog\u2009x = 0 \u21d2 x = 1 = 1+ log3 3 log3 24 Hints and Explanation \t\t(iii)\tUse logb a = 1 logb a n 1 = 1+ log3(3 \u00d7 8) \t\t(iv)\tx = a log\u2009n \u21d2 ax = n. \t\t 1+ 1 \t35.\t Find x. = log3 3 + log3 8 \t36.\t Apply laws of logarithm. = 1 + 1 log3 3\t 7.\t Find a. 1 + 23 \t38.\t Substitute the value of alog\u2009b. = 1 + 1 + 3 1 2 log3 \t39.\t Apply laws of logarithm. = 1 +3 log3 2 + 1 = 2 + 3x . 1+ 3 log3 2 1 + 3x 4\t 0.\t (i)\tlog(a) = log(b) \u21d2 a = b \t\t(ii)\tlog(x + y) = log\u2009xy \t47.\t log \uf8eb 1\uf8f6 + log \uf8eb 2\uf8f6 + + log \uf8eb 99 \uf8f6 \uf8ed\uf8ec 2\uf8f7\uf8f8 \uf8ed\uf8ec 3 \uf8f7\uf8f8 \uf8ec\uf8ed 100 \uf8f8\uf8f7 \t\t(iii)\t\u21d2 x + y = xy. 4\t 1.\t (i)\talog\u2009b = blog\u2009a = log 1 \u2212 log 2 + log 2 \u2212 log 3 + \t\t(ii)\t5log\u20092 = 2log\u20095 + log 99 \u2212 log 100 \t\t(iii)\tam = an \u21d2 m = n. \t42.\t (i)\talog\u2009b = blog\u2009a = log 1 \u2212 log 100 = 0 \u2212 log 102 \t\t(ii)\txm = xn \u21d2 m = n. = \u22122 log 10 = \u22122. 4\t 8.\t Given x2 \u2212 y2 = 1 \t\tApplying log on both the sides, we get \t\t(x2 \u2212 y2) = log\u20091 \t43.\t Apply laws of logarithm and solve for x. \t\t \u21d2 log[(x + y)(x \u2212 y)] = 0","Logarithms 2.19 \t\t \u21d2 log(x + y) + log(x \u2212 y) = 0 \t\tApplying log on both sides \t\t \u21d2 log(x + y) = \u2212log(x \u2212 y) \t\tlog(x \u2212 y)2 = log\u2009xy log(x + y) = \u22121 \t\t2log(x \u2212 y) = log\u2009xy \t\tlog(x \u2212 y) \u21d2 log(x \u2212 y) = 1 \t\t \u21d2 log(x\u2009\u2212\u2009y) (x + y) = \u22121. log xy 2 4\t 9.\t 3log 5 + 5logx 3 = 8 \t\t\u21d2 logxy (x \u2212 y) = 1 . 3 2 \u21d2 5 + 5logx 3 = 8 \t52.\t Let x = 3100 \u21d2 5logx 3 = 3 \t\t \u21d2 log\u2009x = log3100 3logx 5 = 3 \t\t \u21d2 log\u2009x = 100 log3 \u21d2 logx 5 = 1 \t\t \u21d2 log\u2009x = 100 (0.4771) \u21d2 x = 5. 5\t 0.\t As log21 = 0, log21 \u22c5 log32 \u22c5 log43 ... log199200 = 0 \t\t \u21d2 log\u2009x = 47.71 5\t 1.\t Given x2 + y2 \u2212 3xy = 0 \t\tThe characteristic is 47 \t\tx2 + y2 \u2212 2xy = xy \t\tThe number of digits in 3100 is 48. Level 3 \t53.\t Apply laws of logarithm. = 1 abc = logabc ac 1+ 1 a + \t54.\t (i)\tlogax = b \u21d2 ab = x. logac logbc Hints and Explanation \t\t(ii)\tWrite x2 \u2212 14x + 49 = 16 + 1 logbc 1 a = 1 abc logabc bc bc + logbc logbc \t\t \t = 17 and solve for x. \t\tHence the value of the required expression \t\t(iii)\tThen verify for what values of x, log\u2009f(x) is defined. \t\t = logabcab + logabcac + logabcbc \t55.\t (i)\tFind the value of p, q, r then find pqr. \t\t = logabc[(ab)\u2009(ac)\u2009(bc)] = logabc(abc)2 = 2. \t\t(ii)\tExpress p, q and r in terms of k. 5\t 9.\t Given that, x2 + y2 = z2 \t\t(iii)\tSubstitute the above values in pqr = 100 and \t\t \u21d2 z2 \u2212 x2 = y2\b (1) find k. 1 1 \t56.\t (i)\tExpress as product of powers of 2 and 3 then \u2234 logz+x y + log(z\u2212x ) y use the values of log\u20092 and log\u20093 \t\t \t\t(ii)\tlog\u2009150 = 2log\u20095 + log2 + log\u20093. \t\t = logy(z + x) + logy(z \u2212 x) \t\t(iii)\tAlso, log\u20095 = log\u200910 \u2212 log\u20092. \t\t = logy(z2 - x2) = logyy2 {from (1)} \t57.\t (i)\tApply laws of logarithm. \t\t = 2. \uf8eb a \uf8f6 \t60.\t Let x = 21024 \uf8ec\uf8ed b \uf8f7\uf8f8 \t\t(ii)\t Use log = log\u2009a \u2212 log\u2009b and log\u2009am = mlog\u2009a. \t\t \u21d2 log\u2009x = log\u200921024 5\t 8.\t 1+ 1 c = logab 1 logab c = 1 \t\t = 1024 log\u20092 = 1024(0.301) logab ab + logababc \t\t \u21d2 log\u2009x = 308.22 = logabc ab 1 + 1 b = logac ac 1 logac b \t\t \u21d2 The characteristic is 308 logac + \t\t\\\\ The number of digits in 21024 is 309.","2.20 Chapter 1 \t61.\t Given x2 - y2 = 1 \t64.\t logax = m, logbx = n \t\tApplying log on both the sides, we get log(x2 \u2212 y2) log\uf8eb a x = log x = log x x = log\u20091 b a logx a \uf8ec\uf8ed \uf8f6 log \uf8eb b \uf8f6 \uf8eb b \uf8f6 \uf8f7\uf8f8 \uf8ec\uf8ed \uf8f8\uf8f7 \uf8ed\uf8ec \uf8f7\uf8f8 \t\t \t\t \u21d2 log[(x + y)(x \u2212 y)] = 0 \t\t \u21d2 log(x + y) + log(x \u2212 y) = 0 = 1 logxa \u2212 logxb \t\t \u21d2 log(x + y) = \u2212log(x \u2212 y) 1 \u2212 log(x + y) = 1 1 log(x \u2212 y) logxa logbx = \u22121 \t\t\u21d2 log(x\u2212y)(x + y) = 1. = 1 1 1 = \uf8eb 1 \uf8f6 = mn . \t\t m \u2212 n \uf8ed\uf8ec n\u2212m \uf8f7\uf8f8 n\u2212m mn \t62.\t Given, x2 + y2 - 3xy = 0 \t\tx2 + y2 \u2212 2xy = xy \t65.\t log56 1 = log log56 = log56 5) \t\tApplying log on both sides, log5 2 + 52 + log55 log5(2 \u00d7 \t\tlog(x \u2212 y)2 = log\u2009xy = log56 log106 . log510 \t\t2log(x \u2212 y) = log\u2009xy log(x \u2212 y) 1 \t66.\t x = log327 y = log927 log xy 2 \u21d2 = \u21d2 x = log333 y = log32 33 \u21d2 logxy (x \u2212 y) = 1 . \u21d2 x=3 y = 3 \t\t 2 2 Hints and Explanation \t63.\t 2log39 + 25log93 = 8logx9 \u2234 1 + 1 = 1 + \uf8eb 1 \uf8f6 = 1 + 2 + 3 = 1. \u21d2 2log332 + 25log323 = 8logx9 x y 3 3 3 3 3 \u21d2 22 + 251\/2 = 8logx9 \uf8ed\uf8ec 2 \uf8f7\uf8f8 \u21d2 9 = 8logx9 \u21d2 logx9 = log89 6\t 7.\t log6x + 2log36x + 3log216x = 9 \u21d2 x = 8. \t\t \u21d2 log6x + log6x + log6x = 9 \t\t \u21d2 3log6x = 9 \u21d2 log6x = 3 \t\t \u21d2 x = 63 \t\t\\\\ x = 216.","132CChhaapptteerr Polynomials and Figure 1.1 SKqiunaermeaRtiocosts of Algebraic Expressions REMEMBER Before beginning this chapter, you should be able to: \u2022 Apply basic operations on polynomials \u2022 Solve basic factorization of polynomials KEY IDEAS After completing this chapter, you should be able to: \u2022 State the types of polynomials and operations on polynomials \u2022 Find the factorization of polynomials and obtain HCF and LCM of polynomials \u2022 Understand different methods for finding square roots of algebraic expressions \u2022 Learn about homogeneous, symmetric and cyclic expressions","3.2 Chapter 3 INTRODUCTION Before learning the meaning and scope of polynomials, the terms like, constants, variables, algebraic expressions etc., have to be understood. Constant A number having a fixed numerical value is called a constant. Example: 7, 1 , 4.7, 16.5, etc. 2 Variable A number which can take various numerical values is known as variable. Example: x, y, z, a, b, c, etc. A variable raised to any non-zero real number is also a variable. Example: x5, y10\/3, z0.9, etc. A number which is the product of a constant and a variable is also a variable. Example: 8x3, \u22127x5, 4x10, etc. A combination of two or more variables separated by a (+) sign or a (-) sign is also a variable. Example: x2 \u2212 y3 + z, x3 \u2212 y3, etc. Algebraic Expression A combination of constants and variables connected by +, -, \u00d7 and \u00f7 signs is known as an algebraic expression. Example: 8x + 7, 11x2 \u2212 13x, 5x5 + 8x2y, etc. Terms The parts of an algebraic expression separated by + or - signs are called the terms of the expression. Example: In the expression 3x + 4y - 7, we call 3x, 4y and -7 as terms. Coefficient of a Term Consider the term 8x2. In this case, 8 is called the numerical coefficient and x2 is said to be the literal coefficient. In case of 9xy, we have the numerical coefficient as 9 and the literal coefficient as xy. Like Terms Terms having the same literal coefficients are called like terms. Examples: 1.\t 15x2, \u221219x2 and 35x2 are all like terms. 2.\t 8x2y, 5x2y and \u22127x2y are all like terms. Unlike Terms Terms having different literal coefficients are called unlike terms. Example: 5x2, \u221210x and 15x3 are unlike terms.","Polynomials and Square Roots of Algebraic Expressions 3.3 POLYNOMIAL An algebraic expression in which the variables involved have only non-negative integral powers is called a polynomial. Example: 5x2 \u2212 8x + 7, 3x3 + 5x2 \u2212 9, 3y2 \u2212 5y + z, etc. The expression 3x5 \u2212 8x + 4 + 11 x5\/2 is not a polynomial. Since the exponents of x are x negative integers and fractions. A polynomial with one variable is known as a polynomial in that variable. Example: 5x4 + 7x3 + 3x \u2212 9 is a polynomial in the variable x. 3y3 + y2 + y is a polynomial in the variable y. 4x2y2 + 3xy2 \u2212 7xy is a polynomial in variables x and y. Degree of a Polynomial in One Variable The highest index of the variable in a polynomial of one variable is called the degree of the polynomial. Examples: 1.\t 11x3 \u2212 7x2 + 5x + 2 is a polynomial of degree 3. 2.\t 15x6 \u2212 8x + 7 is a polynomial of degree 6. Types of Polynomials with Respect to Degree 1.\t Linear polynomial: A polynomial of degree one is called a linear polynomial. Example: 11x \u2212 5, 10y + 7 and 13z + 4 are polynomials of degree one and hence they can be called as linear polynomials. 2. \t Quadratic polynomial: A polynomial of degree two is called a quadratic polynomial. Example: 5x2 \u2212 8x + 3 and 13y2 \u2212 8y + 3 are polynomials of degree two and hence can be called as quadratic polynomials. 3.\t Cubic polynomial: A polynomial of degree three is called a cubic polynomial. Example: 5x3 + 6x2 + 7x + 8 and 4y3 \u2212 9y2 + 3 are polynomials of degree three and hence can be called as cubic polynomials. 4.\t \u0007Biquadratic polynomial: A polynomial of degree four is called a biquadratic polynomial. Example: 3x4 \u2212 x3 + 7x2 \u2212 2x + 1 and 5x4 \u2212 2x + 7 are polynomials of degree four and hence can be called as biquadratic polynomials. 5.\t \u0007Constant polynomial: A polynomial having only one term which is a constant is called a constant polynomial. Degree of a constant polynomial is 0. Example: 10, \u221211 are constant polynomials. Types of Polynomials with Respect to Number of Terms 1.\t Monomial: An expression containing only one term is called a monomial. Example: 8x, \u221211x2y, \u221215x2y3z2, etc.","3.4 Chapter 3 2.\t Binomial: An expression containing two terms is called a binomial. Example: 3x \u2212 8y, 4xy \u2212 5x, 9x + 5x2, etc. 3.\t Trinomial: An expression containing three terms is called a trinomial. Example: 5x - 2y + 3z, x2 + 2xy - 5z, etc. Addition of Polynomials The sum of two or more polynomials can be obtained by arranging the terms and then adding the like terms. Example 3.1 Add 7x2 \u2212 8x + 5, 3x2 \u2212 8x + 5 and \u22126x2 + 15x \u2212 5. Solution 7x2 \u2212\t8x\t + 5 3x2 \u2212\t8x\t + 5 \u22126x2 + 15x\t \u2212 5 4x2 \u2212\tx\t+ 5 \u2234 The required sum is 4x2 \u2212 x + 5. Subtraction of Polynomials The difference of two polynomials can be obtained by arranging the terms and subtracting the like terms. Example 3.2 Subtract 11x3 \u2212 7x2 + 10x from 16x3 + 4x2 \u2212 11x. Solution 16x3 + 4x2 \u2212 11x 11x3 \u2212 7x2 + 10x \u2212 + \u2009\u2009 \u2009\u2009- 5x3 + 11x2 \u2212 21x \u2234 The required difference is 5x3 + 11x2 \u2212 21x. Multiplication of Two Polynomials The result of multiplication of two polynomials is obtained by multiplying each term of the polynomial by each term of the other polynomial and then taking the algebraic sum of these products.","Polynomials and Square Roots of Algebraic Expressions 3.5 Example 3.3 Multiply (5x2 \u2212 8x + 7) with (2x \u2212 5). Solution 5x2 \u2212 8x + 7 2x \u2212 5 10x3 \u2212 16x2 + 14x \u2009\u2212 25x2 + 40x - 35 10x3 \u2212 41x2 + 54x - 35 \u2234 The required product is 10x3 \u2212 41x2 + 54x - 35. This is true for all real values of x, such equations are called algebraic identities. Division of a Polynomial by a Monomial To divide a polynomial by a monomial, we need to divide each term of the polynomial by the monomial. Example 3.4 Divide 18x4 \u2212 15x3 + 24x2 + 9x by 3x. Solution 18x4 \u2212 15x3 + 24x2 + 9x 3x 18x4 15x3 24x2 9x = 3x \u2212 3x + 3x + 3x = 6x3 \u2212 5x2 + 8x + 3 \u2234 The required result is 6x3\u2009\u2212 5x2 + 8x + 3. Division of a Polynomial by a Polynomial Factor Method In this method, we factorize the polynomial to be divided so that one or more of the factors is equal to the polynomial by which we wish to divide. Example 3.5 Divide 4x2 + 7x \u2212 15 by x + 3. Solution 4x2 + 7x \u2212\u200915 = 4x2 + 12x \u2212 5x - 15 4x(x + 3) \u2212 5(x + 3) = (4x \u2212 5)(x + 3) \u2234 4x2 + 7x \u2212 15 = (4x \u2212 5)(x + 3) = 4x \u2212 5. x +3 x+3","3.6 Chapter 3 \u2009\u2009Note\u2002 \u2002 The factor method for division of polynomials is used only when the remainder is zero. Long Division Method Step 1: First arrange the terms of the dividend and the divisor in the descending order of their degrees. Step 2: Now the first term of the quotient is obtained by dividing the first term of the dividend by the first term of the divisor. Step 3: Then multiply all the terms of the divisor by the first term of the quotient and subtract the result from the dividend. Step 4: Consider the remainder as new dividend and proceed as before. Step 5: Repeat this process till we obtain a remainder which is either 0 or a polynomial of degree less than that of the divisor. Example 3.6 Divide 2x3 + 9x2 + 4x \u2212 15 by 2x + 5. Solution 2x + 5 ) 2x3 + 9x2 + 4x \u2212 15 (x2 + 2x - 3 \u20092x3 + 5x2 \u2009 (-) (-) \u2009 \u2009 4x2 + 4x \u2009 4x2 + 10x \u2009(-) (-) \u22126x \u2212\u200915 -6x -\u200915 (+) (+) \u2009\u2009 0 \u2234 (2x3 + 9x2 + 4x \u2212 15) \u00f7 (2x + 5) = x2 + 2x \u2212 3. Horner\u2019s Method of Synthetic Division Example 3.7 Divide 27x3 \u2212 81x2 + 45x + 23 by (x \u2212 2). Solution x=2 27 \u221281 45 23 0 54 \u221254 \u221218 27 \u221227 \u22129 5 remainder coefficients of quotient","Polynomials and Square Roots of Algebraic Expressions 3.7 Step 1: We first write the coefficients of the dividend arranging them in descending powers of x with zero as the coefficient for missing power of x. Step 2: To divide by x \u2212 2, we write x = 2. Step 3: Bring down the leading coefficient of the dividend, multiply by 2 and add the second coefficient which results \u221227. Step 4: Now multiply \u221227 by 2 and add this to the third coefficient to get \u22129. Step 5: This process is continued until the final sum. Step 6: Thus, we get the quotient as 27x2 \u2212 27x \u2212 9 and the remainder as 5. Factorization Factorization is expressing a given polynomial as a product of two or more polynomials. Example: x3 \u2212 15x2 = x2(x \u2212 15) \u21d2 x2 and x \u2212\u200915 are the factors of x3 \u2212 15x2. 1.\t Factorization of polynomials of the form x2 - y2. x2 - y2 = (x + y)(x - y) \u21d2 x + y and x - y are the factors of x2 - y2. Example 3.8 Factorize 81x2 \u2212 225y2 Solution Let a = 9x and b = 15y a2 \u2212 b2 = (a + b)(a \u2212 b) \u2234 81x2 \u2212 225y2 = (9x)2 \u2212 (15y)2 = (9x + 15y)(9x \u2212 15y) \u2234 9x + 15y and 9x \u2212 15y are the factors of 81x2 \u2212 225y2. 2.\t \u0007Factorization of polynomials by grouping of terms: In this method we group the terms of the polynomials in such a way that we get a common factor out of them. Example 3.9 (a)\t Factorize a2 \u2212 (b \u2212 8)a \u2212 8b Solution \u21d2 a2 \u2212 (b \u2212 8)a \u2212 8b = a2 \u2212 ab + 8a \u2212 8b = a(a \u2212 b) + 8(a \u2212 b)","3.8 Chapter 3 = (a + 8)(a \u2212 b) \u2234 a2 \u2212 (b \u2212 8)a \u2212 8b = (a + 8)(a \u2212 b). (b)\t Factorize 4x3 \u2212 10y3 \u2212 8x2y + 5xy2 Solution 4x3 \u2212 8x2y + 5xy2 \u2212 10y3 = 4x2(x \u2212 2y) + 5y2(x \u2212 2y) = (4x2 + 5y2)(x \u2212 2y). 3.\t \u0007Factorization of a trinomial that is a perfect square. A trinomial of the form x2 \u00b1 2xy + y2 is equivalent to (x \u00b1 y)2. This identity can be used to factorize perfect square trinomials. Example 3.10 (a)\t Factorize 49x2 + 9y2 + 42xy Solution 49x2 + 9y2 + 42xy = (7x)2 + (3y)2 + 2(7x)(3y) = (7x + 3y)2. (b)\t Factorize 16x 2 + 1 2 \u22122 16x Solution 16x 2 + 1 2 \u22122 16x = (4x )2 + \uf8eb 1\uf8f6 \u2212 2(4x ) \uf8eb 1\uf8f6 \uf8ec\uf8ed 4x \uf8f7\uf8f8 \uf8ed\uf8ec 4x \uf8f8\uf8f7 \uf8eb 1\uf8f6 \uf8eb 1\uf8f6 2 \uf8ec\uf8ed 4x \uf8f7\uf8f8 \uf8ed\uf8ec 4x \uf8f7\uf8f8 = (4x )2 \u2212 2(4x ) + \uf8eb 1 \uf8f6 2 \uf8ed\uf8ec 4x \uf8f8\uf8f7 = 4x \u2212 \u2234 16x 2 + 1 \u22122 = \uf8eb 4x \u2212 1 \uf8f6 2 16x 2 \uf8ec\uf8ed 4x \uf8f7\uf8f8 . 4.\t Factorization of a polynomial of the form x2 + (a + b)x + ab. As we have already seen, (x + a)(x + b) = x2 + (a + b)x + ab \u2234 x2 + (a + b)x + ab can be fatcorized as (x + a)(x + b).","Polynomials and Square Roots of Algebraic Expressions 3.9 Example 3.11 (a)\t Factorize x2 + 25x + 144 Solution Here, the constant term is 144 = (16 \u00d7 9) and the coefficient of x is 25 = (16 + 9) \u2234 x2 + 25x + 144 = x2 + 16x + 9x + 144 = x(x + 16) + 9(x + 16) = (x + 16)(x + 9). (b)\t Factorize x2 \u2212 8x + 15 Solution Here, the constant term is 15 = (\u22125)(\u22123) and the coefficient of x is \u22128 = \u22125 \u2212 3. \u21d2 x2 \u2212 8x + 15 = x2 \u2212 5x \u2212 3x + 15 = x(x \u2212 5) \u22123(x \u2212 5) = (x \u2212 3)(x \u2212 5) \u2234 x2 \u2212 8x + 15 = (x \u2212 3)(x \u2212 5). (c)\t Factorize x2 \u2212 5x \u2212 14 Solution Constant term is \u221214 = (\u22127)(2) Coefficient of x is \u22125 = \u22127 + 2 \u21d2 x2 \u2212 5x \u2212 14 = x2 \u2212 7x + 2x \u2212 14 = x(x \u2212 7) + 2(x \u2212 7) = (x + 2)(x \u2212 7). 5.\t Factorization of polynomials of the form ax2 + bx + c. Step 1: Take the product of the constant term and the coefficient of x2, i.e., ac. Step 2: Now this product ac is to split into two factors m and n such that m + n is equal to the coefficient of x, i.e., b. Step 3: Then we pair one of them, say mx, with ax2 and the other nx, with c and factorize. Example 3.12 (a)\t 6x2 + 19x + 15 Solution Here, 6 \u00d7 15 = 90 = 10 \u00d7 9 and 10 + 9 = 19","3.10 Chapter 3 \u2234 6x2 + 19x + 15 = 6x2 + 10x + 9x + 15 = 2x(3x + 5) + 3(3x + 5) = (2x + 3)(3x + 5). (b)\t Factorize 7 \u2212 17x \u2212 12x2 Solution Here, (7)(\u221212) = \u221284 = (\u221221)(4) and \u221217 = \u221221 + 4 7 \u2212 17x \u2212 12x2 = 7 \u2212 21x + 4x \u2212 12x2 = 7(1 \u2212 3x) + 4x(1 \u2212 3x) = (1 \u2212 3x)(7 + 4x). 6.\t Factorization of expressions of the form x3 + y3 (or) x3 - y3. x3 + y3 = (x + y) (x2 - xy + y2) \u21d2 x3 + y3 has factors (x + y) and (x2 - xy + y2) x3 - y3 = (x - y) (x2 + xy + y2) \u21d2 x3 - y3 has factors (x - y) and (x2 + xy + y2). Example 3.13 (a)\t Factorize 27a3 + 125x3 Solution 27a3 + 125x3 = (3a)3 + (5x)3 = (3a + 5x){(3a)2 + (5x)2 \u2212 (3a)(5x)} = (3a + 5x)(9a2 + 25x2 \u2212 15ax). (b)\t Factorize 216x3 \u2212 64y3 Solution 216x3 \u2212 64y3 = (6x)3 \u2212 (4y)3 = (6x \u2212 4y){(6x)2 + (4y)2 + (6x)(4y)} = (6x \u2212 4y)(36x2 + 16y2 + 24xy). 7.\t Factorization of expressions of the form x3 + y3 + z3 when x + y + z = 0. (Given x + y + z = 0) As x + y + z = 0, z = -(x + y)","Polynomials and Square Roots of Algebraic Expressions 3.11 x3 + y3 + z3 = x3 + y3 + {-(x + y)}3 = x3 + y3 - (x + y)3 = x3 + y3 - {x3 + y3 + 3xy(x + y)} = -3xy(x + y) = -3xy(-z) {Since x + y = -z} = 3xyz \u2234 If x + y + z = 0, then x3 + y3 + z3 = 3xyz. HCF of Given Polynomials For two given polynomials, f(x) and g(x), r(x) can be taken as the highest common factor, if 1.\t r(x) is a common factor of f(x) and g(x) and 2.\t every common factor of f(x) and g(x) is also a factor of r(x). Highest common factor is generally referred to as HCF. Method for Finding HCF of the Given Polynomials Step 1: Express each polynomial as a product of powers of irreducible factors which also requires the numerical factors to be expressed as the product of the powers of primes. Step 2: If there is no common factor then HCF is 1 and if there are common irreducible factors, we find the least exponent of these irreducible factors in the factorized form of the given polynomials. Step 3: Raise the common irreducible factors to the smallest or the least exponents found in step 2 and take their product to get the HCF. Example 3.14 (a)\t Find the HCF of 48x5y2 and 112x3y. Solution Let f(x) = 48x5y2 and g(x) = 112x3y Writing f(x) and g(x) as a product of powers of irreducible factors. \tf(x) = 24 . 3 . x5 . y2 \tg(x) = 24 . 7 . x3 . y The common factors with the least exponents are 24, x3 and y \u2234 HCF = 16x3y. (b)\t Find the HCF of 51x2(x + 3)3(x \u2212 2)2 and 34x(x \u2212 1)5(x \u2212 2)3. Solution Let f(x) = 51x2(x + 3)3(x \u2212 2)2 and g(x) = 34x(x \u2212 1)5(x \u2212 2)3 Writing f(x) and g(x) as the product of the powers of irreducible factors.","3.12 Chapter 3 \tf(x) = 17 . 3 . x2(x + 3)3 . (x \u2212 2)2 \tg(x) = 17 . 2 . x(x \u2212 1)5 . (x \u2212 2)3 The common factors with the least exponents are 17, x and (x \u2212 2)2 \u2234 The HCF of the given polynomials = 17 . x . (x \u2212 2)2 = 17x(x \u2212 2)2. LCM of the Given Polynomials Least Common Multiple or the Lowest Common Multiple is the product of all the factors (taken once) of the polynomials given with their highest exponents respectively. Method to Calculate LCM of the Given Polynomials Step 1: First express each polynomial as a product of powers of irreducible factors. Step 2: Consider all the irreducible factors (only once) occurring in the given polynomials. For each of these factors, consider the greatest exponent in the factorized form of the given polynomials. Step 3: Now raise each irreducible factor to the greatest exponent and multiply them to get the LCM. Example 3.15 (a)\t Find the LCM of 18x3y2 and 45x5y2z3. Solution Let f(x) = 18x3y2 and g(x) = 45x5y2z3 Writing f(x) and g(x) as the product of the powers of irreducible factors. \u2009f(x) = 2 . 32 . x3 . y2 g(x) = 32 . 5 . x5 . y2 . z3 Now all the factors (taken only once) with the highest exponents are 2, 32, 5, x5, y2 and z3. \u2234 The LCM of the given polynomials = 2 \u00b7 32 \u00b7 5 \u00b7 x5 \u00b7 y2 \u00b7 z3 = 90x5y2z3. (b)\t Find the LCM of 51x2(x + 3)3 (x \u2212 2)2 and 34x(x \u2212 1)5 (x \u2212 2)3 Solution Writing f(x) and g(x) as the product of powers of irreducible factors. f(x) = 17 . 3x2(x + 3)3 \u00b7 (x \u2212 2)2 g(x) = 17 . 2(x \u2212 1)5 \u00b7 (x \u2212 2)3 Now all the factors (taken only once) with the highest exponents are 2, 3, 17, x2(x \u2212 1)5, (x\u00a0\u2212\u00a02)3 and (x + 3)3. \u2234 The LCM of the given polynomials = 2 . 3 . 17 . x2(x \u2212 1)5 \u00b7 (x \u2212 2)3 \u00b7 (x + 3)3 = 102x2(x \u2212 2)3(x \u2212 1)5(x + 3)3.","Polynomials and Square Roots of Algebraic Expressions 3.13 Relation between the HCF, the LCM and the Product of Polynomials If f(x) and g(x) are two polynomials then we have the relation, (HCF of f(x) and g(x)) \u00d7 (LCM of f(x) and g(x)) = \u00b1(\u2009\u2009f(x) \u00d7 g(x)). Example: Let f(x) = (x + 5)2(x \u2212 7)(x + 8) and g(x) = (x + 5)(x \u2212 7)2(x \u2212 8) be two polynomials. The common factors with the least exponents are x + 5 and x \u2212 7. \u2234 HCF = (x + 5) (x \u2212 7) All the factors (taken only once) with the highest exponents are (x + 5)2, (x \u2212 7)2, (x \u2212 8) and (x + 8). \u21d2 LCM = (x + 5)2(x \u2212 7)2(x \u2212 8) (x + 8) Now f(x) \u00b7 g(x) = (x + 5)2(x \u2212 7)(x + 8)(x + 5)(x \u2212 7)2(x \u2212 8) = (x + 5)3(x \u2212 7)3(x + 8)(x \u2212 8) LCM \u00d7 HCF = (x + 5)2(x \u2212 7)2(x \u2212 8)(x + 8) \u00d7 (x + 5)(x \u2212 7) = (x + 5)3(x \u2212 7)3(x \u2212 8) (x + 8) Thus, we say (LCM of two polynomials) \u00d7 (HCF of two polynomials) = Product of the two polynomials. Concept of Square Roots If x is any variable, then x2 is called the square of the variable and for x2, x is called the square root. Square root of x2 can be denoted as x2 \u22c5 x and -x can both be considered as the square roots of x2 because (x) \u00b7 (x) = x2 and (-x)(-x) = x2. In this study we restrict x2 to x, i.e., positive value of x. Square Root of Monomials The square root of a monomial can be directly calculated by finding the square roots of the numerical coefficient and that of the literal coefficients and then multiplying them. Example 3.16 Find the square root of 1296b4. Solution Now, the given monomial is 1296b4. Square root of 1296b4 = 1296b4 = 1296 \u00d7 b4 = (36)2 \u00d7 (b2 ) = 36 \u00d7 b2 \u21d2 1296b4 = 36b2.","3.14 Chapter 3 Example 3.17 Find the square root of 81b2a4 . 36x 2 y 6 Solution Square root of 81b2a4 81b2a4 36x2y6 = 36x2y6 = 81b2a4 36x 2 y 6 = 81 \u00d7 b2a4 36 \u00d7 x2y6 92 \u00d7 (ba2 )2 = 62 \u00d7 (xy3 )2 = 9ba2 = 3ba2 6xy3 2xy3 \u21d2 Square root of 81b2a4 = 3ba2 . 36x 2 y 6 2xy3 Methods of Finding the Square Roots of Algebraic Expressions Other than Monomials We have four methods to find the square root of an algebraic expression which is not a monomial. They are 1.\t Method of inspection (using algebraic identities). 2.\t Method of factorization 3.\t Method of division 4.\t Method of undetermined coefficients Method of Inspection\u2002 In this method, the square root of the given algebraic expression is found by using relevant basic algebraic identities after proper inspection. Example 3.18 Find the square root of x2 + 12xy + 36y2. Solution x2 + 12xy + 36y2 = (x)2 + 2(x)(6y) + (6y)2 We know that a2 + 2ab + b2 = (a + b)2","Polynomials and Square Roots of Algebraic Expressions 3.15 Now, x2 + 12xy + 36y2 = (x)2 + 2(x)(6y) + (6y)2 = (x + 6y)2 = (x + 6y) \u2234 x2 + 12xy + 36y2 = x + 6y. Example 3.19 Find the square root of a2x2 - 2ayx2 + x2y2. Solution x2(a2 \u2212 2ay + y2 ) = x2(a \u2212 y)2 a2x2 - 2ayx2 + x2y2 = x2(a2 - 2ay +y2) We know that a2 - 2ab + b2 = (a - b)2 Now, a2x2 \u2212 2ayx2 + x2y2 = = [x(a \u2212 y)]2 = x(a \u2212 y) \u21d2 a2x2 \u2212 2ayx2 + x2y2 = x(a \u2212 y). Method of Factorization Example 3.20 (a)\t Find the square root of (x2 \u2212 8x + 15)(2x2 \u2212 11x + 5)(2x2 \u2212 7x + 3). Solution Step 1: Factorize each expression in the given product, i.e., \t x2 \u2212 8x + 15 = x2 \u2212 5x \u2212 3x + 15 \t = x(x \u2212 5) \u2212 3(x \u2212 5) \t = (x \u2212 5)(x \u2212 3) \t2x2 \u2212 11x + 5 = 2x2 \u2212 10x \u2212 x + 5 \t = 2x(x \u2212 5) \u2212 1(x \u2212 5) \t = (2x \u2212 1)(x \u2212 5) \t2x2 \u2212 7x + 3 = 2x2 \u2212 6x \u2212 x + 3 \t = 2x(x \u2212 3) \u2212 1(x \u2212 3) \t = (x \u2212 3)(2x \u2212 1) Step 2: Write all the factors in a row. \u2234 The given expression is (x \u2212 5)(x \u2212 3)(2x \u2212 1)(x \u2212 5)(x \u2212 3)(2x \u2212 1), i.e., (x \u2212 3)2(x \u2212 5)2(2x \u2212 1)2 Step 3: Evaluate the square root. Hence, the square root of the given expression is (x \u2212 3)(x \u2212 5)(2x \u2212 1).","3.16 Chapter 3 (b)\t Find the square root of (x \u2212 1)(x \u2212 2)(x \u2212 3)(x \u2212 4) + 1. Solution Step 1: First we select the terms in such a way that, they have a common expression in their product. In the given expression, we consider terms (x \u2212 1) and (x \u2212 4) and (x \u2212 2)(x \u2212 3). Step 2: Find their product, i.e., x2 \u2212 5x + 4 and x2 \u2212 5x + 6 = (x2 \u2212 5x + 4)(x2 \u2212 5x + 6) Step 3: Take the common expression in the products as a (= x2 \u2212 5x) and substitute in the given expression. \u2234 The given expression becomes (a + 4)(a + 6) + 1 Step 4: Factorize the resultant expression and find its square root, i.e., a2 + 10a + 25 = (a + 5)2 \u2234 (a + 5)2 = a + 5. Step 5: Resubstitute the value of a, which is the required square root. \u2234 The square root of the given expression is x2 \u2212 5x + 5. Method of Division\u2002 We discuss the method of division to find the square root of an algebraic expression using the following example. Example 3.21 Find the square root of x2 - 18x + 81. Solution (x ) x-9 ( \u22129) x x2 \u2212 18x + 81 x2 2x \u2212 9 \u221218x + 81 \u221218x + 81 0 \u2234 x2 \u2212 18x + 81 = x \u2212 9. Step 1: First the given expression is arranged in the descending powers of x. Step 2: Then the square root of the first term in the expression is calculated. In the above problem first term is x2 whose square root is x. This is now the first term of the square root of the expression. Step 3: Then the square of x, i.e., x2 is written below the first term of the expression and subtracted. The difference is zero. Then the next two terms in the expression -18x + 81 are brought down as the dividend for the next step. Double the first term of the square root and put it down as the first term of the next divisor, i.e., 2(x) = 2x is to be written as the first term of the next divisor. Now the first term -18x of the dividend -18x + 81 is to be divided by the","Polynomials and Square Roots of Algebraic Expressions 3.17 first term 2x (of the new divisor). Here we get -9 which is the second term of the square root of the given expression and the second term of the new divisor. Step 4: Thus the new divisor becomes 2x - 9. Multiply (2x - 9) by (-9) and the product -18x\u00a0+ 81 is to be brought down under the second dividend -18x + 81 and subtracted where we get 0. Step 5: Thus x - 9 is the square root of the given expression x2 - 18x + 81. Example 3.22 Find the square root of 4x6 - 12x5 + 9x4 + 8x3 - 12x2 + 4. Solution Follow the steps indicated in the previous example. 2x3 \u2212 3x2 + 2 (2x3 ) 2x3 4x6 \u2212 12x5 + 9x4 + 8x3 \u2212 12x2 + 4 (\u22123x2 ) (+2) 4x6 4x3 \u2212 3x2 \u2212 \u221212x5 + 9x4 \u221212x5 + 9x4 +\u2212 4x3 \u2212 6x2 + 2 8x3 \u2212 12x2 + 4 8x3 \u2212 12x2 + 4 +\u2212 0 \u2234 4x6 \u2212 12x5 + 9x4 + 8x3 \u2212 12x2 + 4 = 2x3 \u2212 3x2 + 2. Method of Undetermined Coefficients\u2002 The method of undetermined coefficients to find the square root of an algebraic expression is explained in the following examples. Example 3.23 (a)\t Find the square root of x4 + 4x3 + 10x2 + 12x + 9. Solution The degree of the given expression is 4, its square root will hence be an expression of degree 2. Let us assume the square root to be ax2 + bx + c. \u21d2 x4 + 4x3 + 10x2 + 12x + 9 = (ax2 + bx + c)2 We know that (p + q + r)2 = p2 + q2 + r2 + 2pq + 2qr + 2rp","3.18 Chapter 3 Here, p = ax2, q = bx, r = c \u21d2 x4 + 4x3 + 10x2 + 12x + 9 = (ax2)2 + (bx)2 + c2 + 2(ax2)(bx) + 2(bx)(c) + 2(c)(ax2) \u21d2 x4 + 4x3 + 10x2 + 12x + 9 = a2x4 + b2x2 + 2abx3 + 2cax2 + 2bcx + c2 Now equating the like terms on either sides of the equality sign, we have x4 = a2x4 \u21d2 a2 = 1 \u21d2 a = 1 4x3 = 2abx3 \u21d2 2ab = 4 \u21d2 ab = 2, but a = 1 \u21d2 b = 2 b2 + 2ca = 10 \u21d2 22 + 2c = 10 \u21d2 2c = 6 \u21d2 c = 3 \u2234 The square root of the given expression is ax2 + bx + c, i.e., x2 + 2x + 3. (b)\t Find the square root of 4x4 - 4x3 + 5x2 - 2x + 1. Solution The degree of the given expression is 4, its square root will hence be an expression in degree 2. Let 4x4 \u2212 4x3 + 5x2 \u2212 2x + 1 = ax2 + bx + c \u21d2 (4x4 - 4x3 + 5x2 - 2x + 1) = (ax2 + bx + c)2 \u21d2 4x4 - 4x3 + 5x2 - 2x + 1 = (ax2)2 + (bx)2 + c2 + 2(ax2)(bx) + 2(bx)(c) + 2(c)(ax2) \u21d2 4x4 - 4x3 + 5x2 - 2x + 1 = a2x4 + 2abx3 + (b2 + 2ac)x2 + 2bcx + c2 Now equating the like terms on either sides of the equation, we have 4x4 = a2x4 \u21d2 a2 = 4 \u21d2 a = 2 c2 = 1 \u21d2 c = 1 2bcx = -2x \u21d2 2bc = -2 \u21d2 bc = -1 \u21d2 b = \u22121 \u21d2 b = \u22121 (\u2234 c = 1) c \u2234 The square root of the given expression is ax2 + bx + c, i.e., 2x2 - x + 1. Rational Integral Function of x A polynomial in x, the exponents in powers of x are non-negative integers and the coefficients of the various powers of x are integers. Example: 11x2 \u2212 8x + 3, 4x2 \u2212 5x + 1, 8x5 \u2212 7x3 + 8x2 + 4x + 5, etc. Remainder Theorem q(x) is a rational integral function of x. If q(x) is divided by x \u2212 a, then the remainder is q(a).","Polynomials and Square Roots of Algebraic Expressions 3.19 Example 3.24 Find the remainder when x3 \u2212 8x2 + 5x + 1 is divided by x \u2212 1. Solution Let q(x) = x3 \u2212 8x2 + 5x + 1 If q(x) is divided by x \u2212 1, then the remainder is q(1). \u2234 q(1) = (1)3 \u2212 8(1)2 + 5(1) + 1 = 1 \u2212 8 + 5 + 1 q(1) = \u22121. \u2002Note\u2002\u2002 If q(x) is divided by ax \u2212 b, then the remainder is q \uf8eb b \uf8f6 . \uf8ec\uf8ed a \uf8f7\uf8f8 Example 3.25 Find the remainder when x2 \u2212 8x + 6 is divided by 2x \u2212 1. Solution Let q(x) = x2 \u2212 8x + 6 \u2234 Remainder = q \uf8eb 1 \uf8f6 \uf8ed\uf8ec 2 \uf8f8\uf8f7 i.e., q \uf8eb 1 \uf8f6 = \uf8eb 1 \uf8f62 \u2212 8 \uf8eb 1 \uf8f6 + 6 \uf8ed\uf8ec 2 \uf8f8\uf8f7 \uf8ec\uf8ed 2 \uf8f8\uf8f7 \uf8ed\uf8ec 2 \uf8f8\uf8f7 = 1 \u22124 + 6 = 1 + 2 4 4 \u2234 q \uf8eb 1 \uf8f6 = 9 . \uf8ec\uf8ed 2 \uf8f8\uf8f7 4 Factor Theorem q(x) is a rational integral function of x and if q(\u03b1) = 0, then x \u2212 \u03b1 is the factor of q(x). Example 3.26 (a)\t Is x \u2212 2 a factor of x3 + x2 \u2212 4x \u2212 4? Solution Let q(x) = x3 + x2 \u2212 4x - 4 q(2) = 8 + 4 \u2212 8 \u2212 4 = 0 \u2234 x \u2212 2 is a factor of q(x).","3.20 Chapter 3 (b)\t Find the value of m, if x + 2 is a factor of x3 \u2212 4x2 + 3x \u2212 5m. Solution Let q(x) = x3 \u2212 4x2 + 3x \u2212 5m. Given x + 2 is a factor of q(x). \u2234 q(\u22122) = 0 \t \u21d2 (\u22122)3 \u2212 4(\u22122)2 + 3(\u22122) \u2212 5m = 0 \t \u22128 \u221216 \u2212 6 \u2212 5m = 0 \t-5m = 30 \tm = -6. (c)\t Factorize x3 \u2212 2x2 \u2212 5x + 6. Solution Let f(x) = x3 \u2212 2x2 \u2212 5x + 6 Step 1: First we find one of the factors of f(x) by substituting the value of x as \u00b11, \u00b12 and so on till the remainder is zero. Here, \tf(1) = (1)3 \u2212 2(1)2 \u2212 5(1) + 6 \t =1\u22122\u22125+6 \tf(1) = 0 \u2234 x \u2212 1 is a factor of f(x). Step 2: To find the other two factors, we use synthetic division. 1 \u22122 \u22125 6 x=1 0 1 \u22121 \u22126 1 \u22121 \u22126 0 \u2234 The other two factors are x2 \u2212 x \u2212 6, i.e., (x \u2212 3)(x + 2). \u2234 The factorization of the given expression is (x \u2212 1)(x \u2212 3)(x + 2). (d)\t Factorize x4 \u2212 x3 \u2212 11x2 + 9x + 18. Solution Let f(x) = x4 \u2212 x3 \u2212 11x2 + 9x + 18 f(\u22121) = (\u22121)4 \u2212 (\u22121)3 \u2212 11(\u22121)2 + 9(\u22121) + 18 = 1 + 1 \u2212 11 \u2212 9 + 18 = 20 \u2212 20 = 0 \u2234 x + 1 is a factor of f(x). Now, f(2) = (2)4 \u2212 23 \u2212 11(2)2 + 9(2) + 18 = 16 \u2212 8 \u2212 44 + 18 + 18 = 52 \u2212 52 = 0 \u2234 x - 2 is a factor of f(x).","Polynomials and Square Roots of Algebraic Expressions 3.21 1 \u22121 \u221211 9 18 0 x = -1 1 \u22121 2 9 \u221218 x=2 0 1 \u22122 \u22129 18 0 2 0 -18 0 \u22129 0 \u2234 The other two factors of f(x) are (x + 3) and (x \u2212 3). Hence, the factorization of the given expression, is (x + 1)(x \u2212 2)(x \u2212 3)(x + 3). Homogeneous Expression An algebraic expression in which, the degree of all the terms is equal is a homogeneous expression. Example: bx + ay is a first degree homogeneous expression. ax2 + bxy + cy2 is a second degree homogeneous expression. \u2009\u2009Notes\u2002 1.\t A homogeneous expression is complete if it contains all the possible terms in it. 2.\t The product of two homogeneous expressions is a homogeneous expression. 3.\t T\u0007 he degree of the product of two or more homogeneous expressions is the sum of degrees of all the expressions involved in product. Symmetric Expressions f(x, y) is an expression in variables x and y. If f(x, y) = f(y, x), then f(x, y), is called a symmetric expression. i.e., If an expression remains same after interchanging the variables x and y is said to be a symmetric expression. Example 3.27 Consider the expressions given below and find if the expressions are symmetric or not: (a)\t ax + ay + b (b)\t ax2 + bxy + ay2 Solution (a)\t Let f(x, y) = ax + ay + b f(y, x) = ay + ax + b = ax + ay + b \u21d2 f(y, x) = f(x, y) \u2234 ax + ay + b is symmetric.","3.22 Chapter 3 (b)\t f(x, y) = ax2 + bxy + ay2 f(y, x) = ay2 + byx + ax2 = ax2 + bxy + ay2 \u2234 f(y, x) = f(x, y) Hence, ax2 + bxy + ay2 is symmetric. \u2009\u2009Notes\u2002 1.\t A\u0007 n expression which is homogeneous and symmetric is called a homogeneous symmetric expression. Example: ax + ay, ax2 + bxy + ay2 2.\t T\u0007 he sum, difference, product and quotient of two symmetric expressions is always symmetric. Cyclic Expressions f(x, y, z) is an expression in variables x, y and z. If f(x, y, z) = f(y, z, x), then f(x, y, z) is cyclic. Example: a2(a \u2212 b) + b2(b \u2212 c) + c2(c \u2212 a) Let f(a, b, c) = a2(a \u2212 b) + b2(b \u2212 c) + c2(c \u2212 a) Now, f(b, c, a) = b2(b \u2212 c) + c2(c \u2212 a) + a2(a \u2212 b) \t = a2(a \u2212 b) + b2(b \u2212 c) + c2(c \u2212 a) \tf(b, c, a) = f(a, b, c) \u2234 f is cyclic. Cyclic expressions are lengthy to write, so we use symbols \u2211 (read as sigma) and \u03c0(pi) to abbreviate them. \u2211 is used for sum of terms and \u03c0 is used for product of terms. Example: \u2211x2(y2 \u2212 z2) + y2(z2 \u2212 x2) + z2(x2 \u2212 y2) can be represented as x2(y2 \u2212 z2 ) x, y, z \u2234 \u2211x2(y2 \u2212 z2) = x2(y2 \u2212 z2) + y2(z2 \u2212 x2) + z2(x2 \u2212 y2) Example: \u2211(x2 + y3)(y2 + z3)(z2 + x3) can be represented as (x2 + y3 ) = (x2 + y3)(y2 + z3)(z2 + x3) x, y, z Example 3.28 Factorize a(b2 \u2212 c2) + b(c2 \u2212 a2) + c(a2 \u2212 b2). Solution Let f(a) = a(b2 \u2212 c2) + b(c2 \u2212 a2) + c(a2 \u2212 b2) f(b) = b(b2 \u2212 c2) + b(c2 \u2212 b2) + c(b2 \u2212 b2) = b(b2 \u2212 c2) + b(c2 \u2212 b2) f(b) = 0","Polynomials and Square Roots of Algebraic Expressions 3.23 \u2234 By remainder theorem, (a \u2212 b) is a factor of the given expression. The given expression is cyclic, so the other two factors will also be cyclic. \u2234 The other two factors are (b \u2212 c) and (c \u2212 a). The given expression may have a constant factor which is non-zero. Let it be m. \u2234 a(b2 \u2212 c2) + b(c2 \u2212 a2) + c(a2 \u2212 b2) = m(a \u2212 b)(b \u2212 c)(c \u2212 a) Put a = 0, b = 1, c = \u22121 in the above equation. i.e., 0(12 \u2212 (\u22121)2) + 1((\u22121)2 \u2212 0) + (\u22121)(0 \u2212 12) = m(0 \u2212 1)(1 \u2212 (\u22121))(\u22121 \u2212 0) \u21d2 0 + 1 + 1 = m(\u22121)(2)(\u22121) \u21d2m=1 \u2234 The factorization of the given cyclic expressions is (a \u2212 b)(b \u2212 c)(c \u2212 a). Example 3.29 Factorize a(b3 \u2212 c3) + b(c3 \u2212 a3) + c(a3 \u2212 b3) Solution Let f(a) = a(b3 \u2212 c3) + b(c3 \u2212a3) + c(a3 \u2212 b3) f(b) = b(b3 \u2212 c3) + b(c3 \u2212 b3) + c(b3 \u2212 b3) = b(b3 \u2212 c3) \u2212 b(b3 \u2212 c3) + 0 f(b) = 0 \u2234 By remainder theorem, a \u2212 b is the factor of the given expression. The given expression is cyclic, so the other two factors are also cyclic. \u2234 The factors are (a \u2212 b)(b \u2212 c)(c \u2212 a). The given expression is of degree 4 but the degree of factor is 3, hence a first degree cyclic expression is the another factor. Let m(a + b + c) be the factor (m \u2260 0). \u2234 a(b3 \u2212 c3) + b(c3 \u2212 a3) + c(a3 \u2212 b3) = m(a + b + c)(a \u2212 b)(b \u2212 c)(c \u2212 a) Put a = 0, b = 1 and c = 2 in the above equation. 0(13 \u2212 (2)3) + 1((2)3 \u2212 0) + 2(0 \u2212 13) = m(0 + 1 + 2)(0 \u2212 1)(1 \u2212 2)(2 \u2212 0) \u21d2 0 + 1(8) + 2(\u22121) = m(3)(\u22121)(\u22121)(2) 6 = 6m \u21d2m=1 \u2234 The factorization of the given cyclic expression is (a \u2212 b)(b \u2212 c)(c \u2212 a)(a + b + c).","3.24 Chapter 3 TEST YOUR CONCEPTS Very Short Answer Type Questions \t1.\t 11x2 \u2212 88x3 + 14x4 is called a _____ polynomial. \t17.\t The product of two symmetric expressions is a\/an \t2.\t The degree of the polynomial 7x3y10z2 is _____. _____ expression. \t3.\t The expression is a polynomial. (True\/False) \t4.\t If A = 3x2 + 5x \u2212 3 and B = 5x2 \u2212 7, then 2A \u2212 B 1\t 8.\t The square root of am2 \u22c5 bn2 is _____. is _____. \t19.\t The value of a if x3 \u2212 8x2 + 2x + a is divisible by \t5.\t If a + b + c = 0, then a3 + b3 + c3 = _____. x \u2212 2 is _____. \t6.\t Factors of x6 \u2212 y6 is _____. \t7.\t The LCM of 2x, 8x7y2 is _____. \t20.\t Factorize a5b \u2212 ab5. \t8.\t The HCF of 44a3 and 66bpa4 is 22a3, then p can be 2\t 1.\t The degree of a polynomial A is 7 and that of _____. polynomial AB is 56, then find the degree of poly- nomial B. \t9.\t One of the factor of x3 - x2 + x - 1 is _____ . 1\t 0.\t The quotient of 8x3 \u2212 7x2 + 5x + 8 when divided 2\t 2.\t If A = x3, B = 4x2 + x \u2212 1, then find AB. by 2x is _____. \t23.\t Factorize m7 + m4. 1\t 1.\t The remainder obtained when 80x3 + 55x2 + 20x \t24.\t Factorize 1 a2 \u2212 a + 4 . + 172 is divided by x + 2 is _____. 6 3 \t12.\t Factorize 6x2 + x \u2212 2. 2\t 5.\t If 3x2 + 8ax + 3 is a perfect square, then find the \t13.\t Find the LCM and HCF of the polynomials value of a. 15x2y3z, 3x3yz2. \t26.\t The factors of a3 + b3 + c3 \u2212 3abc are _____. 1\t 4.\t Find the remainder when x15 is divided by x \u2212 2. 2\t 7.\t The HCF of (a2 + 1)(a + 11) and (a2 + 1)2 (a + 11)2 1\t 5.\t Find the remainder if x5\u2212 3x3 + 5x + 1 is divided is _____. by 2x \u2212 1. 2\t 8.\t The value of 813 \u2212 1003 + 193 is _____. \t16.\t a + b \u2212 2 ab is _____ where a > b. PRACTICE QUESTIONS 2\t 9.\t If A = 4x3 \u2212 8x2, B = 7x3 \u2212 5x + 3 and C = 3x3 + x \u2212 11, then find (A + C) \u2212 B. \t30.\t 8x2 + 11xy +by2 is a symmetric expression, then b = _____. Short Answer Type Questions \t31.\t The HCF of (a \u2212 1)(a3 + m) and (a + 1)(a3 \u2212 n) and \t37.\t Find the quotient and remainder when x4 + 4x3 (a + 1)(a2 \u2212 n) is a2 \u2212 1, then the values of m and n \u2212 31x2 - 94x + 120 is divided by x2 + 3x \u2212 4. are _____. x3 3\t 8.\t Factorize a3 + 3ax + 64 \u2212 1 . \t32.\t Expand \u03c0 a2(b + c ). 8 8 a,b,c 3\t 9.\t Find the LCM and HCF of the following 3\t 3.\t The factors of (a \u2212 b)3 + (b \u2212 c)3 + (c \u2212 a)3 polynomials. is _____. \t\t36(x + 2)2 (x \u2212 1)3 (x + 3)5, 45(x + 2)5 (x \u2212 1)2 3\t 4.\t Expand \u2211c2(a2 \u2212 b2). (x + 3)5 and 63(x \u2212 1)5 (x + 2)5 (x + 3)4. 3\t 5.\t If A = 4x3 \u2212 8x2, B = 7x3 \u2212 5x + 3 and C = 3x3 + 4\t 0.\t The LCM of the polynomials (x2 + x \u2212 2) x \u2212 11, then find 2A \u2212 3B + 4C. (x2 + x \u2212 a) and (x2 + x \u2212 b)(x2 + 5x +a) is (x \u2212 1) (x + 2)2 (x + 3), then find the values of a and b. 3\t 6.\t If A = x3, B = 4x2 + x \u2212 1, C = x + 1, then find (A \u2212 B)(A \u2212 C). \t41.\t Find the remainder when x23 is divided by x2 \u2212 3x + 2.","Polynomials and Square Roots of Algebraic Expressions 3.25 \t42.\t If lmx2 + mnx + ln is a perfect square then prove \t44.\t If 125a6b4 c 2 = x, then find x2 . that, 4l2 = mn. 5a4b2 abc \t43.\t Find the value of \t45.\t Find the square root of (x2 + 6x + 8)(x2 (a + b + c )2 + (a + b \u2212 c )2 + 2(c2 \u2212 a2 \u2212 b2 \u2212 2ab). + 5x + 6) (x2 + 7x + 12). Essay Type Questions \t46.\t Factorize 6x4 \u2212 5x3 \u2212 38x2 \u2212 5x + 6. 4\t 9.\t Find the quadratic polynomial when divided by x, x \u2212 1 and x \u2212 2 leaves remainders 1, 2 and 9 \t47.\t For what values of p and q, the expression, respectively. x4 - 14x3 + 71x2 + px +q is a perfect square? 5\t 0.\t Find the factors of a2(b + c) + b2(c + a) + c2(a + b) 4\t 8.\t Find the square root of 16x6 - 24x5 + 25x4 - 20x3 \u22122b2c. + 10x2 - 4x + 1 by the method of division. CONCEPT APPLICATION Level 1 \t1.\t If the degree of a polynomial AB is 15 and the \t6.\t \u03a3x(y3 \u2212 z3) = _____. degree of polynomial B is 5, then the degree of \t\t(a) (x \u2212 y)(y \u2212 z)(z \u2212 x)(x + y + z) polynomial A is \t\t(b) (x \u2212 y)(y - z)(x \u2212 z)(x \u2212 y \u2212 z) \t\t(c) (x + y)(y + z)(z + x)(x + y + z) \t\t(a) 3\t\t (b) 8 \t\t(d) (x + y)(y + z)(z + z)(z \u2212 y \u2212 z) \t\t(c) 4\t\t (d) 10 \t2.\t The expression 21x2 + 11x \u2212 2 equals to \t7.\t The remainder when f(x) = 4x3 \u2212 3x2 + 2x \u2212 1 is divided by 2x + 1 is _____. \t\t(a) (x \u2212 2)(7x + 1)\t\t (b) (7x + 1)(3x \u2212 2) \u22123 PRACTICE QUESTIONS \t\t(c) (7x \u2212 1)(3x \u2212 2)\t\t (d) (7x \u2212 1)(3x + 2) \t\t(a) 1\t\t (b) 4 \t3.\t If the LCM and HCF of two polynomials are \t\t(c) \u221213 \t\t (d) \u22127 90 m5a6b3x2 and m3 a5 respectively and also one 4 4 of the monomial is 18 m5a6x2, then the other \t8.\t The HCF of the polynomials 12a3b4c2, 18a4b3c3 monomial is and 24a6b2c4 is _____. \t\t(a)\t5 m3a5b3\t (b) 15 m5a3b2 \t\t(a) 12a3b2c2\t\t (b) 6a6b4c4 \t\t(c)\t5 m5a3b5\t (d) 15 m3a5b4 \t \t (c) 6a3b2c2\t\t (d) 48a6b4c4 \t4.\t The remainder when x3 \u2212 3x2 + 5x \u2212 1 is divided \t 9.\t Find the value of a, if (x + 2) is a factor of the poly- by x + 1 is _____. nomial f(x) = x3 + 13x2 + ax + 20. \t\t(a) \u22128\t\t (b) \u221212 \t\t(a) \u221215\t\t (b) 20 \t\t(c) \u221210\t\t (d) \u22129 \t\t(c) 25\t\t (d) 32 \t5.\t Which of the following is a homogeneous expression? 1\t 0.\t The polynomial x3 \u2212 4x2 + x - 4 on factorization gives \t\t(a) 4x2 \u2212 5xy + 5x2y + 10y2 \t \t (b) 5x + 10y + 100 \t\t(a) (x \u2212 4)(x2 \u2212 1) \t\t(c) 14x3 + 15x2y + 16y2x + 24y3 \t\t(b) (x \u2212 4)(x2 + 4) \t \t (d) x2 + y2 + x + y + 1 \t\t(c) (x + 4)(x2 + 1) \t\t(d) (x \u2212 4)(x2 + 1)","3.26 Chapter 3 \t11.\t If the expression ax3 + 2x2y \u2212 bxy2 \u2212 2y3 is \t18.\t If xn + 1 is divisible by x + 1, n must be symmetric, then (a, b) = \t\t(a) any natural number \t\t(b) an odd natural number \t\t(a) (2, 2)\t\t (b) (\u22122, 2) \t\t(c) an even natural number \t\t(d) None of these \t\t(c) (\u22122, \u22122)\t\t (d) (2, \u22122) \t12.\t The square root of y2 + 1 +2 is 1\t 9.\t What is the first degree expression to be subtracted y2 from x6 + 8x4 + 2x3 + 16x2 + 4x + 5 in order to make it a perfect square? \t\t(a) y+ 1 \t\t (b) y\u2212 1 y y \t\t(a) -4x - 4\t\t (b) 4x + 4 \t\t(c) y2 + 1 \t\t (d) y2 \u2212 1 \t\t(c) 4x - 4\t\t (d) -4x + 4 y2 y2 2\t 0.\t Find the square root of mn2 nm2 a(m+n) . 1\t 3.\t The product of the polynomials 2x3 \u2212 3x2 + 6 and (m + n)(m+n)2 x2 \u2212 x is _____. m+n n2 m2 m+n \t\t(a) 2x6 \u2212 5x4 + 3x3 + 6x2 \u2212 6x \t \t (b) 2x5 \u2212 x4 + 3x3 - 6x2 + 6x \t\t(a) mnnma 2 \t(b) m2n 2 a 2 \t \t (c) 2x5 \u2212 5x4 + 3x3 + 6x2 \u2212 6x \t \t (d) None of these (m+n )2 (m + n) 2 \t\t(c) mnnma m+n \t (d) None of these (m + n)(m+n) 1\t 4.\t The LCM of x2 \u2212 16 and 2x2 \u2212 9x + 4 is 2\t 1.\t What is the first degree expression to be added to \t\t(a) (2x + 1)(x + 4)(x \u2212 4) 16x6 + 8x4 - 2x3+ x2 + 2x + 1 in order to make it \t\t(b) (x2 + 16)(2x + 1) a perfect square? \t\t(c) 2(1 \u2212 2x)(x + 4)(x \u2212 4) \t\t(d) (2x \u2212 1)(x + 4)(x \u2212 4) \t\t(a) 5 x + 15 \t(b) \u2212 5 x \u2212 15 2 16 2 16 PRACTICE QUESTIONS 1\t 5.\t If P = 3x3 \u2212 5x + 9, Q = 4x3 + 5x2 \u2212 11 and \t\t(c) \u2212 5 x + 15 \t(d) + 2 x \u2212 15 R = 5x3 + 4x2 \u2212 3x + 7, then P \u2212 2Q + R is 2 16 2 16 \t\t(a) 2(3x2 + 4x \u2212 19) 2\t 2.\t Factorize the polynomial 8x3 \u2212 1 . \t\t(b) \u22126x2 \u2212 5x + 38 64 \t\t(c) \u22122(3x2 + 4x + 19) \uf8eb 1 \uf8f6 \uf8eb x 1\uf8f6 \t\t(d) \u22122(3x2 + 4x - 19) \t\t(a) \uf8ed\uf8ec 2x \u2212 4 \uf8f7\uf8f8 \uf8ed\uf8ec 4x 2 \u2212 2 + 16 \uf8f7\uf8f8 \t16.\t If g(x) = 3ax + 7a2b \u2212 13ab2 + 9by is a homogeneous \t\t(b) \uf8eb 2x \u2212 1 \uf8f6 \uf8eb 4x 2 + x \u2212 16 \uf8f6 expression in terms of a and b, then the values of x \uf8ed\uf8ec 8 \uf8f7\uf8f8 \uf8ec\uf8ed 2 \uf8f7\uf8f8 and y respectively are _____. \t\t(a) 2, 2\t\t (b) 2, 1 \t\t(c) \uf8eb 2x \u2212 1 \uf8f6 \uf8eb 4x 2 + 1 + x\uf8f6 \uf8ec\uf8ed 4 \uf8f7\uf8f8 \uf8ed\uf8ec 16 2 \uf8f8\uf8f7 \t\t(c) 3, 2\t\t (d) 3, 3 1\t 7.\t The polynomial 11a2 \u2212 12 2 a + 2 on factoriza- \t\t(d) \uf8eb 2x \u2212 1 \uf8f6 \uf8eb 4x 2 + x \u2212 16 \uf8f6 tion gives \uf8ed\uf8ec 4 \uf8f7\uf8f8 \uf8ec\uf8ed 2 \uf8f8\uf8f7 \t\t(a) (11a + 2 )(a \u2212 2 ) 2\t 3.\t The product of polynomials 3x3 \u2212 4x2 + 7 and x2 + 1 is \t\t(b) (a \u2212 2 )(11a \u2212 2 ) \t\t(a) 3x5 \u2212 4x4 + 3x3 + 3x2 + 7 \t\t(c) (a + 11)(a + 2 ) \t\t(b) x5 + 4x2 \u2212 2x + 3 \t\t(c) 3x5 \u2212 4x4 \u2212 3x3 + 4x + 8 \t\t(d) (11a \u2212 2 )(a + 2 ) \t\t(d) 3x5 \u2212 5x4 + 8x2 + 2x + 1","Polynomials and Square Roots of Algebraic Expressions 3.27 \t24.\t The LCM and HCF of two monomials is \t\t(a) 5x3 \u2212 2x2 + x + 4 60x4y5a6b6 and 5x2y3 respectively. If one of the two monomials is 15x4y3a6, then the other monomial is \t\t(b) \u22125x3 + 10x2 \u2212 5x - 9 \t\t(a) 12x2y3a6b6\t (b) 20x4y5b6 \t\t(c) x3 + 10x2 \u2212 5x + 9 \t\t(c) 20x2y5b6\t\t (d) 15x2y5b6 \t\t(d) 5x3 \u2212 8x2 + x - 1 \t25.\t Which of the following is a factor of the polynomial 2\t 8.\t The square root of xm2 \u2212n2 \u22c5 xn2 +2mn \u22c5 xn2 is f(x) = 2x3 \u2212 5x2 + x + 2? \t\t(a) xm\u2009+\u2009n\t \t(b) x(m+n)2 \t\t(a) x + 1\t\t (b) x + 2 \t\t(c) x(m\u2009+\u2009n)\/2\t\t (d) x 12(m+n )2 \t\t(c) 2x + 1\t\t (d) 2x \u2212 1 2\t 6.\t If 3x \u2212 1 is a factor of the polynomial 81x3 \u2212 45x2 \t29.\t x831 + y831 is always divisible by + 3a \u2212 6, then a is _____. \t\t(a) x - y\t\t (b) x2 + y2 \t\t(a) 8 \t\t (b) \u22127 \t\t(c) x + y\t\t (d) None of these 3 3 \t30.\t If (x + 1)(x + 2)(x + 3)(x + k) + 1 is a perfect \t\t(c) \u221210 \t\t (d) 11 square, then the value of k is 3 3 \t\t(a) 4\t\t (b) 5 \t27.\t If A = 4x3 \u2212 5x + 7, B = 2x3 \u2212 x2 + 3 and C = 5x3 \u2212 8x2 + 10, then A \u2212 2B \u2212 C is \t\t(c) 6\t\t (d) 7 Level 2 \t31.\t If A = 6x4 + 5x3 \u2212 14x2 + 2x + 2 and B = 3x2 \t\t(a) 2x\t\t (b) \u2212x \u2212 2x \u2212 1, then the remainder when A \u00f7 B is \t \t (c) 0\t\t (d) x \t\t(a) x\t\t (b) 2x \t \t (c) 3x\t\t (d) 4x \u2211\t36.\tFactorize a2(b4 \u2212 c4 ). a,b,c 3\t 2.\tThe polynomial x5 \u2212 a2x3 \u2212 x2y3 + a2y3 on PRACTICE QUESTIONS factorization gives \t\t(a) (a \u2212 b)2(b \u2212 c)2(c \u2212 a)2 \t \t (b) (a \u2212 b)(a + b)(b \u2212 c)(b + c)(c \u2212 a)(c + a) \t\t(a) (x \u2212 y)(x \u2212 a)(x + a)(x2 + y2 + xy) \t\t(c) (a + b)2(b + c)2(c + a)2 \t\t(b) (x + a)(x \u2212 y)(x \u2212 a)(x2 \u2212 y2 + xy) \t \t (d) None of these \t\t(c) (x + a)(x + y)(x \u2212 a)(x2 + y2 + xy) \t\t(d) None of these 3\t 7.\t The polynomial 6y4 \u2212 19y3 \u2212 23y2 + 10y + 8 on factorization gives \t33.\t The HCF of the polynomials x4 + 6x2 + 25, x3 \u2212 3x2 + 7x \u2212 5 and x2 + 5 \u2212 2x is \t\t(a) (y + 1)(y \u2212 4)(3y + 2)(2y + 1) \t\t(b) (y + 1)(y \u2212 4)(3y \u2212 2)(2y \u2212 1) \t\t(a) x2 \u2212 2x - 5\t (b) x2 \u2212 2x + 5 \t\t(c) (y + 1)(y \u2212 4)(3y \u2212 2)(2y + 1) \t\t(d) (y + 1)(y \u2212 4)(3y + 2)(2y \u2212 1) \t\t(c) x \u2212 1\t\t (d) 3x + 2 \t34.\t The HCF of the polynomials (2x \u2212 1)(5x2 \u2212 ax + 3) and (x \u2212 3)(2x2 + x + b) is (2x \u2212 1)(x \u2212 3). \t\tThen the values of a and b respectively 3\t 8.\t If the LCM of the polynomials (y \u2212 3)a(2y + 1)b are _____. (y + 13)7 and (y - 3)4(2y + 1)9(y + 13)c is (y \u2212 3)6 (2y + 1)10(y + 13)7, then the least value of \t\t(a) 16, \u22121\t\t (b) \u221216, 1 a + b + c is \t\t(c) \u221216, -1\t\t (d) 16, 1 \t\t(a) 23\t\t (b) 3 \t35.\t The remainder when x45 is divided by x2 \u2212 \t\t(d) 10\t\t (d) 16 1 is","3.28 Chapter 3 \t39.\t The LCM of the polynomials 195(x + 3)2(x \u2212 2) \t\t(a) \uf8eb 3x 2 \u2212 1 + 2 \uf8f6 \uf8eb 3x 2 + 1 + 2 \uf8f6\uf8f7\uf8f8 (x + 1)2 and 221(x + 1)3(x + 3)(x + 4) is _____. \uf8ed\uf8ec x2 \uf8f8\uf8f7 \uf8ec\uf8ed x2 \t\t(a) 221(x + 3)2 (x + 1)2(x \u2212 2)(x \u2212 14) \t\t(b) \uf8eb 3x 2 \u2212 1 \u2212 2 \uf8f6 \uf8eb 3x 2 + 1 + 2 \uf8f6 \t\t(b) 13(x + 3)(x + 1)2 \uf8ec\uf8ed x2 \uf8f7\uf8f8 \uf8ed\uf8ec x2 \uf8f8\uf8f7 \t\t(c) 3315 (x + 3)2 (x + 1)3(x \u2212 2)(x + 4) \t\t(d) None of these \t\t(c) \uf8eb 3x 2 \u2212 1 + 2 \uf8f6 \uf8eb 3x 2 \u2212 1 + 2 \uf8f6 \uf8ed\uf8ec x2 \uf8f7\uf8f8 \uf8ed\uf8ec x2 \uf8f7\uf8f8 4\t 0.\t For what value of k the HCF of x2 + x + (5k \u2212 1) \t\t(d) \uf8eb 3x 2 + 1 + 2 \uf8f6 \uf8eb 3x 2 + 1 \u2212 2\uf8f6\uf8f8\uf8f7 and x2 \u2212 6x + (3k + 11) is (x \u2212 2)? \uf8ec\uf8ed x2 \uf8f7\uf8f8 \uf8ed\uf8ec x2 \t\t(a) 2\t\t (b) 2 \t46.\t The following are the steps involved in factorizing 64x6 \u2212 y6. Arrange them in sequential order. \t\t(c) \u22122\t\t (d) \u22121 \t41.\t The HCF of the polynomials 9(x + a)p(x \u2212 b)q \t\t(A) {(2x)3 + y3} {(2x)3 \u2212 y3} (x + c)r and 12(x + a)p\u2009+\u20093(x \u2212 b)q\u2009\u2212\u20093(x + c)r\u2009+\u20092 is 3(x + a)6(x \u2212 b)6(x + c)6, then the value of \t\t(B) (8x3)2 \u2212 (y3)2 p + q \u2212 r is \t \t (C) (8x3 + y3)(8x3 \u2212 y3) \t\t(a) 21\t\t (b) 9 \t\t(D) (2\u0007x + y)(4x2 \u2212 2xy + y2)(2x \u2212 y)(4x2 + 2xy + y2) \t\t(c) 15\t\t (d) 6 \t\t(a) BADC\t\t (b) BDAC 4\t 2.\t The remainders obtained when the polynomial x3\u00a0+ x2 \u2212 9x \u2212 9 divided by x, x + 1 and x + 2 \t\t(c) BCAD\t\t (d) BACD respectively are _____. \t47.\t If a + b + c = 0, show that a3 + b3 + c3 = 3abc. \t\t(a) \u22129, 0, \u221215\t(b) \u22129, \u221216, 5 \t\tThe following are the steps involved in showing \t\t(c) 0, 0, 5\t\t (d) \u22129, 0, 5 the above result. Arrange them in sequential order. \t43.\t Find the value of \t\t(A) a3 + b3 + 3ab(\u2212c) = \u2212c3 \t \t (B) (a + b)3 = (\u2212c)3 (b (a + b)2 a ) + (a (b + c )2 a ) + (a (c + a)2 c ) . \t \t (C) a + b + c = 0 \u21d2 a + b = \u2212c \u2212 c )(c \u2212 \u2212 b)(c \u2212 \u2212 b)(b \u2212 \t \t (D) a3 + b3 + 3ab(a + b) = \u2212c3 PRACTICE QUESTIONS \t\t(E) a3 + b3 + c3 = 3abc \t\t(a) -1\t\t (b) 0 \t\t(c) 1\t\t (d) 2 \t \t (a) ABDCE\t\t (b) BCDAE \t44.\t Find the square root of the expression \t\t(c) CBDAE\t\t (d) CADBE 1 (x2 + y2 + z2 ) + 2 \uf8eb 1 + 1 + 1 \uf8f7\uf8f6 . \t48.\t If the HCF of 8x3y\u2009a and 12xby2 is 4xayb, then find xyz \uf8ec x y z \uf8f8 the maximum value of a + b. \uf8ed x+y+z \t\t(a) 2\t\t (b) 4 xyz \t\t(a) \t\t(c) 6\t\t (d) Cannot be determined \t\t(b) yz + zx + xy \t49.\t The polynomial 5x5 \u2212 3x3 + 2x2 \u2212 k gives a remain- x y z der 1, when divided by x + 1. Find the value of k. \t\t(a) 5\t\t (b) \u22121 \t\t(c) x + y + z \t\t(c) 2\t\t (d) 1 \t\t(d) x + y + z 5\t 0.\t Factorize: a3 + b3 + 3ab \u2212 1. yz xz xy \t\t(a) (a + b \u2212 1)(a2 + b2 + a + b + 1 - ab) \t\t(b) (a + b \u2212 1)(a2 + b2 + a + b - 1 + ab) \t45.\t Factorize the expression 9x4 + 1 + 2. \t\t(c) (a + b \u2212 1)(a2 + b2 - a - b + 1 + ab) x4 \t\t(d) None of these","Polynomials and Square Roots of Algebraic Expressions 3.29 \t51.\t If f and g are two polynomials of degrees 3 and 4 \t\t(a) \uf8ee \u2211 x \u2212 \u2211 \uf8f9 respectively, then what is the degree of f - g? 2\uf8ef xy \uf8fa \t\t(a) 1 \uf8ef\uf8f0x,y,z x,y,z \uf8fa\uf8fb \t\t(b) 3 \uf8ee \uf8f9 3\uf8ef x\uf8fa \t\t(c) 4 \t\t(b) \u2211 x2 \u2212 \u2211 \t\t(d) Cannot be determined \uf8ef\uf8f0x,y,z x,y,z \uf8fa\uf8fb 5\t 2.\t Find the square root of x2 9 \u2212 x \u2212 3 + 5 . \t\t(c) \uf8ee xy \u2212 \u2211 x 2 \uf8f9 9 + 4x2 3 2x 4 \uf8fa 2\uf8ef \u2211 \uf8ef\uf8f0x,y,z x,y,z \uf8fa\uf8fb 2x 3 1 x 3 \t\t(a) 3 + 2x \u2212 2 \t\t (b) 3 \u2212 2x +1 \uf8ee \uf8f9 3\uf8ef x\uf8fa \t\t(d) \u2211 x2 \u2212 \u2211 \t\t(c) 3 + 2 \u2212 1 \t\t (d) x + 3 \u2212 1 \uf8ef\uf8f0x,y,z x,y,z \uf8fa\uf8fb x 3x 2 3 2x 2 \uf8eb \uf8f62 \uf8eb \uf8f6 \t53.\t The square root of (xy + xz \u2212 yz)2 \u2212 4xyz(x \u2212 y) is \u2211 \u2211\t55.\t\uf8ec\uf8ed\uf8ec x,y,zx\uf8f8\uf8f7\uf8f7\u2212 \uf8ed\uf8ec\uf8ec x2 \uf8f8\uf8f7\uf8f7 = ______ . _____. x,y,z \t\t(a) xy + yz \u2212 2xyz \t \t (b) (x + y \u2212 2xy) \t\t(a) \u2211 x \t\t \u2211(b) \uf8eb\uf8f6 \t\t(c) (xy + 3 \u2212 y) x,y,z 2 \uf8ec\uf8ec\uf8ed x,y,z xy \uf8f8\uf8f7\uf8f7 \t\t(d) (xy + yz \u2212 zx) \u03c4 \t\t(c) \u03c0 xy \t\t \u2211(d) \uf8eb + \uf8f6 x,y,z \u03c4 2 \uf8ed\uf8ec\uf8ec x,y,z x y \uf8f7\uf8f7\uf8f8 \u2211 \u2211\t54.\t\uf8eb (x + 1)2 \uf8f6 \u2212 \uf8eb \uf8f62 \u2212 3 = _____ . \uf8ec\uf8ec\uf8ed \uf8f7\uf8f8\uf8f7 \uf8ec\uf8ec\uf8ed (x ) \uf8f7\uf8f7\uf8f8 x,y,z x,y,z Level 3 \t56.\t If 4x4 + 12x3 + 25x2 + 24x + 16 = ax2 + bx + c, \t58.\t If the each of algebraic expressions lx2 + mx + n, PRACTICE QUESTIONS then which of the following is true? mx2 + nx + l and nx2 + lx + m are prefect squares, \t\t(a) 2b = a - c \t \t (b) 2a = b + c then l +m = _____ . \t \t (c) 2b = a + c n \t \t (d) 2b = c - a \t\t(a) \u22124\t\t (b) 6 \t\t(c) \u22128\t\t (d) None of these \t57.\tFind the square root of the algebraic expression 5\t 9.\t Which of the following is to be added to make which is the average of the following expressions x6 \u2212 6x4 + 4x3 + 8x2 \u2212 10x + 3 a perfect square? x2 + 1 , \u2212 2 \uf8eb x \u2212 1\uf8f6 and \u2212 1. \t\t(a) (x \u2212 1)2\t\t (b) (x \u2212 2)2 x2 \uf8ed\uf8ec x \uf8f7\uf8f8 \t \t (c) (2x \u2212 3)2\t (d) (2x + 1)2 x\u2212 1 1 \t\t(a) 3 3 + x \uf8eb \uf8f63 \uf8ec\uf8ec\uf8ed x,y,z x \uf8f7\uf8f7\uf8f8 \u2211 \u2211\t60.\tResolve \u2212 x3. \t\t(b) x + 1 + x into factors: 3 x,y,z \t\t(a) (x + y)(y + z)(z + x) \t\t(c) 1 \uf8eb x \u2212 1 \u2212 1 \uf8f6 \t\t(b) -(x + y)(y + z)(z + x) 3 \uf8ed\uf8ec x \uf8f8\uf8f7 \t\t(c) 3(x + y)(y + z)(z + x) \t\t(d) None of these \t\t(d) -3(x + y)(y + z)(z + x)"]