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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.
| 351. |
The approximate value of `(7.995)^(1//3)` correct to four decimal places , isA. 1.9995B. 1.9996C. 1.999.0D. 1.9991 |
| Answer» Correct Answer - B | |
| 352. |
If `(1-x-x^(2))^(20) = sum_(r=0)^(40)a_(r).x^(r )`, then value of `a_(1) + 3a_(3) + 5a_(5) + "….." + 39a_(39)` is |
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Answer» Correct Answer - 40 `(1-x-x^(2))^(20) = underset(r=0)overset(40)suma_(r).x^(r)` Differentiating both sides w.r.t.x, we get `20 (1-x-x^(2))^(19)(-1-2x) = underset(r=0)overset(40)suma_(r).x^(r-1)"........."(1)` Putting `x = 1` into `(1)`, we get `20(1-4)^(19)(-1-2) = underset(r=0)overset(40)sumra_(r)` `rArr 60 = a_(1) + 2a_(2) + 3a_(3) + "...." + 40a_(40) "......."(2)` Putting `x = - 1` into `(1)`, we get `20(1+1-1)^(19)(-1+2) = underset(r=0)overset(40)sumra_(r)(-1)^(r-1)` `rArr 20 = a_(1) - 2a_(2) + 3a_(3) + "......" - 40 a_(40) "......."(3)` Adding `(2)` and `(3)`, we get `80 = 2a_() + 6a_(3) + "....." + 78a_(39)` `:. a_(1) + 3a_(3) + "....." + 39a_(39) = 40` |
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| 353. |
The remainder when `27^(10)+7^(51)` is divided by `10`A. `4`B. `6`C. `9`D. `2` |
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Answer» Correct Answer - D `(d)` `(27)^(10)+7^(51)=(30-3)^(10)+(10-3)^(51)` `=3^(10)-3^(51)+10lambda` `=(10-1)^(5)-3(10-1)^(25)+10lambda` `=-1+3+10lambda_(1)` `=2+10lambda_(1)` `:.` Remainder is `2` |
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| 354. |
The coefficient of `x^4` in the expansion of `(1+x+x^2+x^3)^n` is `.^n C_4+^n C_2+^n C_1xx^n C_2`A. `""^(n)C_(4)`B. `""^(n)C_(4) + ""^(n)C_(2)`C. `""^(n)C_(4) + ""^(n)C_(1) + ""^(n)C_(4) xx""^(n)C_(2) `D. `""^(n)C_(4) + ""^(n)C_(1) + ""^(n)C_(1) xx""^(n)C_(2)` |
| Answer» Correct Answer - D | |
| 355. |
If `x^n`occurs in the expansion `(x+1//x^2)^n`, then the coefficient of `x^m`is`((2n)!)/((m)!(2n-m)!)`b. `((2n)!3!3!)/((2n-m)!)`c. `((2n)!)/(((2n-m)/3)!((4n+m)/3)!)`d. none of theseA. `((2n)!)/(m!(2n-m)!)`B. `((2n)!3!3!)/((2n-m)!)`C. `((2n)!)/(((2n-m)/(3))!((4n-m)/(3))!)`D. none of these |
| Answer» Correct Answer - C | |
| 356. |
Find the remainder when `32^(32^32)` is divided by 7A. 1B. 2C. 3D. 4 |
| Answer» Correct Answer - D | |
| 357. |
If `(1+px+x^(2))^(n)=1+a_(1)x+a_(2)x^(2)+…+a_(2n)x^(2n)`. The remainder obtained when `a_(1)+5a_(2)+9a_(3)+13a_(4)+…+(8n-3)a_(2n)` is divided by `(p+2)` isA. `1`B. `2`C. `3`D. `0` |
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Answer» Correct Answer - C `(c )` `a_(1)+5a_(2)+9a_(3)+…+(8n-3)a_(2n)=sum_(r=1)^(2n)(4r-3)a_(r )` `=4sum_(r=1)^(2n)ra_(r )-3sum_(r=1)^(2n)a_(r )` `(1+px+x^(2))^(n)=1+a_(1)x+a_(2)x^(2)+….+a_(2n)X^(2n)` so, `sum_(r=1)^(2n)a_(r )=(p+2)^(n)-1` Differentiating the expansion and substituting `x=1` `sum_(r=1)^(2n)rar_(r)=n(p+2)^(n)` `:.sum_(r=1)^(2n)(4r-3)a_(r )=4n(p+2)^(n)-3((p+2)^(n)-1)` `=(4n-3)(p+2)^(n)+3` |
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| 358. |
Write the coefficient of `x^(7) y^(2)` in the expansion of `(x+2y)^(9)`.A. Least value `-1/4`B. Least value `-9/4`C. Greatest value `1/4`D. Greatest value `9/4` |
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Answer» Correct Answer - 144 `T_(r+1)= .^(9) C _(r)x^(9-r) * (2y)^(r) = .^(9)C_(r) xx 2^(r) xx x^((9-r)) xx y^(r).` Putting ` r = 2, " we get the coefficient of " x^(7) y^(2) " as " .^(9) C_(2) xx 2^(2) = 144.` |
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| 359. |
The remainder when `1!+2!+3!+4!+......+1000!` is divided by `10` is |
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Answer» Given expression is = `1!+2!+3!+4!+5!...+1000!` `1! = 1` `2! = 2**1=2` `3! = 3**2**1 = 6` `4! = 4**3**2**1 = 24` `5! = 5**4**3**2**1 = 120` So, `5!` is divisible by `10`. Now, `6! = 6**5!` So, `6!` will also be divisible by `10`.Similarly, every term greater than `5!` in the given expression will be divisible by `10`. So, sum of the terms that are not divisible by `10` is, `1!+2!+3!+4! = 1+2+6+24 = 33` `:.` Remainder of `33` when divided by `10` will be `3` which is the required answer. |
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| 360. |
Which is larger,`(1.01)^(1000000) or 10000`? |
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Answer» We have `(1.01)^(1000000)-10000` `(1.01)^(p)-10000, " where " 1000000 =p` `(1+0.01)^(p)-10000` `=.^(p)C_(0) + ^(p) C_(1) xx 0.01 + .^(p)C_(2) xx (0.01)^(2)+ ...+ .^(p)C_(p) xx(0.01)^(p)-10000` `=1+ (p xx 0.01)+[.^(p)C_(2) xx(0.01)^(2) +...+(1.01)^(p)]-10000` `1+(1000000 xx0.01) +[.^(1000000)C_(2) xx(0.01)^(2)+...+(1.01)^(1000000)]-10000` `1+ 10000 + " [a positive real number] "-10000` `1+" (1 positive real number) "gt0` Hence, `(1.01)^(1000000) gt 10000`. |
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| 361. |
Write the coefficient of the middle term in the expansion of `(1+x)^(2n)`. |
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Answer» Correct Answer - `.^(2n)C_(n)` Middle term `=((2n)/2 +1)th " term " = (n+1)th " term " = T_(n+1).` Now , `T_(r+1) = .^(2n)C_(r)x^(r) rArr T_(n+1) = .^(2n)C_(n)x^(n). " It coefficient is " .^(2n)C_(n).` |
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| 362. |
The sum of the coefficients of the first three terms in the expansion of`(x-3/(x^2))^m ,x!=0,`m being a natural number, is 559. Find the term of the expansion containing `x^3`. |
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Answer» representing as sum of first three terms `.^mC_0(-3)^0 + .^mC_1(-3)^1 + .^mC_2(-3)^2 = 559` `1 + m(-3) + (m(m-1))/2*9 = 559` `-3m + (9m^2-9m)/2 = 558` `3m^2/2- 5m/2 = 186` `3m^2 - 5m -372 = 0` `m = (5+- sqrt(25+12*372))/6` `= (5 + sqrt(4464+25))/6` `= (5+67)/6 72/6 = 12` to find the term for `x^3` so,`m-3r=3` `r=3` so fourth term will be taken`t_4 = .^12C_3(-3)x^3` `= 12*11*10/(3*2)* (-27x^3)` `= -5940x^3` |
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| 363. |
Using binomial theorem, indicate which number is larger `(1. 1)^(10000)`or 1000. |
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Answer» Using Binomial theorem,`(1.1)^10000 = (1+0.1)^100000` ` = 10000_(C_0)*1*(0.1)^0+10000_(C_1)*1*(0.1)^1+....`Here, we can see that first two terms are more than 1000. and rest are all positive values. So, we can say that , `(1.1)^10000 > 1000` |
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| 364. |
If the coefficient of the middle term in the expansion of `(1+x)^(2n+2)i salpha`and the coefficients of middle terms in the expansion of `(1+x)^(2n+1)`are `beta`and `gamma`then relate `alpha,betaa n dgammadot` |
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Answer» Since `(n+2)` th term is the middle term in the expansion of `(1+x)^(2n+n)`, we have `alpha = .^(2n+2)C_(n+1)`. Since `(n+1)` th and `(n+)` th terms are middle term in the expansion of `(1+x)^(2n+1)`. We have `beta = .^(2n+1)C_(n)` and `gamma = .^(2n+1)C_(n+1)` But `.^(2n+1)C_(n) + .^(2n+1)C_(n+1) = .^(2n+2)C_(n+1)` or `beta + gamma = alpha` |
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| 365. |
Find the term independent of x in the expansion of `(3/2x^2-1/(3x))^6`. |
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Answer» `(r+1)^(th)` term in given expression, `T_(r+1) = C(6,r)((3x^2)/2)^(6-r)(-1/(3x))^r` `= C(6,r)(3/2)^(6-r)(-3)^(-r)(x^2)^((6-r)-r)` `T_(r+1)= C(6,r)(3/2)^(6-r)(-3)^(-r)(x)^(12-3r)->(1)` For `T_(r+1)` to be independent of `x`,`(12-3r) = 0` `=>r = 4` So, fifth term of given expression is independent of `x`. `:. T_5 = C(6,4)(3/2)^2(-3)^(-4)` (putting value of r in (1)) ` = 15**9/4**1/81 = 5/12` |
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| 366. |
Find the `r^(t h)`term from the end in the expansion of `(x+a)^n`. |
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Answer» We have to find `r^(th)` term from end in the expansion of `(x+a)^n` It wil be same as `r^(th)` term from beginning in the expansion of `(a+x)^n` From Binomial theorem, we have, `T_r = n_(C_(r-1))*a^(n-(r-1))*x^(r-1)` `T_r = n_(C_(r-1))*a^(n-r+1)*x^(r-1)` |
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| 367. |
Show that the coefficient of the middle term in the expansion of `(1+x)^(2n)`is equal to the sum of the coefficients of two middle terms in the expansion of `(1+x)^(2n-1)`. |
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Answer» Middle terms in expansion `(1+x)^(2n-1)` are `T_n` and `T_(n+1).` So, sum of coefficients of these two middle terms will be, `C(2n-1,n-1)+C(2n-1,n) = ((2n-1)!)/((n-1)!n!) + ((2n-1)!)/(n!(n-1)!)` `=2((2n-1)!)/(n!(n-1)!) = (2n((2n-1)!))/(n(n-1)!(n!))` `=(2n!)/((n!)(n!))->(1)` Now, we will find the coefficient of middle term `T_(n+1)` in expansion `(1+x)^(2n)`. Coefficient of `T_(n+1)` in `(1+x)^(2n)` `= C(2n,n) = (2n!)/((n!)(n!))->(2)` From `(1)` and `(2)`, we can see that both values are equal. |
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| 368. |
Find the middle term (terms) in the expansion of (i) `((x)/(a) - (a)/(x))^(10)` (ii) `(3x - (x^(3))/(6))^(9)` |
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Answer» (i) Given expansion is `((x)/(a) - (a)/(x))^(10)` Here, the power of Binomial i.e., `n = 10` is even Since, it has one middle term `((10)/(2) + 1)` th terms i.e., 6 th term `:. T_(6) = T_(5 + 1) = .^(10)C_(5) ((x)/(a))^(10 - 5) ((-a)/(x))^(5)` `= -.^(10)C_(5) ((x)/(a))^(5) ((a)/(x))^(5)` `= - (10 xx 9 xx 8 xx 7 xx 6 xx 5!)/(5! xx 5 xx 4 xx 3 xx 2 xx 1) ((x)/(a))^(5) ((x)/(a))^(-5)` `= - 9 xx 4 xx 7 = - 252` (ii) given expansion is `(3x - (x^(3))/(6))^(9)` Here, `n = 9` Since, the Binomial expansion has two middle terms i.e., `((9 + 1)/(2)) th` and `((9 + 1)/(2) + 1)th` i.e., 5th term and 6th term. `:. T_(5) = T_((4 + 1)) = .^(9)C_(4) (3x)^(9 - 4) (-(x^(3))/(6))^(4)` `= (9 xx 8 xx 7 xx 6 xx 5!)/(4 xx 3 xx 2 xx 1 xx 5!) 3^(5) x^(5) x^(12) 6^(-4)` `= (7 xx 6 xx 3 xx 3^(1))/(2^(4)) x^(17) = (189)/(8) x^(17)` `:. T_(6) = T_(5) = .^(9)C_(5) (3x)^(9 - 5) (- (x^(3))/(6))^(5)` `= - (9 xx 8 xx 7 xx 6 xx 5!)/(5! xx 4 xx 3 xx 2 xx 1) . 3^(4). x^(4). x^(15). 6^(-5)` `= (-21 xx 6)/(3 xx 2^(5)) x^(19) = (-21)/(16) x^(19)` |
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| 369. |
If `(1+x)^(n) = C_(0) + C_(1)x + C_(2)x^(2) + "….." + C_(n)x^(n)`, then `C_(0) - (C_(0) + C_(1)) +(C_(0) + C_(1) + C_(2)) - (C_(0) + C_(1) + C_(2) + C_(3))+"….."(-1)^(n-1) (C_(0) + C_(1) + "……" + C_(n-1))` is (where n is even integer and `C_(r) = .^(n)C_(r)`)A. a positive valueB. a negative valueC. divisible by `2^(n-1)`D. divisible by `2^(n)` |
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Answer» Correct Answer - B::C For `n = 2m`, the given expression is `C_(0)-(C_(0)+C_(1))+(C_(0)+C_(1)+C_(2))-(C_(0)+C_(1)+C_(2)+C_(3))+"....."(-1)^(n-1)(C_(0)+C_(1)+"....."+C_(n-1))` `= C_(0)-(C_(0)+C_(1))+(C_(0)+C_(1)+C_(2))-(C_(0)+C_(1)+C_(2)+C_(3))+"....."-(C_(0)+C_(1)+"...."+C_(2m-1))` `= - (C_(1)+C_(3)+C_(5)+"...."+C_(2m-1))` `= - (C_(1)+C_(3)+C_(5)+"......."+C_(n-1)) = -2^(n-1)` |
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| 370. |
In the expansion of `(a+b)^(n)`, if two consecutive terms are equal, then which of the following is/are always integer ?A. `((n+1)b)/(a+b)`B. `((n+1)a)/(a+b)`C. `(na)/(a-b)`D. `(na)/(a+b)` |
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Answer» Correct Answer - A::B We have `(a+b)^(n)` `(T_(r+1))/(T_(r)) = (n-r+1)/(r ) . b/a = 1` `:. (n+1)b-r = ar` `:. r = -((n+1)b)/(a+b)` is integer Also, considering `(b+a)^(n)`. `(T_(r+1))/(T_(r)) = (n-r+1)/(r ) . a/b = 1` `:. R = ((n+1)a)/(a+b)` is integer |
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| 371. |
Find the value of `^20 C_0-(^(20)C_1)/2+(^(20)C_2)/3-(^(20)C_3)/4+dot` |
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Answer» Correct Answer - `1/21` `.^(20)C_(0) - (.^(20)C_(1))/(2) + (.^(20)C_(2))/(3) - (.^(20)C_(3))/(4) + "....."` `= underset(r=0)overset(20)sum(.^(20)C_(r))/(r+1)(-1)^(r)` `= underset(r=0)overset(20)sum(.^(21)C_(r+1))/(20+1)(-1)^(r)` ` = -1/21 underset(r=0)overset(20)sum.^(21)C_(r+1)(-1)^(r+1)` `= - 1/21[-.^(21)C_(1) + .^(21)C_(2)-.^(21)C_(3) + "...."]` `= - 1/(21)[(.^(21)C_(0) - .^(21)C_(1) + .^(21)C_(2) - .^(21)C_(3) + ".....")-.^(21)C_(0) ]` `= -(1)/(21) [(1-1)^(21) - 1]` `= 1/21` |
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| 372. |
If `f(m) = sum_(i=0)_(m) ({:(30),(30-i):})({:(20),(m-i):})` where `({:(p),(q):})= ""^(p)C_(q)`, thenA. maximum value of `f(m)` is `.^(50)C_(25)`B. `f(0) + f(1)+"….."+f(50) = 2^(50)`C. `f(m)` is always divisible by `50(1 le m le 49)`D. The value of `underset(m=0)overset(50)sum(f(m))^(2) = .^(100)C_(50)` |
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Answer» Correct Answer - A::B::D `f(m)= underset(i=0)overset(m)sum({:(" "30 ),(30-i):})({:(" "30 ),(m-i):})= underset(i=0)overset(m)sum({:(" "30 ),(" "i):})({:(" "20 ),(m-i):})= .^(50)C_(m)` `f(m)` is greatest when `m = 25`. Also, `f(0)+f(1)+"….." +f(50)= .^(50)C_(0) + .^(50)C_(1)+.^(50)C_(2)+"....."+.^(50)C_(50) = 2^(50)` Also, `.^(50)C_(m)`is not divisible by 50 for all m as 50 is not a prime number `underset(m=0)overset(50)sum(f(m))^(2)=(.^(50)C_(0))^(2)+(.^(50)C_(1))^(2)+(.^(50)C_(2))^(2)+"......."+(.^(50)C_(50))^(2)` `= .^(100)C_(50)` |
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| 373. |
If for `z`as real or complex, `(1+z^2+z^4)^8=C_0+C1z2+C2z4++C_(16)z^(32)t h e n``C_0-C_1+C_2-C_3++C_(16)=1``C_0+C_3+C_6+C_9+C_(12)+C_(15)=3^7``C_2+C_5+C_6+C_(11)+C_(14)=3^6``C_1+C_4+C_7+C_(10)+C_(13)+C_(16)=3^7`A. `C_(0) - C_(1) + C_(2) - C_(3) + "….." + C_(16) = 1`B. `C_(0) + C_(3) + C_(6) + C_(12) + C_(15) = 3^(7)`C. `C_(2) + C_(5) + C_(8) + C_(11) + C_(14) = 3^(6)`D. `C_(1) + C_(4) + C_(7) + C_(10) + C_(13) + C_(16) = 3^(7)` |
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Answer» Correct Answer - A::B::D `(1+z^(2)+z^(4))^(8) = C_(0) + C_(1)z^(2) + X_(2)z^(4) + "….." + C_(16)z^(32) " "(1)` Putting `x = i`, where `i = sqrt(-1)`. `(1-1+1)^(8) = C_(0) - C_(1) + C_(2) - C_(3) + "….." + C_(16)` or `C_(0) - C_(1) + C_(2) - C_(3) + "……" + C_(16) = 1` Also, putting `z = omega` `(1+omega^(2)+omega^(4))^(8)= C_(0) + C_(1)omega^(2) + C_(2)omega^(4) + "....." + C_(16)omega^(32)` or `C_(0) + C_(1)omega^(2) + C_(2)omega + C_(3) + "...." + C_(16)omega^(2) = 0 " "(3)` Putting `x = omega^(2)`. `(1+omega^(4)+omega^(8))^(8) = C_(0) + C_(1)omega^(4) + C_(2)omega^(8) + "......" + C_(16)omega^(64)` or `C_(0) + C_(1)omega + C_(2)omega^(2) + "....." + C_(16)omega = 0 " "(3)` Putting `x = 1`, `3^(8) = C_(0) + C_(1) + C_(2) + "....."+C_(16) " "(4)` Adding (2), (3) and (4), we have `3(C_(0) + C_(3) + "......" + C_(15)) = 3^(8)` or `C_(0) + C_(3) + "......" + C_(15) = 3^(7)` Similarly, first multiplying (1) by x and then putting `1 omega, omega^(2)` and adding, we get `C_(1) + C_(4) + C_(7) + C_(10) + C_(13)+ C_(16) = 3^(7)` Multiplying (1) by `z^(2)` and then putting `1, omega, omega^(2)` and adding, we get `C_(2)+C_(5)+C_(8)+C_(11)+C_(14)= 3^(7)` |
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| 374. |
Find `(x+1)^6+(x-1)^6`. Hence or otherwise evaluate `(sqrt(2)+1)^6+(sqrt(2)-1)^6`.A. 184B. 192C. 198D. 202 |
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Answer» Correct Answer - C N/a |
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| 375. |
Find the sum of the series `sum_(r=0)^(n) (-1)^(r ) ""^(n)C_(r ) [(1)/(2^(r )) + (3^(r ))/(2^(2r)) + (7^(r ))/(2^(3 r)) + (15^(r ))/(2^(4r)) …. "upto m terms"]` |
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Answer» `underset (r=0)overset(n)Sigma (-1)^(n) C_(r)[(1)/(2^(r ))+(3^(r ))/(2^(2r))+(7^(r ))/(2^(3r))+(15^(r ))/(2^(4r))+…"upto m terms"]` `=underset(r=0)overset(n)Sigma(-1)^(r ) C_(r)(1/2)^(r )+ underset(r=0)overset(n)Sigma^(r )C_(r )(3/4)^(r )+underset(r=0)overset(n)Sigma(-1)^(r )C_(r )(7/8)^(r )+....` upto m terms `=(1-1/2)^(n)+(1-3/4)^(n)+(1-7/8)^(r) +...` up to m terms `[using underset(r=0)overset(n)Sigma (-1)^(r )C_(r )x^(r )=(1-x)^(n)]` `=(1/2)^(n)+(1/4)^(n)+(1/8)^(n)+...` up to m terms `=(1/2)^(n)[(1=-(1/2)^(m))/(1-1/2)]=(2^(mn)-1)/(2^(mn)(2^(n)-1)`. |
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| 376. |
If `f(x) = x^100-2x+1` is divided by `x^2-1` then the remainder is equal toA. `0`B. `2x+2`C. `2x-2`D. `-2x+2` |
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Answer» Correct Answer - D |
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| 377. |
Using binomial theorem, evaluate : `(96)^3` |
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Answer» `(96)^3=(100-4)^3` `(x+a)^n=sum_(r=0)^(n).^nC_ra^rr^(n-r)` `(100-4)^3= .^3C_0(100)^3+.^3C_1(100)^2(-4)^(1).+^3C_2(100)^1(-4)^(2)+.^3C_3(100)^0(-4)^(3)` =>`1000000-120000+4800-64` =>`884736`. |
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| 378. |
In the expansion of `(1+ x + 7/x)^11`find the term not containing x.A. `underset(r=0)overset(5)sum.^(11)C_(r).^(11-r)C_(2r)7^(r)`B. `underset(r=0)overset(5)sum.^(11)C_(r).^(11-r)C_(11-2r)7^(r)`C. `underset(r=0)overset(5)sum.^(11)C_(r).^(11-r)C_(11-2r)7^(r)`D. none of these |
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Answer» Correct Answer - C `(1+x+7/x)^(11) = ((7+x+x^(2))/(x))^(11)` So, we have to find the coefficient of `x^(11)` in `(7+x+x^(2))^(11)` Now, `(7+x+x^(2))^(11)` `=[((7+x)+x^(2))]^(11)` `= .^(11)C_(0)(7+x)^(11)+.^(11)C_(1)(7+x)^(10)x^(2)+.^(11)C_(2)(7+x)^(9)x^(4)"....."+"...."` `:.` Required coefficient `= .^(11)C_(0)+.^(11)C_(1)xx.^(10)C_(9)xx7+.^(11)C_(2)xx.^(9)C_(7)` `xx7^(2)+.^(11)C_(3)xx.^(8)C_(5) xx7^(3)+.^(11)C_(4)` `xx.^(7)C_(3)xx7^(4)+.^(11)C_(5)xx.^(6)C_(1)xx7^(5)` `= underset(r=0)overset(5)sum.^(11)C_(r).^(11-r)C_(11-2r)7^(r)` |
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| 379. |
Prove that there is no term containing `x^(10)` in the expansion of `(x^(2) - 2/x)^(18).` |
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Answer» The general term in the given expansion is given by `T _(r+1)= (-1)^(r) xx.^(18)C_(r) xx (x^(2))^((18-r)) xx (2/x)^(r)` `rArr T_((r+1))=(-1)^(r) xx.^(18)C_(r) xx 2^ (r) xxx^((36-3r))`. Let `T_(r+1)" contain " x^(10)`. Then, `36-3r = 10 rArr 3r = 26 rArr r=26/3` Since the value of r cannot be a fraction , so there is no term containing `x^(10)`. |
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| 380. |
Find the coefficient of (i) `x^(5)" in the expansion of "(x+3)^(8)` (ii) `x^(6) " in the expansion of "(3x^(2) - a/(3x))^(9)` (iii) `x^(-15) " in the expansion of " (3x^(2) - a/(3x^(3)))^(10).` (iv) `a^(7)b^(5)" in the expansion of "(a-2b)^(12)`. |
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Answer» Correct Answer - (i) 1512 (ii) 378 (iii) `(-40)/27a^(7)` (iv)-25344 (iii) `T_(r+1)= ( - 1)^(r) xx .^(10)C_(r) ( 3x^(2))^((10-r)) xx (a/(3x^(3)))^(r)` `rArr T_(r+1) = ( -1)^(r) xx . ^(10)C_(r) xx 3 ^((10-r)) xx x^((20 xx 5r)) xx a^(r).` Now, `20 - 5r = -15 rArr 5r = -15 rArr 5r = 35 rArr r = 7.` (iv) `T_(r+1) =(-1)^(r) xx .^(12) C _(r) xxa^((12-r)) xx ( 2b)^(r)` ` rArr T_(r+1) = (-1)^(r) xx .^(12)C_(r) xx 2^(r) xx a^((12-r)) b^(r).` Put`r=5 and " get " T_(6).` |
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| 381. |
Show that the term containing to does not exist in the expansion of `(3x-1/(2x))^8` |
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Answer» `T_(r+1) = (-1)^(r)*^(8)C_(r)*(3x)^((8-r)) .(1/(2x))^(r)` `rArr T_(r+1) = (-1)^(r)*^(8) C_(r)*3^(8-r)*1/2^(r) *x ^((8-2r)).` Now, `8-2r = 3 rArr r=5/2`, which is a fraction. |
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| 382. |
Write the general term in the expansion of `(x^(2)-y)^(6)`. |
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Answer» Correct Answer - `T_(r+1) = (-1)^(r)xx.^(6)C_(r) xx x^(12-2r)xxy^(r)` `T_(r+1)=(-1)^(r) xx .^(6)C_(r) xx x^((12-2r)) xx y^(r).` |
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| 383. |
Write the general term in the expansion of `(x^2-y)^6` |
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Answer» General term in `(a+b)^n` is given by, `T_(r+1) = C(n,r)a^(n-r)b^r` So, in given expression, `(x^2-y)^6)` `T_(r+1) = C(6,r)(x^2)^(6-r)(-y)^r` `T_(r+1)=(-1)^r*C(6,r)(x)^(12-2r)(y)^r` |
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| 384. |
Using binomial theorem, evaluate : `(101)^4` |
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Answer» `(101)^4 = (100+1)^4` `=4_(C_0)(100^4)+4_(C_1)(100^3)+4_(C_2)(100^2)+4_(C_3)(100^1)+4_(C_4)(100^0)` `100000000+4000000+60000+400+1 = 104060401` |
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