{"id":17482,"date":"2021-07-19T11:46:48","date_gmt":"2021-07-19T06:16:48","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=17482"},"modified":"2022-03-02T10:26:05","modified_gmt":"2022-03-02T04:56:05","slug":"mcq-questions-for-class-11-maths-chapter-8","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/mcq-questions-for-class-11-maths-chapter-8\/","title":{"rendered":"MCQ Questions for Class 11 Maths Chapter 8 Binomial Theorem with Answers"},"content":{"rendered":"

Students can access the NCERT MCQ Questions for Class 11 Maths Chapter 8 Binomial Theorem with Answers Pdf free download aids in your exam preparation and you can get a good hold of the chapter. Use\u00a0MCQ Questions for Class 11 Maths with Answers<\/a> during preparation and score maximum marks in the exam. Students can download the Binomial Theorem Class 11 MCQs Questions with Answers from here and test their problem-solving skills. Clear all the fundamentals and prepare thoroughly for the exam taking help from Class 11 Maths Chapter 8 Binomial Theorem Objective Questions.<\/p>\n

Binomial Theorem Class 11 MCQs Questions with Answers<\/h2>\n

Students are advised to solve the Binomial Theorem Multiple Choice Questions of Class 11 Maths to know different concepts. Practicing the MCQ Questions on Binomial Theorem Class 11 with answers will boost your confidence thereby helping you score well in the exam.<\/p>\n

Explore numerous MCQ Questions of Binomial Theorem Class 11 with answers provided with detailed solutions by looking below.<\/p>\n

Binomial Theorem Calculator<\/a> online with solution and steps. Detailed step by step solutions to your Binomial Theorem problems online with our math solver.<\/p>\n

Question 1.
\nThe number (101)100<\/sup> – 1 is divisible by
\n(a) 100
\n(b) 1000
\n(c) 10000
\n(d) All the above<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) All the above
\nGiven, (101)100<\/sup> – 1 = (1 + 100)100<\/sup> – 1
\n= [100<\/sup>C0<\/sub> + 100<\/sup>C1<\/sub> \u00d7 100 + 100<\/sup>C2<\/sub> \u00d7 (100)\u00b2 + ……….+ 100<\/sup>C100<\/sub> \u00d7 (100)100<\/sup>] – 1
\n= 1 + [100<\/sup>C1<\/sub> \u00d7 100 + 100<\/sup>C2<\/sub> \u00d7 (100)\u00b2 + ……….+ 100<\/sup>C100<\/sub> \u00d7 (100)100<\/sup>] – 1
\n= 100<\/sup>C1<\/sub> \u00d7 100 + 100<\/sup>C2<\/sub> \u00d7 (100)\u00b2 + ……….+ 100<\/sup>C100<\/sub> \u00d7 (100)100<\/sup>
\n= 100 \u00d7 100 + 100<\/sup>C2<\/sub> \u00d7 (100)\u00b2 + ……….+ 100<\/sup>C100<\/sub> \u00d7 (100)100<\/sup>
\n= (100)\u00b2 + 100<\/sup>C2<\/sub> \u00d7 (100)\u00b2 + ……….+ 100<\/sup>C100<\/sub> \u00d7 (100)100<\/sup>
\n= (100)\u00b2 [1 + 100<\/sup>C2<\/sub> + ……….+ 100<\/sup>C100<\/sub> \u00d7 (100)98<\/sup>]
\nWhich is divisible by 100, 1000 and 10000<\/p>\n<\/details>\n

\n

Question 2.
\nThe value of -1\u00b0 is
\n(a) 1
\n(b) -1
\n(c) 0
\n(d) None of these<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) -1
\nFirst we find 10
\nSo, 10 = 1
\nNow, -10 = -1<\/p>\n<\/details>\n

\n

Question 3.
\nIf the fourth term in the expansion (ax + 1\/x)\u207f is 5\/2, then the value of x is
\n(a) 4
\n(b) 6
\n(c) 8
\n(d) 5<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) 6
\nGiven, T4<\/sub> = 5\/2
\n\u21d2 T3+1<\/sub> = 5\/2
\n\u21d2 \u207fC3<\/sub> \u00d7 (ax)n-3<\/sup> \u00d7 (1\/x)\u00b3 = 5\/2
\n\u21d2 \u207fC3<\/sub> \u00d7 an-3<\/sup> \u00d7 xn-3<\/sup> \u00d7 (1\/x)\u00b2 = 5\/2
\nClearly, RHS is independent of x,
\nSo, n – 6 = 0
\n\u21d2 n = 6<\/p>\n<\/details>\n

\n

Question 4.
\nThe number 111111 \u2026\u2026\u2026\u2026.. 1 (91 times) is
\n(a) not an odd number
\n(b) none of these
\n(c) not a prime
\n(d) an even number<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) not a prime
\n111111 \u2026\u2026\u2026\u2026.. 1 (91 times) = 91 \u00d7 1 = 91, which is divisible by 7 and 13.
\nSo, it is not a prime number.<\/p>\n<\/details>\n

\n

Question 5.
\nIn the expansion of (a + b)\u207f, if n is even then the middle term is
\n(a) (n\/2 + 1)th<\/sup> term
\n(b) (n\/2)th<\/sup> term
\n(c) nth<\/sup> term
\n(d) (n\/2 – 1)th<\/sup> term<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) (n\/2 + 1)th<\/sup> term
\nIn the expansion of (a + b)\u207f
\nif n is even then the middle term is (n\/2 + 1)th<\/sup> term<\/p>\n<\/details>\n

\n

Question 6.
\nThe number of terms in the expansion (2x + 3y – 4z)\u207f is
\n(a) n + 1
\n(b) n + 3
\n(c) {(n + 1) \u00d7 (n + 2)}\/2
\n(d) None of these<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) {(n + 1) \u00d7 (n + 2)}\/2
\nTotal number of terms in (2x + 3y – 4z)\u207f is
\n= n+3-1<\/sup>C3-1<\/sub>
\n= n+2<\/sup>C2<\/sub>
\n= {(n + 1) \u00d7 (n + 2)}\/2<\/p>\n<\/details>\n

\n

Question 7.
\nIf A and B are the coefficient of x\u207f in the expansion (1 + x)2n<\/sup> and (1 + x)2n-1<\/sup> respectively, then A\/B equals
\n(a) 1
\n(b) 2
\n(c) 1\/2
\n(d) 1\/n<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) 2
\nA\/B = \u00b2\u207fCn<\/sub>\/ 2n-1<\/sup>Cn<\/sub>
\n= {(2n)!\/(n! \u00d7 n!)}\/{(2n – 1)!\/(n! \u00d7 (n – 1!))}
\n= {2n(2n – 1)!\/(n(n – 1)! \u00d7 n!)}\/{(2n – 1)!\/(n! \u00d7 (n – 1!))}
\n= 2
\nSo, A\/B = 2<\/p>\n<\/details>\n

\n

Question 8.
\nThe coefficient of y in the expansion of (y\u00b2 + c\/y)5<\/sup> is
\n(a) 29c
\n(b) 10c
\n(c) 10c\u00b3
\n(d) 20c\u00b2<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) 10c\u00b3
\nWe have,
\nTr+1<\/sub> = 5<\/sup>Cr<\/sub> \u00d7(y\u00b2)5-r<\/sup> \u00d7 (c\/y)r<\/sup>
\n\u21d2 Tr+1<\/sub> = 5<\/sup>Cr<\/sub> \u00d7 y10-3r<\/sup> \u00d7 cr<\/sup>
\nFor finding the coefficient of y,
\n\u21d2 10 – 3r = 1
\n\u21d2 33r = 9
\n\u21d2 r = 3
\nSo, the coefficient of y = 5<\/sup>C3<\/sub> \u00d7 c\u00b3
\n= 10c\u00b3<\/p>\n<\/details>\n

\n

Question 9.
\nThe coefficient of x-4<\/sup> in (3\/2 – 3\/x\u00b2)10<\/sup> is
\n(a) 405\/226
\n(b) 504\/289
\n(c) 450\/263
\n(d) None of these<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) None of these
\nLet x-4<\/sup> occurs in (r + 1)th term.
\nNow, Tr+1<\/sub> = 10<\/sup>Cr<\/sub> \u00d7 (3\/2)10-r<\/sup> \u00d7(-3\/x\u00b2)r<\/sup>
\n\u21d2 Tr+1<\/sub> = 10<\/sup>Cr<\/sub> \u00d7 (3\/2)10-r<\/sup> \u00d7(-3)r<\/sup> \u00d7 (x)-2r<\/sup>
\nNow, we have to find the coefficient of x-4<\/sup>
\nSo, -2r = -4
\n\u21d2 r = 2
\nNow, the coefficient of x-4<\/sup> = 10<\/sup>C2<\/sub> \u00d7 (3\/2)10-2<\/sup> \u00d7 (-3)2<\/sup>
\n= 10<\/sup>C2<\/sub> \u00d7 (3\/2)8<\/sup> \u00d7 (-3)2<\/sup>
\n= 45 \u00d7 (3\/2)8<\/sup> \u00d7 9
\n= (312<\/sup> \u00d7 5)\/28<\/sup><\/p>\n<\/details>\n

\n

Question 10.
\nIf n is a positive integer, then 9n+1<\/sup> – 8n – 9 is divisible by
\n(a) 8
\n(b) 16
\n(c) 32
\n(d) 64<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) 64
\nLet n = 1, then
\n9n+1<\/sup> – 8n – 9 = 91+1<\/sup> – 8 \u00d7 1 – 9 = 9\u00b2 – 8 – 9 = 81 – 17 = 64
\nwhich is divisible by 64
\nLet n = 2, then
\n9n+1<\/sup> – 8n – 9 = 92+1<\/sup> – 8 \u00d7 2 – 9 = 9\u00b3 – 16 – 9 = 729 – 25 = 704 = 11 \u00d7 64
\nwhich is divisible by 64
\nSo, for any value of n, 9n+1<\/sup> – 8n – 9 is divisible by 64<\/p>\n<\/details>\n

\n

Question 11.
\nThe general term of the expansion (a + b)\u207f is
\n(a) Tr+1<\/sub> = \u207fCr<\/sub> \u00d7 ar<\/sup> \u00d7 br<\/sup>
\n(b) Tr+1<\/sub> = \u207fCr<\/sub> \u00d7 ar<\/sup> \u00d7 bn-r<\/sup>
\n(c) Tr+1<\/sub> = \u207fCr<\/sub> \u00d7 an-r<\/sup>\u00d7 bn-r<\/sup>
\n(d) Tr+1<\/sub> = \u207fCr<\/sub> \u00d7 an-r<\/sup> \u00d7 br<\/sup><\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) Tr+1<\/sub> = \u207fCr<\/sub> \u00d7 an-r<\/sup> \u00d7 br<\/sup>
\nThe general term of the expansion (a + b)\u207f is
\nTr+1<\/sub> = \u207fCr<\/sub> \u00d7 an-r<\/sup> \u00d7 br<\/sup><\/p>\n<\/details>\n

\n

Question 12.
\nIn the expansion of (a + b)\u207f, if n is even then the middle term is
\n(a) (n\/2 + 1)th<\/sup>\u00a0term
\n(b) (n\/2)th<\/sup> term
\n(c) nth<\/sup> term
\n(d) (n\/2 – 1)th<\/sup> term<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) (n\/2 + 1)th<\/sup>\u00a0term
\nIn the expansion of (a + b)\u207f,
\nif n is even then the middle term is (n\/2 + 1)th<\/sup> term<\/p>\n<\/details>\n

\n

Question 13.
\nThe smallest positive integer for which the statement 3n+1<\/sup> < 4\u207f is true for all
\n(a) 4
\n(b) 3
\n(c) 1
\n(d) 2<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) 4
\nGiven statement is: 3n+1<\/sup> < 4\u207f is
\nLet n = 1, then
\n31+1<\/sup> < 41<\/sup> = 3\u00b2 < 4 = 9 < 4 is false
\nLet n = 2, then
\n32+1<\/sup> < 4\u00b2 = 3\u00b3 < 4\u00b2 = 27 < 16 is false
\nLet n = 3, then
\n33+1<\/sup> < 4\u00b3 = 34<\/sup> < 4\u00b3 = 81 < 64 is false
\nLet n = 4, then
\n34+1<\/sup> < 44<\/sup> = 35<\/sup> < 44<\/sup> = 243 < 256 is true.
\nSo, the smallest positive number is 4<\/p>\n<\/details>\n

\n

Question 14.
\nThe number of ordered triplets of positive integers which are solution of the equation x + y + z = 100 is
\n(a) 4815
\n(b) 4851
\n(c) 8451
\n(d) 8415<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) 4851
\nGiven, x + y + z = 100
\nwhere x \u2265 1, y \u2265 1, z \u2265 1
\nLet u = x – 1, v = y – 1, w = z – 1
\nwhere u \u2265 0, v \u2265 0, w \u2265 0
\nNow, equation becomes
\nu + v + w = 97
\nSo, the total number of solution = 97+3-1<\/sup>C3-1<\/sub>
\n= 99<\/sup>C2<\/sub>
\n= (99 \u00d7 98)\/2
\n= 4851<\/p>\n<\/details>\n

\n

Question 15.
\nif n is a positive ineger then 2\u00b3\u207f – 7n – 1 is divisible by
\n(a) 7
\n(b) 9
\n(c) 49
\n(d) 81<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) 49
\nGiven, 2\u00b3\u207f – 7n – 1 = 23\u00d7n<\/sup> – 7n – 1
\n= 8\u207f – 7n – 1
\n= (1 + 7)\u207f – 7n – 1
\n= {\u207fC0<\/sub> + \u207fC1<\/sub> 7 + \u207fC2<\/sub> 7\u00b2 + …….. + \u207fCn<\/sub> 7\u207f} – 7n – 1
\n= {1 + 7n + \u207fC2<\/sub> 7\u00b2 + …….. + \u207fCn<\/sub> 7\u207f} – 7n – 1
\n= \u207fC2<\/sub> 7\u00b2 + …….. + \u207fCn<\/sub> 7\u207f
\n= 49(\u207fC2<\/sub> + …….. + \u207fCn<\/sub> 7n-2<\/sup>)
\nwhich is divisible by 49
\nSo, 2\u00b3\u207f – 7n – 1 is divisible by 49<\/p>\n<\/details>\n

\n

Question 16.
\nThe greatest coefficient in the expansion of (1 + x)10<\/sup> is
\n(a) 10!\/(5!)
\n(b) 10!\/(5!)\u00b2
\n(c) 10!\/(5! \u00d7 4!)\u00b2
\n(d) 10!\/(5! \u00d7 4!)<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) 10!\/(5!)\u00b2
\nThe coefficient of xr in the expansion of (1 + x)10<\/sup> is 10<\/sup>Cr<\/sub> and 10<\/sup>Cr<\/sub> is maximum for r = 10\/ = 5
\nHence, the greatest coefficient = 10<\/sup>C5<\/sub>
\n= 10!\/(5!)\u00b2<\/p>\n<\/details>\n

\n

Question 17.
\nIf A and B are the coefficient of xn in the expansion (1 + x)2n<\/sup> and (1 + x)2n-1<\/sup> respectively, then A\/B equals
\n(a) 1
\n(b) 2
\n(c) 1\/2
\n(d) 1\/n<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) 2
\nA\/B = \u00b2\u207fCn<\/sub>\/2n-1<\/sup>Cn<\/sub>
\n= {(2n)!\/(n! \u00d7 n!)}\/{(2n – 1)!\/(n! \u00d7 (n – 1!))}
\n= {2n(2n – 1)!\/(n(n – 1)! \u00d7 n!)}\/{(2n – 1)!\/(n! \u00d7 (n – 1!))}
\n= 2
\nSo, A\/B = 2<\/p>\n<\/details>\n

\n

Question 18.
\n(1.1)10000<\/sup> is _____ 1000
\n(a) greater than
\n(b) less than
\n(c) equal to
\n(d) None of these<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) greater than
\nGiven, (1.1)10000<\/sup> = (1 + 0.1)10000<\/sup>
\n= 10000<\/sup>C0<\/sub> + 10000<\/sup>C1<\/sub> \u00d7(0.1) + 10000<\/sup>C2<\/sub> \u00d7 (0.1)\u00b2 + other +ve terms
\n= 1 + 10000 \u00d7 (0.1) + other +ve terms
\n= 1 + 1000 + other +ve terms
\n> 1000
\nSo, (1.1)10000<\/sup> is greater than 1000<\/p>\n<\/details>\n

\n

Question 19.
\nIf n is a positive integer, then (\u221a3+1)\u00b2\u207f + (\u221a3\u22121)\u00b2\u207f is
\n(a) an odd positive integer
\n(b) none of these
\n(c) an even positive integer
\n(d) not an integer<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) an even positive integer
\nSince n is a positive integer, assume n = 1
\n(\u221a3 + 1)\u00b2 + (\u221a3 – 1)\u00b2
\n= (3 + 2\u221a3 + 1) + (3 – 2\u221a3 + 1) {since (x + y)\u00b2 = x\u00b2 + 2xy + y\u00b2}
\n= 8, which is an even positive number.<\/p>\n<\/details>\n

\n

Question 20.
\nif y = 3x + 6x\u00b2 + 10x\u00b3 + ………. then x =
\n(a) 4\/3 – {(1 \u00d7 4)\/(3\u00b2 \u00d7 2)}y\u00b2 + {(1 \u00d7 4 \u00d7 7)\/(3\u00b2 \u00d73)}y\u00b3 – ………..
\n(b) -4\/3 + {(1 \u00d7 4)\/(3\u00b2 \u00d7 2)}y\u00b2 – {(1 \u00d7 4 \u00d7 7)\/(3\u00b2 \u00d73)}y\u00b3 + ………..
\n(c) 4\/3 + {(1 \u00d7 4)\/(3\u00b2 \u00d7 2)}y\u00b2 + {(1 \u00d7 4 \u00d7 7)\/(3\u00b2 \u00d73)}y\u00b3 + ………..
\n(d) None of these<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) None of these
\nGiven, y = 3x + 6x\u00b2 + 10x\u00b3 + ……….
\n\u21d2 1 + y = 1 + 3x + 6x\u00b2 + 10x\u00b3 + ……….
\n\u21d2 1 + y = (1 – x)-3<\/sup>
\n\u21d2 1 – x = (1 + y)-1\/3<\/sup>
\n\u21d2 x = 1 – (1 + y)-1\/3<\/sup>
\n\u21d2 x = (1\/3)y – {(1 \u00d7 4)\/(3\u00b2 \u00d7 2)}y\u00b2 + {(1 \u00d7 4 \u00d7 7)\/(3\u00b2 \u00d7 3!)}y\u00b3 – ………..<\/p>\n<\/details>\n

\n

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Students can access the NCERT MCQ Questions for Class 11 Maths Chapter 8 Binomial Theorem with Answers Pdf free download aids in your exam preparation and you can get a good hold of the chapter. Use\u00a0MCQ Questions for Class 11 Maths with Answers during preparation and score maximum marks in the exam. Students can download …<\/p>\n

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