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Q.1. The greatest positive integer k, for which 49^{k} + 1 is a factor of the sum 49^{125} + 49^{124 }+ ...+ 49^{2} + 49 + 1, is (2020)
(1) 32
(2) 63
(3) 60
(4) 65
Ans. (2)
Solution. We have
1 + 49 + 49^{2} + .... + 49^{125}
Hence, the greatest positive integer value of k is 63.
Q.2. If the sum of the coefficients of all even powers of x in the product (1 + x + x^{2} + ... + x^{2n})(1  x + x^{2}  x^{3} + ... + x^{2n}) is 61, then n is equal to ____.
Ans. (30.00)
Solution. We have
(1 + x + x^{2} + ... + x^{2n})(1  x + x^{2}  x^{3} + ... + x^{2n}) = a_{0} + a_{1}x + a_{2}x^{2} + a_{3}x^{3} + ... + a_{4n}x^{4n}
Substituting x = 1, we get
a_{0} + a_{1} + a_{2 }+ ... + a_{4n} = 2n + 1 (1)
Substituting x = 1 here, we get
a_{0}  a_{1} + a_{2}  a_{3}_{ }+ ... + a_{4n} = 2n + 1 (2)
From Eqs. (1) and (2), we get
a_{0} + a_{2} + a_{4} + ... + a_{4n} = 2n + 1 ...(3)
Now, 2n + 1 = 61 ⇒ n = 30
Q.3. The coefficient of x^{7} in the expression (1 + x)^{10} + x(1 +x)^{9} + x^{2}(1 +x)^{8} + ... + x^{10} is (2020)
(1) 210
(2) 330
(3) 120
(4) 420
Ans. (2)
Solution. We have
(1 + x)^{10} + x(1 +x)^{9} + x^{2}(1 +x)^{8} + ... + x^{10}
Therefore, the coefficient of x^{7} in the expression is
Q.4. If α and β be the coefficients of x^{4} and x^{2} respectively in the expansion of then (2020)
(1) α + β = 60
(2) α + β = 30
(3) α  β = 60
(4) α  β = 132
Ans. (4)
Solution. We have
(x + a)^{n} + (x  a)^{n} = 2(T_{1} + T_{3} + T_{5} + ...)
= 2[^{6}C_{0}x^{6} + ^{6}C_{2}x^{4}(x^{2}  1) + ^{6}C_{4}x^{2}(x^{2}  1)^{2} + ^{6}C_{6}(x^{2}  1)^{3}]
= 64x^{6}  96x^{4} + 36x^{2}
Hence, α  β =  96  36 = 132
Q.5. The coefficient of x^{4} in the expansion of (1 + x + x^{2})^{10} is ____. (2020)
Ans. (615.00)
Solution. The general term of the expansion of (1 + x + x^{2})^{10} is
For coefficient of x^{4}, β + 2γ = 4. So,
...(1)
...(2)
...(3)
Hence, the coefficient of x^{4} in the expansion of (1 + x + x^{2})^{10} is 210 + 360 + 45 = 615.
Q.6. In the expansion of if l_{1} is the least value of the term independent of x when and l_{2} is the least value of the term independent of x when then the ratio l_{2} : l_{1} is equal to (2020)
(1) 1 : 8
(2) 1 : 16
(3) 8 : 1
(4) 16 : 1
Ans. (4)
Solution. The term independent from x in expression is
If So, l_{1} = ^{16}C_{8} . 2^{8} ...(1)
If So, ...(2)
Hence,
Q.7. The coefficient of t^{4} in the expansion of (2019)
(1) 14
(2) 15
(3) 10
(4) 12
Ans. (2)
Solution. Consider the expression
Hence, the coefficient of
= 15
Q.8. If then k equals: (2019)
(1) 400
(2) 50
(3) 200
(4) 100
Ans. (4)
Solution. Consider the expression,
∴ k=100
Q.9. If the third term in the binomial expansion of equals 2560, then a possible value of x is: (2019)
(1) 1/4
(2) 4√2
(3) 1/8
(4) 2√2
Ans. (1)
Solution. Third term of
Q.10. The positive value of λ for which the coefficient of x^{2} in the expression is 720, is: (2019)
(1) 4
(2) 2√2
(3) √5
(4) 3
Ans. (1)
Solution. Since, coefficient of x^{2} in the expression is a constant term, then Coefficient of x^{2} in x^{2}= coefficient of constant term in
Then, for constant term,
Coefficient of x^{2} in expression = ^{10}C_{2}λ^{2} = 720
⇒λ = 4
Hence, required value of λ is 4.
Q.11. If then K is equal to: (2019)
(1) (25)^{2}
(2) 2^{25}  1
(3) 2^{24}
(4) 2^{25 }
Ans. (4)
Solution.
Then, by comparison, K = 2^{25 }
Q.12. The value of r for which is maximum, is: (2019)
(1) 15
(2) 20
(3) 11
(4) 10
Ans. (2)
Solution. Consider the expression
For maximum value of above expression r should be equal to 20.
Which is the maximum value of the expression,
So, r = 20.
Q.13. The sum of the real values of x for which the middle term in the binomial expansion of equals 5670 is: (2019)
(1) 0
(2) 6
(3) 4
(4) 8
Ans. (1)
Solution. Middle Term, term in the binomial expansion of
⇒ x^{8}  81 = 0
∴ sum of all values of x = sum of roots of equation (x^{8}  81 = 0)
Q.14. Let S_{n} = 1 + q + q^{2} +.... + q^{n} and
where q is a real number and q ≠ 1. If
^{101}C_{1} + ^{101}C_{2}·S_{1} + .... + ^{101}C_{100}·S_{100} = αT_{100}, then α is equal to: (2019)
(1) 2^{99}
(2) 202
(3) 200
(4) 2^{100 }
Ans. (4)
Solution.
Q.15. Let(x+10)^{50} + (x10)^{50} = a_{o} + a_{1}x+a_{2}x^{2} + .... + a_{50}x^{50}, for all x ∈ R; then is equal to: (2019)
(1) 12.50
(2) 12.00
(3) 12.25
(4) 12.75
Ans. (3)
Solution.
= 12.25
Q.16. A ratio of the 5^{th} term from the beginning to the 5^{th} term from the end in the binomial expansion of is: (2019)
(1)
(2)
(3)
(4)
Ans. (3)
Solution.
5^{th} term from beginning
and 5^{th} term from end T_{115+1}
∴ T_{5} : T_{7} =
Q.17. The total number is irrational terms in the binomial expansion of is: (2019)
(1) 55
(2) 49
(3) 48
(4) 54
Ans. (4)
Solution. Let the general term of the expansion
Then, for getting rational terms, r should be multiple of L.C.M. of (5, 10)
Then, r can be 0, 10, 20, 30, 40, 50, 60.
Since, total number of terms = 61
Hence, total irrational terms = 61  7 = 54
Q.18. The sum of the coefficients of all even degree terms in x in the expansion of is equal to: (2019)
(1) 29
(2) 32
(3) 26
(4) 24
Ans. (4)
Solution.
Hence, the sum of coefficients of even powers of
x = 2[1  15 + 15 + 15  3 1] = 24
Q.19. The sum of the series 2^{.20}C_{0} + 5^{.20}C_{1} + 8^{.20}C_{2} + 11^{.20}C_{3} + ... + 62^{.20}C_{20} is equal to: (2019)
(1) 2^{26}
(2) 2^{25}
(3) 2^{23}
(4) 2^{24 }
Ans. (2)
Solution.
Q.20. If the fourth term in the binomial expansion of is equal to 200, and x > 1, then the value of x is: (2019)
(1) 100
(2) 10
(3) 10^{3}
(4) 10^{4 }
Ans. (2)
Solution. Fourth term is equal to 200.
Taking log_{10} on both sides and putting log_{10} x = t
According to the question x > 1, ∴ x = 10.
Q.21. If the fourth term in the Binomial expansion of is 20 x 8^{7}, then a value of x is: (2019)
(1) 8^{3}^{ }
(2) 8^{2}
(3) 8
(4) 8^{2 }
Ans. (2)
Solution.
Now, take log_{8} on both sides, we get
Q.22. If some three consecutive coefficients in the binomial expansion of (x+1)^{n} in powers of x are in the ratio 2:15:70, then the average of these three coefficients is: (2019)
(1) 964
(2) 232
(3) 227
(4) 625
Ans. (2)
Solution.
Q.23. If the coefficients of x^{2} and x^{3} are both zero, in the expansion of the expression (1 + ax + bx^{2}) (13x)^{15} in powers of x, then the ordered pair (a, b) is equal to: (2019)
(1) (28,861)
(2) (54,315)
(3) (28,315)
(4) (21,714)
Ans. (3)
Solution. Given expression is
Coefficient of x^{2} = 0
Q.24. The smallest natural number n, such that the coefficient of x in the expansion of (2019)
(1) 38
(2) 58
(3) 23
(4) 35
Ans. (1)
Solution.
To find coefficient of x, 2n  5r = 1
Given ^{n}C_{r} = ^{n}C_{23} ⇒ r = 23 or n  r = 23
∴ n = 58 or n = 38
Minimum value is n = 38
Q.25. The coefficient of x^{18} in the product (1+x)(1x)^{10}(1+x+x^{2})^{9 }is: (2019)
(1) 84
(2) 126
(3) 84
(4) 126
Ans. (1)
Solution. Given expression,
Q.26. If ^{20}C_{1} + (2^{2}) ^{20}C_{2} +(3^{2}) ^{20}C_{3}+ ...... + (20^{2}) ^{20}C_{20}= A(2^{β}), then the ordered pair (A, P) is equal to: (2019)
(1) (420, 19)
(2) (420, 18)
(3) (380,18)
(4) (380, 19)
Ans. (2)
Solution.
=420 x 2^{18}
Now, compare it with R.H.S., A = 420 and β = 18
Q.27. The term independent of x in the expansion of is equal to: (2019)
(1)  72
(2) 36
(3)  36
(4)  108
Ans. (4)
Solution. Given expression is,
Term independent of x,
= 72+ 36 = 36
Q.28. The sum of the coefficients of all odd degree terms in the expansion of (2018)
(1) 1
(2) 0
(3) 1
(4) 2
Ans. (4)
Solution.
Sum of coefficient of odd powers = 2(1  10 + 10) = 2.
Q.29. If n is the degree of the polynomial and m is the coefficient of x^{n} in it, then the ordered pair (n,m) is (2018)
(1) (12, (20)^{4})
(2) (8, 5(10)^{4})
(3) (24, (10)^{8})
(4) (12, 8(10)^{4})
Ans. (4)
Solution.
After rationalising the polynomial,we get
So,the degree of the polynomial is 12,
Now coefficient of x^{12} is
Q.30. The coefficient of x^{2} in the expansion of the product (2  x^{2}) · ((1 + 2x + 3x^{2})^{6} + (1  4x^{2})^{6}) is: (2018)
(1) 107
(2) 108
(3) 155
(4) 106
Ans. (4)
Solution. (a) Let a = ((1 + 2x + 3x^{2})^{6} + (1  4x^{2})^{6})
∴ Coefficient of x^{2} in the expansion of the product
(2  x^{2}) ((1 + 2x + 3x^{2})^{6}+ (1  4x^{2})^{6})
= 2 (Coefficient of x^{2} in a)  1 (Constant of expansion)
In the expansion of ((1 + 2x + 3x^{2})^{6} + (1  4x^{2})^{6}).
Constant = 1 + 1 = 2
Coefficient of x^{2} = [Coefficient of x^{2} in (^{6}C_{0}(1 + 2x)^{6}(3x^{2})^{0})] + [Cofficient of x^{2} in (^{6}C_{1}(1 + 2x)^{5} (3x^{2})^{1})]  [^{6}C_{1} (4x^{2})] = 60 + 6 x 3  24 = 54
∴ The coefficient of x^{2} in (2  x^{2})((1 + 2x + 3x^{2})^{6}+ (1  4x^{2})^{6})
=2 x 54  1 (2)= 108  2 = 106
Q.31. The value of (^{21}C_{1}  ^{10}C_{1}) + (^{21}C_{2}  ^{10}C_{2}) + (^{21}C_{3}  ^{10}C_{3}) + (^{21}C_{4}  ^{10}C_{4}) + ... + (^{21}C_{10}  ^{10}C_{10}) is: (2017)
(1) 2^{20} – 2^{10}
(2) 2^{21} – 2^{11}
(3) 2^{21} – 2^{10}
(4) 2^{20} – 2^{9 }
Ans. (1)
Solution.
Q.32. If (27)^{999} is divided by 7, then the remainder is: (2017)
(1) 3
(2) 1
(3) 6
(4) 2
Ans. (3)
Solution.
Q.33. The coefficient of x^{5} in the binomial expansion of where x ≠ 0, 1 is: (2017)
(1) 1
(2) 4
(3) 1
(4) 4
Ans. (3)
Solution.
Since a^{3}+1 = (a+1)(a^{2}a+1)
Q.34. If the number of terms in the expansion of , x ≠ 0, is 28, then the sum of the coefficients of all the terms in this expansion, is: (2016)
(1) 64
(2) 2187
(3) 243
(4) 729
Ans. (4)
Solution. Total number of terms = ^{n+2}C_{2} = 28
⇒(n+2)(n+1) = 56
⇒ n=6
Sum of coefficients =(12+4)^{n} = 3^{6} = 729
Q.35. The value of is equal to: (2016)
(1) 1085
(2) 560
(3) 680
(4) 1240
Ans. (3)
Solution.
Q.36. For x ∈ R, x ≠ –1, if (1 + x)^{2016} + x (1 + x)^{2015} + x^{2} (1 + x)^{2014} + …….. + x^{2016} =,then a_{17} is equal to: (2016)
Ans. (3)
Solution.
=
Q.37. then n satisfies the equation (2016)
(1) n^{2} + n – 110 = 0
(2) n^{2} + 5n – 84 = 0
(3) n^{2} + 3n – 108 = 0
(4) n^{2} + 2n – 80 = 0
Ans. (3)
Solution.
Q.38. If the coefficients of x^{–2} and x^{–4} in the expansion of are m and n respectively, then m/n is equal to: (2016)
(1) 5/4
(2) 4/5
(3) 27
(4) 182
Ans. (4)
Solution.
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