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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
What is `C(n, r) + 2C(n, r - 1) + C(n, r - 2)` equal to?A. `C(n + 1, r)`B. `C(n -1, r + 1)`C. `C(n, r + 1)`D. `C(n + 2, r)` |
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Answer» Correct Answer - D `C(n,r)+2C(n,r-1)+C(,r-2)` `=.^(n)C_(r)+2(.^(n)C_(r-1))+.^(n)C_(r-2)` `=.^(n)C_(r)+.^(n)C_(r-1)+.^(n)C_(r-1)+.^(n)C_(r-2)" "(because .^(n)C_(r)+.^(n)C_(r-1)=.^(n+1)C_(r))` `=.^(n+1)C_(r)+.^(n+1)C_(r-1)` `=.^(n+2)C_(r)` `=C(n + 2, r)` |
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| 2. |
In the expansion of `(1+x)^(50),`find the sum of coefficients of odd powers of `xdot`A. `2^(26)`B. `2^(49)`C. `2^(50)`D. `2^(51)` |
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Answer» Correct Answer - B Sum of odd terms of expansion `(a+b)^(n) "is" (1)/(2).2^(n)`. `therefore` Sum of odd terms of expansion `(1+x)^(50)"is"(1)/(2).2^(50)`. `=2^(-1).2^(50)=2^(49)` |
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| 3. |
The number of terms in the expansion of `(x+a)^100+(x-a)^100` after simplificationA. 202B. 101C. 51D. 50 |
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Answer» Correct Answer - C `(x+a)^(100)+(x-a)^(100)` Simple logic is we get `.^(n)c_(0),.^(n)c_(2),.^(n)c_(4).....^(n)C_(100)` in this expansion. The number of terms from `.^(n)c_(0) "to" .^(n)c_(100) "are 51"` |
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| 4. |
What is the number of non-zero terms in the expansion of `(1+2 sqrt(3)x)^(11) + (1 - 2 sqrt(3)x)^(11)` (after simplification)?A. 4B. 5C. 6D. 11 |
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Answer» Correct Answer - C We know, in the expansion of `(x+y)^(n)+(x-y)^(n)`, of n = even, then number of non zero terms is `(n)/(2) + 1` n = odd, then number of non zero terms in `(n+1)/(2)`. Here, n = 11 which is odd. `therefore` number of non zero terms = `(11+1)/(2)=6`. |
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| 5. |
What is the number of terms in the expansion of `[(2x - 3y)^(2)(2x+3y)^(2)]^(2)`?A. 4B. 5C. 8D. 16 |
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Answer» Correct Answer - B `[(2x-3y)^(2)(3x+3y)^(2)]^(2)` `=[(4x^(2)-9y^(2))^(2)]^(2)=(4x^(2)-9y^(2))^(4)` `therefore` Number of terms = 4 + 1 = 5 |
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| 6. |
What is the sum of the coefficients in the expansion of `(1+x)^(n)` ?A. `2^(n)`B. `2^(n)-1`C. `2^(n)+1`D. `n+1` |
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Answer» Correct Answer - A Given expansion is `(1+x)^(n)`. Put x = 1, we get Sum of coefficient = `2^(n)`. |
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| 7. |
After simplification, what is the number of terms in the expansion of `[(3x+y)^(5)]^(4)-[(3x-y)^(4)]^(5)` ?A. 4B. 5C. 10D. 11 |
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Answer» Correct Answer - C Given expression is : `[(3x+y)^(5)]^(4)-[(3x-y)^(4)]^(5)=[(3x+y)]^(20)-[(3x-y)]^(20)` First and second expansion will have 21 terms each but odd terms in second expansion be 1st, 3rd, 5th……21st will be equal and opposite to those of first expansion. Thus, the number of terms in the expansion of above expression is 10. |
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| 8. |
Consider the expansion `(x^(2)+(1)/(x))^(15)`. What is the sum of the coefficients of the middle terms in the given expansion ?A. C(15, 9)B. C(16, 9)C. C(16, 8)D. None of these |
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Answer» Correct Answer - C Given `(x^(2)+(1)/(x))^(15)` Since, n is odd. So, it has two middle terms `T_(8) and T_(9)`. `therefore T_(8)+T_(9)=.^(15)C_(7)+.^(15)C_(8)=.^(16)C_(8)" "(because .^(n)C_(r-1)+.^(n)C_(r)=.^(n+1)C_(r))` |
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| 9. |
What is the sum of the coefficients of all the terms in the expansion of `(45x-49)^(4)` ?A. `-256`B. `-100`C. 100D. 256 |
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Answer» Correct Answer - D Given expansion is `(45x-49)^(4).` To find the sum of the coefficients of all the terms in the expansion, we have to put x=1 in the expansion. Thus, required sum of coefficients `=(45-49)^(4)` `=(-4)^(4)=256` |
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| 10. |
In the expansion of `(1 + x)^43` ,the co-efficients of `(2r + 1)th and (r + 2)th` terms are equal. Find `r`.A. 5B. 14C. 21D. 22 |
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Answer» Correct Answer - B Given, in the expression `g(1+x)^(43)`, coefficients of `(2r + 1)^(th)` term and `(r + 2)^(th)` term are equal. Coefficient of `(2r+1)^(th)` term = `n_(C_(2r))` Coefficient of `(r + 2)^(th)` term = `n_(C_(r+1))` `n_(C_(2r))=n_(C_r+1)` `rArr 43_(C_(2r))=43_(C_(r+1))" "(because n = 43)` `rArr 2r + r + 1 = 43` `rArr 3r + 1 = 43` `rArr 3r + 42 rArr r = 14` |
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| 11. |
If `n in N`, then `121^n- 25^n + 1900^n – (-4)^n` is divisible byA. 1904B. 2000C. 2002D. 2006 |
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Answer» Correct Answer - B n `in` N, `121^(n)-25^(n)+1900^(n)-(-4^(n))` Let us substitute n = 1 We get, `(121)^(1)-(25)^(1)+(1900)^(1)-(-4^(1))` `=121 - 25 + 1900 + 4 `=2025 - 25` `=2000` So, given expression is divisible by 2000 |
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| 12. |
In the expansion of `(1+x)^n`, the sum of the coefficients of the terms in even positions is `2^(n-1)`A. `2^(n)`B. `2^(n)-1`C. `2^(n)+1`D. None of these |
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Answer» Correct Answer - B Sum of all binomial coefficients `=(1+1)^(n)=2^(n)` `therefore` Sum of even binomial coefficient `=(2^(n))/(2)=2^(n-1)` |
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| 13. |
If the coefficients of `a^(m) and a^(n)` in the expansion of `(1+a)^(m+n)` are `alpha and beta` then which one of the following is correct ?A. `alpha = 2 beta`B. `alpha = beta`C. `2 alpha = beta`D. `alpha = (m + n)beta` |
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Answer» Correct Answer - B `(1+a_^(m+n)` `alpha` = coefficient of `a^(m)=.^(m+n)C_(m)` `beta` = coefficient of `a^(n) = .^(m+n)C_(n)` We know, `.^(n)C_(r)=.^(n)C_(n-r)` `therefore beta =.^(m+n)C_(n)=.^(m+n)C_(m+n-n)=.^(m+n)C_(m)=alpha` `therefore` alpha = beta` |
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| 14. |
What is the coefficient of `x^(4)` in the expansion of `(1+2x+3x^(2)+4x^(3)+….)^(1//2)` ?A. `1//4`B. `1//16`C. 1D. `1//128` |
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Answer» Correct Answer - C Consider `(1+2x+3x^(2)+4x^(3)+…)^(1//2)=(1-x^(-2))^(1//2)` As we know that `(1-x)^(-2)=1+2x+3x^(2)+4x^(3)+….` `rArr" "(1-x)(-1)=1+x+x^(2)+x^(3)+x^(4)+….` `therefore" Required coefficient of "x^(4)" is 1 "` |
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| 15. |
What is the coefficient of `x^(17)` in the expansion of `(3x-(x^(3))/(6))^(9)` ?A. `(189)/(8)`B. `(567)/(2)`C. `(21)/(16)`D. None of these |
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Answer» Correct Answer - A Given expansion is `(3x-(x^(3))/(6))^(9)" where "a=3x, b=(-x^(3))/(6), n=9` `"Now, General Term "=T_(r+1)=""^(n)C_(r)(a)^(n-r),b^(r )` `=""^(9)C_(r),(3x)^(9-r)((-x^(3))/(6))^(r )=""^(9)C_(r).3^(9-r)x^(9-r).((-1)^(r )x^(3r))/(6^(r ))` `=""^(9)C_(r)3^(9-r)(-1)^(r)(x^(9+2r))/(6^(r))` We can get coeff of `x^(17)` when `9+2r=17` `rArr" "2r=17-9` `rArr" "r=(8)/(2)=4` Hence, required coefficient `=""^(9)C_(4)(3^(5))/(6^(4))=(126xx3)/(16)=(189)/(8)` |
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| 16. |
Find the constant term in the expansion of `(sqrtx+1/(3x^2))^10`.A. 5B. 8C. 45D. 90 |
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Answer» Correct Answer - A Let `r^(th)` term is independent of x. `T_(r) = .^(n)C_(r)x^(r)y^(n-r)` `=.^(10)C_(r)(sqrt(x))^(r)((1)/(3x^(2)))^(10-r)` `=.^(10)C_(r)((1)/(3))^(10-r).(sqrt(x))^(r)((1)/(x^(2)))^(10-r)` Equating the coefficient of x to zero. `rArr x^(r//2).x^(-2(10-r)=x^(0)` `rArr (r)/(2)-20 + 2r = 0` `rArr (5)/(2)r = 20 rArr r = 8` Coefficient `=.^(10)C_(r)((1)/(3))^(10-r)` `=.^(10)C_(8)((1)/(3))^(10-8)=(10xx9)/(2)xx(1)/(9)=5` |
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| 17. |
What is the sum of all the coefficients in the expansion of `(1+x)^(n)` ?A. `2^(n)`B. `2^(n)-1`C. `2^(n)-1`D. `2(n-1)` |
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Answer» Correct Answer - A Given expansion is `(1 + x)^(n)`. Put x = 1, we get Required sum = `(1 + 1)^(n)=2^(n)` |
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| 18. |
The number of terms in the expansion of `(a+b+c)^n ,w h e r en in Ndot`A. `n+1`B. `n+2`C. `n(n+1)`D. `((n+1)(n+2))/(2)` |
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Answer» Correct Answer - D Required number of terms in `(a + b + c)^(n)` `=.^(n+2)C_(2)((n+2)!)/(2!n!)=((n+1)(n+2))/(2)` |
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| 19. |
In the expansion of `(1 + ax)^(n)`, the first three terms are respectively 1, 12x and `64x^(2)`. What is n equal to ?A. 6B. 9C. 10D. 12 |
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Answer» Correct Answer - B The first three terms in expansion of `(1+ax)^(n)` are `.^(n)C_(0), .^(n)C_(1)ax, .^(n)C_(2)a^(2)x^(2)` Given, `.^(n)C_(0)=1, .^(n)C_(1) ax = 12x, .^(n)C_(2)a^(2)x^(2)=64x^(2)` `rArr "nax" = 12x,(n(n-1))/(2)a^(2)=64` `rArr "na"=12 rArr a = (12)/(n)` `therefore (n(n-1))/(2)a^(2)=64 rArr (n(n-1))/(2)xx(144)/(n^(2))=64` `rArr(n-1)/(n)=(64xx2)/(144)=(8)/(9)` `therefore n = 9` |
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| 20. |
The coefficient of `x^99` in `(x-1)(x-2).....(x-100)` isA. 5050B. 5000C. -5050D. -5000 |
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Answer» Correct Answer - C Coefficient of `x^(1)` in `[(x-1)(x-2)or(x^(2)-3x + 2)]` `=-3 =-1 - 2 =- (1+2)` Coefficient of `x^(2)` in `[(x-1)(x-2)(x-3)or(x^(3)-6x^(2)+5x-6)]` `=-6=-[1+2+3]`. Coefficient of `x^(3)` in `[(x-1)(x-2)(x-3)(x-4)or(x^(4)-10x^(3)-29x^(2)-11x + 24)]` `=-10=-[1+2+3+4]` `therefore` Coefficient of `x^(99)` in `[(x-1)(x-2)........(x-100)]` `=-[1+2+3+....+100]=(-100(100+1))/(2)=-5050`. |
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| 21. |
Consider the expansion of `(1 + x)^(2n+1)` The sum of the coefficients of all the terms in the expansion isA. `2^(2n-1)`B. `4^(n-1)`C. `2 xx 4^(n)`D. None of these |
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Answer» Correct Answer - C Sum of all coefficient `=.^((2n+1))C_(0)+.^((2n+1))C_(1)+...+.^((2n+1))C_(2n+1)` `=(1+1)^(2n+1)=2^(2n+1)=2.2^(2n)=2.4^(n)` |
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| 22. |
Consider the expansion of `(1 + x)^(2n+1)` If the coefficients of `x^(r) and x^(r+1)` are equal in the expansion, then r is equal toA. nB. `(2n-1)/(2)`C. `(2n+1)/(2)`D. n + 1 |
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Answer» Correct Answer - A `(1+x)^(2n+1)=.^((2n+1))C_(0)x^(0)+.^((2n+1))C_(1)x^(1)+...+.^((2n+1))C_(2n+1)(x)^(2n+1)` Coefficient of `x^(r)=.^((2n+1))Cr` Coefficient of `x^(r+1)=.^((2n+1))Cr + 1` `.^((2n+1))C_(r)=.^((2n+1))Cr+1` `rArr ((2n+1)!)/(r!(2n+1-r)!)=((2n+1)!)/((r+1)!(2n-r)!)` `rArr ((2n-r)!)/((2n+1-r)(2n-r)!)=(r!)/((r+1)r!)` `rArr (r + 1) = 2n + 1 - r` `rArr r = n` |
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| 23. |
Consider the expansion `(x^(2)+(1)/(x))^(15)`. Consider the following statements: 1. The term containing `x^(2)` does not exist in the given expansion. 2. The sum of the coefficients of all the terms in the given expansion is `2^(15)`. Which of the above statements is/are correct ?A. 2 onlyB. 3 onlyC. Both 1 and 3D. Neither 1 nor 3 |
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Answer» Correct Answer - C 1. For coefficient of `x^(2)`, `30 - 3r = 2 rArr r = (28)/(3), r notin N` So, `x^(2)` does not exist in the expansion Hence, Statement 1 is correct. 2. Now, `(x^(2)+(1)/(4))^(15)=.^(15)C_(0)(x^(2))^(15)+.^(15)C_(1)(x^(2))^(14)((1)/(x))+...+.^(15)C_(15)((1)/(x))^(15)` Put x = 1 both sides, we get `(1+1)^(15)=.^(15)C_(0)+.^(15)C_(1)+...+.^(15)C_(15)` `rArr 2^(15)=.^(15)C_(0)+.^(15)C_(1)+...+.^(15)C_(15)` Hence, Statement 2 is correct |
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| 24. |
What is ` sum_(r=0)^(1) ""^(n+r)C_(n)` equal to ?A. `.^(n+2)C_(1)`B. `.^(n+2)C_(n)`C. `.^(n+3)C_(n)`D. `.^(n+2)C_(n+1)` |
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Answer» Correct Answer - A::D `underset(r=0)overset(1)sum .^(n+r)C_(n)=.^(n)C_(n)+.^(n+1)C_(n)` `=1+((n+1)!)/((n+1-n)!n!)=1+((n+1)(n!))/(n!)` `=1+n+1=n+2` `.^(n+2)C_(n+1)((n+2)!)/((n+2-n-1)!(n+1)!)` `=((n+2)(n+1)!)/((n+1)!)=n+2` `" "`OR `underset(r=0)overset(1)sum .^(n+r)C_(n)=.^(n)C_(n)+.^(n+1)C_(n)` `=1+(n+1)=n+2` Now, `.^(n+2)C_(1)=((n+2)!)/(1!(n+2-1)!)=((n+2)(n+1)!)/((n+1)!)=(n+2)` `therefore` Option (a and d) is correct. |
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| 25. |
The value of the term independent of x in the expansion of `(x^(2)-(1)/(x))^(9)` is :A. 9B. 18C. 48D. 84 |
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Answer» Correct Answer - D `(x^(2)-(1)/(x))^(9)` `t_(r+1)=.^(9)C_(r)(x^(2))^(9-r)((-1)/(x))^(r)` `.^(9)C_(r)x^(18-2r).(-1)^(r).x^(-r)` `=.^(9)C_(r)(x)^(18-3r)(-1)^(r)" "...(1)` Term will be independent of x when 18 - 3r = 0 r = 6 Put r = 6, in [1] `t_(7)=.^(9)C_(6)(-1)^(6)=(9!)/(6!3!)=84` |
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| 26. |
If `x^4`occurs in the rth term in the expansion of `(x^4+1/(x^3))^(15),`then find the value of `rdot`A. 4B. 8C. 9D. 10 |
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Answer» Correct Answer - C In the expansion of `(x^(4)+(1)/(x^(3)))^(15)`, let `T_(r)` is the `n_(th)` term `T_(r)=""^(15)C_(r-1)(x^(4))^(15-r+1)((1)/(x^(3)))^(r-1)` `=15_(C_(r-1))x^(64-4r-3r+3)=15_(C_(r-1))x^(67-7r)` `x^(4)` occurs in this term `rArr" "4=67-7r` `rArr" "7r=63` `rArr" "r=9.` |
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| 27. |
What is the coefficient of `x^(3)y^(4)" in "(2x+3y^(2))^(5)` ?A. 240B. 360C. 720D. 1080 |
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Answer» Correct Answer - C `T_(r)=""^(n)C_(r-1)(2x)^(r-1)(3y^(2))^(n-1+1)` `T_(4) = =""^(5)C_(3)(2x)^(3) (3y^(2))^(2)` `=(5!)/(3!2!)2^(3) .x^(3) .9y^(4)=(5.4)/(2.1)xx8xx9xxx^(3)y^(4)=720 x^(3)y^(4)` `therefore" Coefficient of "x^(3)y^(4)=720` |
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| 28. |
Consider the expansion of `(1 + x)^(2n+1)` The average of the coefficients of the two middle terms in the expansion isA. `.^(2n+1)C_(n+2)`B. `.^(2n+1)C_(n)`C. `.^(2n+1)C_(n-1)`D. `.^(2n)C_(n+1)` |
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Answer» Correct Answer - B Total no. of terms in the expansion is 2n + 2. The middle two terms will be `n^(th),(n+1)^(th)` term. So. Average = `(.^((2n+1))C_(n)+.^((2n+1))C_(n+1))/(2)` `=[((2n+1)!)/(n!(n+1)!)+((2n+1)!)/((n+1)!n!)]//2` `=((2n+1)!)/(n!(n+1)!)=.^((2n+1))C_(n)` |
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| 29. |
If `C(20,n+2)=C(20,n-2)` then `n=`A. 8B. 10C. 12D. 16 |
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Answer» Correct Answer - B Given, `C(20, n+2)=C(20, n-2)` `rArr .^(20)C_(n+2)=.^(20)C_(n-2)` `rArr 20 = n + 2 + n-2" "(because .^(n)c_(r)=.^(n)c_(s) rArr n = r + s)` `rArr 20 = 2n` `rArr n = 10` |
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| 30. |
The value of `[C(7, 0)+C(7, 1)] + [C(7, 1) + C(7, 2)]+ ... + [C(7, 6) + C(7, 7)]`A. 254B. 255C. 256D. 257 |
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Answer» Correct Answer - A `(7_(c_(0)+7_(c_(1))))+(7_(c_(1)+7_(c_(2))))+...+(7_(c_(6)+7_(c_(7))))` We know, `n_(c_(r))+n_(c_(r-1))=.^(n+1)C_(r)` `=8_(c_(1))+8_(c_(2))+...+8_(c_(7))` `=(8_(c_(0))+8_(c_(1))+8_(c_(2))+....+8_(c_(7))+8_(c_(8)))-(8_(c_(0))+8_(c_(8)))=2^(8)-(1+1)` `["Since", n_(c_(0))+ n_(c_(1))+ n_(c_(2))+...+ n_(c_(n))=2^(n)]` = 256 - 2 = 254 |
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| 31. |
The expansion of `(x-y)^(2), n ge 5` is done in the descending power of x. If the sum of the fifth and sixth terms is zero, then `(x)/(y)` is equal toA. `(n-5)/(6)`B. `(n-4)/(5)`C. `(5)/(n-4)`D. `(6)/(n-5)` |
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Answer» Correct Answer - B `(X-Y)^(n), n ge 5` General term, `T_(r+1)=n_(c_(r))x^(n-r)(-y)^(r)`. `T_(5) + T_(6) = 0` `rArr [n_(c_(4))x^(n-4)(-y)^(4)]+[n_(c_(5))x^(n-5)(-y)^(5)]=0` `rArr n_(c_(4))x^(n-4)y^(4)-n_(c_(5))x^(n-5)y^(5)=0` `rArr n_(c_(4))x^(n-4)y^(4)=n_(c_(5))x^(n-5)y^(5)` `rArr (x^(n-4-n+5))/(y)=(n_(c_(5)))/(n_(c_(4)))rArr (x)/(y)=(cancel(n!))/(5!(n-5)!)xx(4!(n-4)!)/(cancel(n!))` `=(cancel(4!)(n-4)cancel((n-5)!))/(5 xx cancel(4!) cancel((n-5)!))=(n-4)/(5)` |
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| 32. |
If `|z-4/z|=2`, then the maximum value of`|Z|`is equal to(1) `sqrt(3)+""1`(2) `sqrt(5)+""1`(3) 2(4) `2""+sqrt(2)`A. `1+sqrt(3)`B. `1+sqrt(5)`C. `1-sqrt(5)`D. `sqrt(5)-1` |
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Answer» Correct Answer - B `|z-(4)/(z)|=2` We know `|a-b|ge|a|-|b|` ` therefore |z-(4)/(z)|ge|z|-|(4)/(z)|` `rArr 2 ge |z|-(4)/(|z|)rArr|z|^(2)-2|z|-4 le 0` `rArr |z|=(2+-sqrt(4-4(1)(-4)))/(2(1))=(2+-sqrt(20))/(2)=(2+-2sqrt(5))/(2)=1+-sqrt(5)` So, `z = 1 + sqrt(5)`. |
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| 33. |
`1.3+2.3^2+3.3^3+..............+n.3^n=((2n-1)3^(n+1)+3)/4`A. n, 2B. n, 3C. n + 1, 2D. n + 1, 3 |
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Answer» Correct Answer - D `1.3 + 2.3^(2)+3.3^(3)+...+n.3^(n)=((2n-1)3^(a)+b)/(4)` Let us put 3 = x. L.H.S: S = `x + 2x^(2)+3x^(3)+...+n.x^(n)" "...(1)` `xs = x^(2)+2x^(3)+3x^(4)+...+n.x^(n+1)" "...(2)` `(1)-(2) rArr S - xS = (x + 2x^(2)+3x^(2)+...+n.x^(n))-(x^(2)+2x^(3)+3x^(4)+...+n.x^(n+1))` `rArr S(1-x)=x+x^(2)+x^(3)+...+x^(n)-nx^(n+1)` `S(1-x)=(x(1-x^(n)))/(1-x)-nx^(n+1)` `rArr S=((1)/(x-1))((-x(x^(n)-1)+nx^(n+1)(x-1))/(x-1))` Put x = 3, `rArr S = (1)/(2)((-3^(n+1)+3+2n.3^(n+1))/(2))=((3^(n+1)(2n-1)+3)/(4))` |
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| 34. |
What is the coefficient of the middle term in the binomial expansion of `(2+3x)^(4)`?A. 6B. 12C. 108D. 216 |
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Answer» Correct Answer - D Middle term in the expansion of `(x + y)^(n)` `=((n+1)/(2))^(th)` term, if n is odd `=((n)/(2)+1)^(th)` term, if n is even Here n = 4 `therefore` Middle term is `((4)/(2)+1)^(th)=3^(rd)` term `4c_(2) xx 2^(2) xx 3^(2) = 6 xx 4 xx 9 = 216` |
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| 35. |
For all `n in N, 2^(4n)-15n-1` is divisible byA. 125B. 225C. 450D. None of these |
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Answer» Correct Answer - B Let `P(n) : 2^(4n) - 15 n - 1` Put n = 2 `P(2)=2^(8)-30-1=225` which is divisible by 225. Let us assume, P(n) is true for n = k is P(K) : `2^(4k)-15k-1` is divisible by 225. `rArr 2^(4k)-15k-1=225 lambda, lambda in R, k in N" "...(i)` To prove for n = k + 1 Consider `2^(4k+4)-15k-15-1=2^(4k).2^(4)-15k-16` `=2^(4)[225 lambda + 1 + 15k]-15k-16" "("from (i)")` `=2^(4).225 lambda .2^(4)+15.2^(4).k-15k-16` `=25.225 lambda + 225 k` `=225[2^(4) lambda + k]` = 225 r where `r = 2^(4)lambda + k` is a constant Hence, `2^(4n)-15n - 1` is divisible by 225. |
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| 36. |
Consider the expansion `(x^(2)+(1)/(x))^(15)`. What is the ratio of coefficient of `x^(15)` to term independent of x in the given expansion ?A. `1//64`B. `1//32`C. `1//16`D. `1//4` |
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Answer» Correct Answer - B Given expansion is `(x^(2)+(2)/(x))^(15)` `T_(r+1)=.^(15)C_(r)(x^(2))^(15-r)((2)/(x))^(r)` `=.^(15)C_(r)x^(30-2r)2^(r)x^(-r)=.^(15)C_(r)x^(30-3r).2^(r)` Now, Above term will be independent of x when `30 - 3r = 0 rArr r = 10` `therefore` Term independent of `x = .^(15)C_(10)2^(10)` Now, coeff of `x^(15)` When `30-3r = 15 rArr r = 5` `therefore` Required coeff `= .^(15)C_(5)2^(5)` Thus, Required Ratio = `(.^(15)C_(5).2^(5))/(.^(15)C_(10).2^(10))` `=((15!)/(5!(10!)))/((15!)/(10!5!)xx2^(5))=(1)/(2^(5))=(1)/(32)` |
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| 37. |
In the expansion of `(x^3-1/(x^2))^n ,n in N`, if the sum of the coefficients of `x^5a n dx^(10)`, then `n`isa. 25 b. 20 c. 15 d. none of theseA. 5005B. 7200C. -5005D. -7200 |
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Answer» Correct Answer - C `(x^(3)-(1)/(x^(2)))^(n)` General term, `T_(r+1)=.^(n)C_(r)(x^(3))^(n-r).(-(1)/(x^(2)))^(r)` `=.^(n)C_(r).x^((3n-3r)).(-1)^(r).x^(-2x)` `=.^(n)C_(r).(-1)^(r).x^((3n-5r))" "...(i)` For the coefficient `x^(5)` Put 3n - 5r = 5 5r = 3n - 5 `therefore r = (3n)/(5)-1` `therefore "Coefficient of" x^(5)=.^(n)C_(((3n)/(5)-1))(-1)^(((3n)/(5)-1))` For the cefficient of `x^(10)` Put 3n - 5r = 10 5r = 3n - 10 `therefore r = (3n)/(5)-2` `therefore "Coefficient of" x^(10)=.^(n)C_(((3n)/(5)-2))(-1)^(((3n)/(5)-2))` The sum of the coefficient of `x^(5) and x^(10)=0` `rArr .^(n)C_(((3n)/(5)-1))(-1)^(((3n)/(5)-1))+.^(n)C_(((3n)/(5)-2))(-1)^(((3n)/(5)-2))=0` `rArr (-1)^((3n)/(5))[.^(n)C_(((3n)/(5)-1)).(-1)^(-1)+.^(n)C_(((3n)/(5)-2)).(-1)^((-2))]=0` `rArr -.^(n)C_(((3n)/(5)-1))+.^(n)C_(((3n)/(5)-2))=0" "...(ii)` For the independent term, put `3n - 5 r = 0 " "["from eq. (i)"]` `rArr 5r = 3n = 3 xx 15` `5r = 3 xx 3 xx 5` r = 9 Putting the value of r in eq. (i), we get `T_(9+1)=.^(15)C_(9).(-1)^(9).x^((3 xx 15 - 5 xx 9))` `rArr T_(10)=-.^(15)C_(9).x^(0)=-.^(15)C_(9)` `rArr T_(10)=-.^(15)C_(6)" "[because .^(n)C_(r)=.^(n)C_(n-r)]` `=(-15!)/(6!9!)" "[because .^(n)C_(r)=(n!)/(r!(n-r)!)]` `=-5005` |
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| 38. |
In the expansion of `(x^3-1/(x^2))^n ,n in N`, if the sum of the coefficients of `x^5a n dx^(10)`, then `n`isa. 25 b. 20 c. 15 d. none of these |
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Answer» Correct Answer - A `(x^(3)-(1)/(x^(2)))^(n)` General term, `T_(r+1)=.^(n)C_(r)(x^(3))^(n-r).(-(1)/(x^(2)))^(r)` `=.^(n)C_(r).x^((3n-3r)).(-1)^(r).x^(-2x)` `=.^(n)C_(r).(-1)^(r).x^((3n-5r))" "...(i)` For the coefficient `x^(5)` Put 3n - 5r = 5 5r = 3n - 5 `therefore r = (3n)/(5)-1` `therefore "Coefficient of" x^(5)=.^(n)C_(((3n)/(5)-1))(-1)^(((3n)/(5)-1))` For the cefficient of `x^(10)` Put 3n - 5r = 10 5r = 3n - 10 `therefore r = (3n)/(5)-2` `therefore "Coefficient of" x^(10)=.^(n)C_(((3n)/(5)-2))(-1)^(((3n)/(5)-2))` The sum of the coefficient of `x^(5) and x^(10)=0` `rArr .^(n)C_(((3n)/(5)-1))(-1)^(((3n)/(5)-1))+.^(n)C_(((3n)/(5)-2))(-1)^(((3n)/(5)-2))=0` `rArr (-1)^((3n)/(5))[.^(n)C_(((3n)/(5)-1)).(-1)^(-1)+.^(n)C_(((3n)/(5)-2)).(-1)^((-2))]=0` `rArr -.^(n)C_(((3n)/(5)-1))+.^(n)C_(((3n)/(5)-2))=0" "...(ii)` n = 15 Total term in the expansion of `(x^(3)-(1)/(x^(2)))^(15)` is 16. `therefore` middle term = `8^(th)` term and `9^(th)` term `T_(8)=T_((7+1))=.^(15)C_(7).(-1)^(7).x^((3 xx 15 - 5 xx 7))` `=-.^(15)C_(7).x^(10)" "("from eq. (i)")` `T_(9)=T_((8+1))=.^(15)C_(8).(-1)^(8).x^((3 xx 15 - 5 xx 8))` `=-.^(15)C_(8).x^(5)" "("from eq. (ii)")` The sum of the coefficients of the two middle terms `=-.^(15)C_(7)+.^(15)C_(8)=-.^(15)C_(7)+.^(15)C_(7)." "[because .^(n)C_(r)=.^(n)C_(n-r)]` = 0 |
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| 39. |
Given that `C(n, r) : C(n,r + 1) = 1 : 2 and C(n,r + 1) : C(n,r + 2) = 2 : 3`. What is n equal to ?A. 11B. 12C. 13D. 14 |
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Answer» Correct Answer - D `(.^(n)C_(r))/(.^(n)C_(r+1))=(1)/(2)` `(|ul(n)|ul(r+1)|ul(n-r-1))/(|ul(r)|ul(n-r).|ul(n))=(1)/(2)` `(r+1)/(n-r)=(1)/(2)rArr 3r -n+2=0" "...(i)` `(.^(n)C_(r+1))/(.^(n)C_(r+2))=(2)/(3)` `(|ul(n)|ul(r+2)|ul(n-r-2))/(|ul(r+1)|ul(n-r-1)|ul(n))=(2)/(3)` `(r+2)/(n-r-1)=(2)/(3)rArr 5r - 2n + 8 = 0` Solving equation (i) and (ii), we get `n = 14, r = 4` |
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| 40. |
Given that `C(n, r) : C(n,r + 1) = 1 : 2 and C(n,r + 1) : C(n,r + 2) = 2 : 3`. What is r equal to ?A. 2B. 3C. 4D. 5 |
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Answer» Correct Answer - C `(.^(n)C_(r))/(.^(n)C_(r+1))=(1)/(2)` `(|ul(n)|ul(r+1)|ul(n-r-1))/(|ul(r)|ul(n-r).|ul(n))=(1)/(2)` `(r+1)/(n-r)=(1)/(2)rArr 3r -n+2=0" "...(i)` `(.^(n)C_(r+1))/(.^(n)C_(r+2))=(2)/(3)` `(|ul(n)|ul(r+2)|ul(n-r-2))/(|ul(r+1)|ul(n-r-1)|ul(n))=(2)/(3)` `(r+2)/(n-r-1)=(2)/(3)rArr 5r - 2n + 8 = 0` Solving equation (i) and (ii), we get `n = 14, r = 4` |
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| 41. |
Given that `C(n, r) : C(n,r + 1) = 1 : 2 and C(n,r + 1) : C(n,r + 2) = 2 : 3`. What is P(n, r) : C(n, r) equal to ?A. 6B. 24C. 120D. 720 |
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Answer» Correct Answer - B `(.^(n)C_(r))/(.^(n)C_(r+1))=(1)/(2)` `(|ul(n)|ul(r+1)|ul(n-r-1))/(|ul(r)|ul(n-r).|ul(n))=(1)/(2)` `(r+1)/(n-r)=(1)/(2)rArr 3r -n+2=0" "...(i)` `(.^(n)C_(r+1))/(.^(n)C_(r+2))=(2)/(3)` `(|ul(n)|ul(r+2)|ul(n-r-2))/(|ul(r+1)|ul(n-r-1)|ul(n))=(2)/(3)` `(r+2)/(n-r-1)=(2)/(3)rArr 5r - 2n + 8 = 0` Solving equation (i) and (ii), we get `n = 14, r = 4` `P(n, r) : C(n, r) = |ul(r) = 24` |
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| 42. |
Consider the expansion `(x^(2)+(1)/(x))^(15)`. Consider the following statements: 1. There are 15 terms in the given expansion. 2. The coefficient of `x^(12)` is equal to that of `x^(3)`. Which of the above statements is/are correct ?A. 1 onlyB. 2 onlyC. Both 1 and 2D. Neither 1 nor 2 |
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Answer» Correct Answer - B 1. W e know that, `(a+b)^(n)` have total (n + 1) number of terms So, `(x^(2)+(1)/(x))^(15)` have 16 terms. Hence, Statement 1 is false. 2. For coefficient of `x^(12)` `30-3r = 12 rArr r = 6 rArr .^(15)C_(6)` and for coefficient of `x^(3)`, `30 - 3r = 3 rArr r = 9 rArr .^(15)C_(9) .^(15)C_(6) = .^(15)C_(9)` Hence, statement 2 is correct. |
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| 43. |
What is `((sqrt(3)+i)/(sqrt(3)-i))^(6)` equal to, where `I = sqrt(-1)` ?A. 1B. `1//6`C. 6D. 2 |
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Answer» Correct Answer - A `((sqrt(3)+i)/(sqrt(3)-i))=(sqrt(3)+i)/(sqrt(3)-i)xx(sqrt(3)+i)/(sqrt(3)+i)` `(3+i^(2)+2 sqrt(3)i)/(3-i^(2))=(3-1+2 sqrt(3)i)/(3+1)` `=(2(1+sqrt(3)i))/(4)=(1)/(2)+i(sqrt(3))/(2)` `=("cos"(pi)/(3)+"i sin"(pi)/(3))=e^(i(pi)/(3)` `therefore ((sqrt(3)+i)/(sqrt(3)-i))^(6)=(e^(i(pi)/(3)))^(6)=e^(i2pi)cos 2 pi + i sin 2 pi` `= 1 + 0.i = 1` |
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| 44. |
Consider the expansion `(x^(2)+(1)/(x))^(15)`. What is the ratio of coefficient of `x^(15)` to term independent of x in the given expansion ?A. 1B. `1//2`C. `2//3`D. `3//4` |
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Answer» Correct Answer - A For coefficient of `x^(15)` 30 - 3r = 15 `rArr r = 5` `therefore` the coefficient of `x^(15)` is `.^(15)C_(5)`. and coefficient of independent of x is 30 - 3r = 0 `rArr r = 10` So, coefficient of independent of x is `.^(15)C_(10)`. `therefore` Required ratio `=(.^(15)C_(5))/(.^(15)C_(10))=(.^(15)C_(5))/(.^(15)C_(5))=1" "(because .^(n)C_(r)=.^(n)C_(n-r))` |
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| 45. |
Consider the expansion `(x^(2)+(1)/(x))^(15)`. What is the independent term in the given expansion ?A. 2103B. 3003C. 4503D. None of these |
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Answer» Correct Answer - B `(x^(2)+(1)/(x))^(15)` `T_(r+1)=.^(15)C_(r)(x^(2))^(15-r)((1)/(x))^(r)` `=.^(15)C_(r)x^(30-2r-r)=.^(15)C_(r)x^(30-3r)` For independent term, `30 - 3r = 0 rArr r = 10` Put r = 10, we get `T_(10+1)=.^(15)C_(10)=(15!)/(10!5!)` `=(15 xx 14 xx 13 xx 12 xx 11 xx 10!)/(10! xx 1 xx 2 xx 3 xx 4 xx 5)=3003` |
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| 46. |
Find the term independent of `x`in the expansion of `(1+x+2x^3)[(3x^2//2)-(1//3)]^9`A. `1//3`B. `17//54`C. `1//4`D. No such term exists in the expansion |
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Answer» Correct Answer - B Given expansion is `(1+x+2x^(3))((3)/(2)x^(2)-(1)/(3x))^(9)` `=(1+x+2x^(2))[((3)/(2)x^(2))^(9)-""^(9)C_(1)((3)/(2)x^(2))^(8).(1)/(3x).....+""^(9)C_(6)((3)/(2)x^(2))^(3)((1)/(3x))^(6)-""^(9)C_(7)((3)/(2)x^(2))^(2)((1)/(3x))^(7).....]` In the second bracket we have to search out terms of `x^(0) and (1)/(x^(3))` which when multipiled with terms 1 and `2x^(3)` in the first bracket will give a term independent of x. The term containing `(1)/(x)` will not occur in the 2nd bracket. `therefore` Term independent of x `=1-[""^(9)C_(6)(3^(3))/(2^(3)).(1)/(3^(6))]-2x^(3)[""^(9)C_(7)(3^(2))/(2^(2)).(1)/(3^(7)).(1)/(x^(3))]` `=[(9.8.7)/(1.2.3).(1)/(8.27)]-2[(9.8)/(1.2).(1)/(4.243)]` `=(7)/(18)-(2)/(27)=(17)/(54)` |
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| 47. |
What is the value of `""^(8)C_(0)-""^(8)C_(1)+""^(8)C_(2)-""^(8)C_(3)+""^(8)C_(4)-""^(8)C_(5)+""^(8)C_(6)-""^(8)C_(7)+""^(8)C_(8)` |
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Answer» Correct Answer - A `(1-x)^(n)=""^(n)C_(0)-""^(n)C_(1)(x)+""^(n)C_(2)x^(2)-""^(n)C_(3)x^(3)+...+(-1)^(n)""^(n)C_(n)` Put x=1 and n=8 `therefore" "(1-1)^(8)=""^(8)C_(0)-""^(8)C_(1)+""^(8)C_(2)-""^(8)C_(3)+...+""^(8)C_(8)` `rArr" "(""^(8)C_(0)-""^(8)C_(1)+""^(8)C_(3)+...+""^(8)C_(8))=0` |
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| 48. |
What is the approximate value of `(1.02)^(8)` ?A. `1.171`B. `1.175`C. `1.177`D. `1.179` |
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Answer» Correct Answer - A `(1.02)^(8)=(1+0.02)^(8)` `(1+x)^(n)=1+nx+(n(n-1))/(2!)x^(2)+(n(n-1)(n-2))/(3!)x^(3)+....` `n=8, x=0.02` `(1+0.02)^(8)` `=1+8xx0.02+(8xx7)/(2!).(0.02)^(2)+(8.7.6)/(3!)(.02)^(3)` Neglecting higher terms `=1+0.16+28xx0.0004+56xx0.000008` `cong 1+0.16+0.0112=1.171` |
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| 49. |
If `t_(1)` is the rth term in the expansion of `(1+x)^(101)`, then what is the rato `(t_(20))/(t_(19))` equal to ?A. `(20x)/(19)`B. 83xC. 19xD. `(83x)/(19)` |
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Answer» Correct Answer - D We find `r_(n)` term : `t_(r)` is the rth term in the expansion of `(1+x)^(101).` `t_(r)=""^(101)C_(r-1).(x)^((r-1))` `therefore(t_(20))/(t_(19))=(""^(101)C_(19))/(""^(101)C_(18)).(x^(19))/(x^(18))=(""^(101)C_(19)x)/(""^(101)C_(18))" "=((101!)/(19!82!))/((101!)/(18!83!))x=(83x)/(19)` |
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| 50. |
What is the middle term in the expansion of `(1-(x)/(2))^(8)` ?A. `(35x^(4))/(8)`B. `(17x^(5))/(8)`C. `(35x^(5))/(8)`D. None of these |
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Answer» Correct Answer - A Since n = 8 is even number therefore middle term `=((n)/(2)+1)^(th)"term"=(4+1)=5^(th)"term"` Hence, `T_(5)=.^(8)C_(4)(1)^(4)(-(x)/(2))^(4)` `=(8!)/(4!4!)xx(x^(4))/(16)=(8xx7xx6xx5)/(4xx3xx2xx1).(x^(4))/(16)=(70x^(4))/(16)=(35x^(4))/(8)` |
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