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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.
| 151. |
Middle term in the expansion of `(a^(3) + ba)^(28)` is ...... |
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Answer» Given expansion is `(a^(3) + ba)^(28)` `:. N = 28` `:.` Middle term `= ((28)/(2) + 1)th` term = 15 th term `:. T_(15) = T_(14 + 1)` `= .^(28)C_(14) (a^(3))^(28 - 14) (ba)^(14)` `= .^(28)C_(14) a^(42) b^(14) a^(14)` `= .^(28)C_(14) a^(56) a^(14)` |
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| 152. |
Prove that `.^(n)C_(0) +5 xx .^(n)C_(1) + 9 xx .^(n)C_(2) + "…." + (4n+1) xx .^(n)C_(n) = (2m+1) 2^(n)`. |
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Answer» `S = .^(n)C_(0) + 5 xx .^(n)C_(1)+9xx.^(n)C_(2)+"....."(4n-3)xx.^(n)C_(n-1)+(4n+1)xx.^(n)C_(n)"......"(1)` `:. S = (4n+1).^(n)C_(n)+(4n-3).^(n)C_(n-1)+"...."+5.^(n)C_(1)+.^(n)C_(n)"....."(2)` Adding (1) and (2), we get `2S = (4n+2)(.^(n)C_(0)+.^(n)C_(1)+"....."+.^(n)C_(n-1)+.^(n)C_(n))` `= (4n+2)2^(n)` `rArr S = (2n+1)2^(n)` |
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| 153. |
If `pa n dq`are ositive, then prove that the coefficients of `x^pa n dx^q`in the expansion of `(1+x)^(p+q)`will be equal. |
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Answer» Given expansion is `(1 + x)^(p + q)` `:.` Coefficient of `x^(p) = .^(p + q)C_(p)` and coefficient of `x^(q) = .^(p + q)C_(q)` `:. (.^(p + q)C_(p))/(.^(p + q)C_(q)) = (.^(p + q)C_(p))/(.^(p + q)C_(p)) = 1 : 1` |
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| 154. |
The number of terms in the expansion of `(1+5sqrt2 x)^9+(1-5sqrt2 x)^9` isA. 5B. 7C. 9D. 10 |
| Answer» Correct Answer - a | |
| 155. |
Prove that `underset(rles)(underset(r=0)overset(s)(sum)underset(s=1)overset(n)(sum))""^(n)C_(s) ""^(s)C_(r)=3^(n)-1`. |
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Answer» `underset(rles)(underset(r=0)overset(s)(sum)underset(s=1)overset(n)(sum)).^(n)C_(s).^(s)C_(r)=underset(s=1)overset(n)sum.^(n)C_(s)(.^(s)C_(0)+.^(s)C_(1)+.^(s)C_(2)+"....."+.^(s)C_(s))` `= underset(s=1)overset(n)sum.^(n)C_(s)2^(s)` `= underset(s=0)overset(n)sum.^(n)C_(s)2^(s)-.^(n)C_(0)2^(0)` `= (1+2)^(n)-1` `= 3^(n) - 1` |
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| 156. |
The value of `(.^(n)C_(0))/(n)+(.^(n)C_(1))/(n+1)+(.^(n)C_(2))/(n+2)+"..."+(.^(n)C_(2))/(2n)`A. `underset(0)overset(1)intx^(n-1)(1-x)^(n)dx`B. `underset(1)overset(2)intx^(n)(x-1)^(n-1)dx`C. `underset(1)overset(2)int(1+x)^(n)dx`D. `underset(0)overset(1)int(1-x)^(n)x^(n-1)dx` |
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Answer» Correct Answer - B Let, `S = (.^(n)C_(0))/(n)+(.^(n)C_(1))/(n+1)+(.^(n)C_(2))/(n+2)+"...."+(.^(n)C_(n))/(2n)` `= .^(n)C_(0)underset(0)overset(1)intx^(n-1)+.^(n)C_(1)underset(0)overset(1)intx^(n)dx+"....."+.^(n)C_(n)underset(0)overset(1)intx^(2n-1)dx` `=underset(0)overset(1)int[.^(n)C_(0)x^(n-1)+.^(n)C_(1)x^(n)+"..."+.^(n)C_(n)x^(2n-1)]dx` `= underset(0)overset(1)intx^(n-1)(1+x)^(n)dx` `= underset(1)overset(2)intx^(n)(x-1)^(n-1)dx` |
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| 157. |
The position of the term independent of `x` in the expansion of `(sqrt((x)/(3)) + (3)/(2x^(2)))^(10)` is ...... |
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Answer» Given expansion is `(sqrt((x)/(3)) + (3)/(2x^(2)))^(10)` Let the constant term be `T_(r + 1)` Then, `T_(r + 1) = .^(10)C_(r) (sqrt((x)/(3)))^(10 - r) ((3)/(2x^(2)))^(r)` `= .^(10)C_(r) . X^((10 - r)/(2)) . 3^((-10 + r)/(2)) .3^(r).2^(-r). x^(-2r)` `= .^(10)C_(r) x^((10 - 5r)/(2)) 3^((-10 + 3r)/(2)) 2^(-r)` For constant term, `10 - 5r = 0 rArr r = 2` Hence, third term is independent of x |
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| 158. |
The sum of the series `underset(r = 0)overset(10)Sigma .^(20)C_(r) " is " 2^(19) + (.^(20)C_(10))/(2)`A. `2^(20)`B. `2^(19)`C. ` 2^(19) + (1)/(2) ""^(20)C_(10)`D. `2^(19)- (1)/(2) ""^(20)C_(10)` |
| Answer» Correct Answer - c | |
| 159. |
`7^9+9^7` is divisible by (A) 16 (B) 24 (C) 64 (D) 72 |
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Answer» Correct Answer - 1 Given expression `= 7^(9) + 9^(7) = (1 + 8)^(7) - (1 - 8)^(9)` `= (.^(7)C_(0) + .^(7)C_(1) 8 + .^(7)C_(2) 8^(2) + ... + .^(7)C_(7) 8^(7) - (.^(9)C_(0) - .^(9)C_(1) 8 + .^(9)C_(2) 8^(2) .... - .^(9)C_(9) 8^(9))` `= (1 + 7 xx 21 xx 8^(2) + ...) - (1 - 9 xx 8 + 36 xx 8^(2) + .... - 8^(9))` `= ( 7 xx 8 + 9 xx 8) + (21 xx 8^(2) - 36 xx 8^(2)) +`.... `= 2 xx 64 + (21 - 36) 64 +`..... Which is divisible by 64 Hence, the statement is true |
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| 160. |
If `25^(15)` is divided by 13, then the remainder is ..... |
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Answer» Let `25^(15) = (26 - 1)^(15)` `= .^(15)C_(0) 26^(15) - .^(15)C_(1) 26^(14) + ... .^(-15)C_(15)` `= .^(15)C_(0) 26^(15) - .^(15)C_(1) 26^(14) + .... -1 - 13 + 13` `= .^(15)C_(0) 26^(15) - .^(15)C_(1) 26^(14) + ..... -13 + 12` It is clear that, when `25^(15)` is divided by 13, then remainder will be 12 |
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| 161. |
Prove that `: ""^(n)C_(0).""^(2n)C_(n)-""^(n)C_(1).""^(2n-2)Cn_(n)+""^(n)C_(2).""^(2n-4)Cn_(n)+......=2^n` |
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Answer» L.H.S = coefficent of `x^n "in" [""^(n)C_(0)(1+x)^(20)-""^(n)C_(1)(1+x)^(2n-2)....]` = cofficient of `x^(n)[(1+x)^2-1]^n` Coffficient of `x^n "in" x^n(X+2)^n=2^n` |
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| 162. |
The sum of the series `Sigma_(r = 0)^(10) ""^(20)C_(r) " is " 2^(19) + (""^(20)C_(10))/(2)` |
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Answer» Given series `= underset(r = 0)overset(10)Sigma .^(20)C_(r) = .^(20)C_(0) + .^(20)C_(1) + .^(20)C_(2) + ... + .^(20)C_(10)` `= .^(20)C_(0) + .^(20)C_(1) +..... + .^(20)C_(10) + .^(20)C_(11) + ... .^(20)C_(20) - (.^(20)C_(11) + .... + .^(20)C_(20))` `= 20^(20) - (.^(20)C_(11) + ... + .^(20)C_(20))` Hence, the given statement is false |
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| 163. |
Find the coffiecient of `x^5` in the expansion of `(2-x+ 3x ^2)^6` |
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Answer» The general term in the expansion of `(2-x+3x^2)^6=(6!)/(r !s! t!) 2 ^r(-x)^3(3x^2)^t` where r+s+ t =6 `(6!)/(r! s!t!)2 r xx (-1)xx(3)^t xxx ^(s+2t)` for the cofficient of `x^5` we must have s+2t =5 . `therefore S= 5-2 t and r=1 +t ` where ` 0 le r, s, t le 6 ` Now t=0 `rarr` r= 1 , s= 5 t=1 `rArr ` , s=3 , t = 2 `rArr` r= 3 , s=1 Thus there are three terms containig `x^5` and coefficient of `x^5` `=(6 !)/(1 ! 5 ! 0 ! )xx2^1xx(-1)^5xx3^0+(6 !)/(2! 3 !1 ! )xx2^2xx(-3)^1+(6!)/(3 ! 1 !2! ) xx2^3 xx (-1)^1 xx 3^2` ltbrge-12- 720 - 4320 = - 5052 |
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| 164. |
If `(1+X+x^2)^n = Sigma_(r=0)^(2n) a_(r) x^r` then prove that (a) `a_(r)=a_(2n-r)` (b ) `Sigma_(r=0)^(n-1) a_(r)=1/2 (3^n-a_n)` |
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Answer» We have ` (1+x+x^2)^n = underset(r=0)overset(2n)Sigma a_(r) x^r` Replace x by 1/x `therefore (1 +1/x+1/(x^2))^n = underset(r=0)overset(2n) Sigma a_(r)((1)/(x))^r` `rArr (x^2+x+1)^n = underset( r=0)overset(2n)Sigma a_rx^(2n-r)` `underset(r=0)overset(2n ) a_(r)x^r =underset(r=0)overset(2n ) a_(r)x^r " " {"Using (A) "}` Equating the cofficient of `x^(2n-r)` on the both sides ,we get `a_(2n-r)= a_(r) " for " 0 le r le 2n` Hence `a_(r)=a_(2n-r)` (b) Putting x=1 in given series , then `a_(0)+a_1 +a_2+......+a_(2n)=(1+1+1)^n` `a_(0)+a_(1)+a_(2)+...... +a_(2n)=3^n` But `a^r= a_(2n-r) for 0 le r le4 2 n ` `therefore ` Series (1) reduces to `2(a_(0)+a_1+a_2+.......+a_(n-1)) + a_n =3^n` `therefore a_0+a_1+a_2+........+a_(n-1)= 1/2 (3^n-a_n)` |
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| 165. |
If `(1+x)^n=C_(0)C_1c+C_(2)x^2+…..+C_(n)x^n` then show that the sum of the products of the `C_(i)` taken two at a time represented by :`Sigma_(0 le I lt) Sigma_( j le n) C_(i)C_(j)` "is equal to " 2^(2n-1)-(2n!)/(2.n! n !)` |
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Answer» Since `(c_0+C_1+C_2 +……+C_(n-1)+C_(n))^2` `=C_(0)^(2)+C_(1)^(2)+C_(2)^(2)+.......C_(n-1)^(2)+C_(n)^2+2(C_(0)C_(1)+C_(0)C_(2)+C_(0)C_(3)+....+C_(0)C_(n)+C_(1)C_(2)+C_1C_3+...C_1C_n+C_2C_3+C_2C_4+...+C_2C_n+...+C_(n-1)C_n)` `(2^)2=""^(2n)C_(n)+2 underset(0 le i lt j le n )(SigmaSigma) C_(i)C_(j)` Hence `underset(0 le i lt j le n )(SigmaSigma) C_(i)C_(j)=2""^(2n-1)-(2n!)/(2.n!n!)` |
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| 166. |
Find the coffiecient of `x^2 y^3 z^4 ` w in the expansion of `(x-y-z+w)^(10)` |
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Answer» `(X-Y-Z +W)^(10)= underset(P+q+r+s=10) Sigma (n!)/(p ! Q! R ! S! )(x)^p(-y)^q(-z)^r (W)^s` we want to get `x^2 y^3 z^4` w this implies that p=2,q = 3 ,r = 4 ,s=1 `therefore "Cofficient of " x^2 y^3 z^4 W is (10!)/(2! 3 ! 4 ! 1 !) (-1 )^3(-1)^4= -12600` |
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| 167. |
Find the coefficient of `x^5` in `(1+2x+3x^2...........)^(-3/2)`A. 21B. 25C. 26D. none of these |
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Answer» Correct Answer - D `(1+2x+3x^(2)+"……")^(-3//2) =[(1-x)^(-2)]^(-3//2)` `= (1-x)^(3)= 1-3x+3x^(2) - x^(3)` Therefore, coefficientof `x^(5)` is `0` |
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| 168. |
If `(1+x)^n=C_(0)+C_(1)x+C_(2)x^2+….+C_(n)x^n` then prove that `(SigmaSigma)_(0 le i lt j le n ) C_(i)C_(j)^2=(n-1)^(2n)C_(n)+2^(2n)` |
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Answer» L.H.S = `underset(0 le i lt j le n )(SigmaSigma) C_(i)C_(j)^2=(n-1)^(2n)C_(n)+2^(2n)` `(C_(0)+C_(1))^2+(C_(0)+C_2)^2+....+(C_(0)+C_(n))^2+(C_1C_2)^2+(C_1+C_3)^2+....+(C_1+C_c)^2+(C_2+C_3)^2+(C_(2)+C_4)^2+....+(C_2+C_n)^2+......+(C_(n-1)+C_n)^2` `=n(C_(0)^2+C_1^2+C_2^2+....+C_n^2)+2 underset(0 le i lt j le n )(SigmaSigma) C_(i)C_(j)` `n.""^(2n)C_(n)+2{2^(2n-1)-(2n !)/(2.n!n!)} " " {"from Illustration 17 "}` `=n.""^(2n)C_n+2^(2n)-""^(2n)C_n= (n-1).""^(2n)C_(n)+2^(2n)= R.H.S` |
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| 169. |
If `|x| |
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Answer» Correct Answer - D Required value of `(1-(2x)/(1+x))^(-n) = ((1+x-2x)/(1+x))^(-n) = ((1-x)/(1+x))^(-n) = ((1+x)/(1-x))^(n)` |
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| 170. |
Find the sum `1-1/8+1/8xx3/(16)-(1xx3xx5)/(8xx16xx24)+` |
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Answer» Comparing the given series to `1+nx + (n(n-1))/(2!) x^(2) + "…." = (1+x)^(n)` we get ` nx = -1/8` and `(n(n-1))/(2!) x^(2) = 3/128` or `x = 1/4, n = - 1/2` Hence, `1-1/(8)+(1)/(8)(3)/(16)-"…."=(1+(1)/(4))^(-1//2) = (2)/(sqrt(5))` |
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| 171. |
If `C_(0), C_(1), C_(2), ..., C_(n)` denote the binomial cefficients in the expansion of `(1 + x )^(n)` , then ` a C_(0) + (a + b) C_(1) + (a + 2b) C_(2) + ...+ (a + nb)C_(n) = `.A. `(a + nb)2^(n)`B. `(a + nb)2 ^(n-1)`C. `(2a + nb) 2^(n-1)`D. `(2a + nb)2^(n)` |
| Answer» Correct Answer - C | |
| 172. |
`1+1/4 + (1xx3)/(4xx8) + (1xx3xx5)/(4xx8xx12) x^(3) + "….."` is equal toA. `sqrt(2)`B. `(1)/(sqrt(2))`C. `sqrt(3)`D. `(1)/(sqrt(3))` |
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Answer» Correct Answer - A Let the given series be identical with `(1+x)^(n) = 1+nx + (n(n-1))/(1xx2) "…." oo` `:. nx = 1/4` or `n^(2)x^(2) = 1/16` Also, `(n(n-1))/(2) x^(2) = 3/(32)` or `(2n)/(n-1) = ((1)/(16))/((3)/(32)) = 2/3` or `3n = n-1` or `2n = - 1` or `n = - 1/2` `rArr x = - 1/2` `:.` Reqired sum `=(1-(1)/(2))^(-1/2)= (1/2)^(-1/2) = (2)^(-1/2) = sqrt(2)` |
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| 173. |
4 The term independent ofx in the expansion of `(x-1/x)^4(x+1/x)^3` isA. -3B. 0C. 1D. 3 |
| Answer» Correct Answer - B | |
| 174. |
The number of terms in the expansion of `(x^2+1+1/x^2)^n, n in N` , is:A. 2nB. 3nC. 2n+1D. 3n+1 |
| Answer» Correct Answer - C | |
| 175. |
The term independent ofx in the expansion of `(1+x)^10*(1+1/x)^10` isA. `""^(22)C_(10)`B. 0C. `""^(22)C_(11)`D. none of these |
| Answer» Correct Answer - A | |
| 176. |
`1+1/3x+(1xx4)/(3xx6)x^2+(1xx4xx7)/(3xx6xx9)x^3+`is equal to`x`b. `(1+x)^(1//3)`c. `(1-x)^(1//3)`d. `(1-x)^(-1//3)`A. xB. `(1+x)^(1//3)`C. `(1-x)^(1//3)`D. `(1-x)^(-1//3)` |
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Answer» Correct Answer - D Let, `(1+y)^(n) = 1+1/3x+(1xx4)/(3xx6)x^(2)+(1xx4xx7)/(3xx6xx9)x^(3)+"…."` `= 1+ny + (n(n-1))/(2!) y^(2) + "….."` Comparing the terms, we get `ny = 1/3 x, (n(n-1))/(2!) y^(2) = (1xx4)/(3xx6) x^(2)` Solving, `n = - 1//3, y = -x` Hence, the given series. |
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| 177. |
In the expansion of `(x^(3) - (1)/(x^(2)))^(15)` , the constant term,isA. `""^(15)C_(6)`B. 0C. `-""^(15)C_(6)`D. 1 |
| Answer» Correct Answer - C | |
| 178. |
The number of terms in the expansion of `(x+a)^100+(x-a)^100` after simplificationA. 50B. 51C. 202D. none of these |
| Answer» Correct Answer - b | |
| 179. |
The sum of the co-efficients of all odd degree terms in the expansion of (x+sqrt(x^3-1))^5+(x-(sqrt(x^3-1))^5`, (x gt 1)`A. -1B. 0C. 1D. 2 |
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Answer» `= (a + b)^(n) + (a - b)^(n)` `= 2 (.^(n)C_(0) a^(n) + .^(n)C_(2) a^(n - 2) b^(2) + .^(n)C_(4) a^(n - 4) b^(4) + …)` We have , `(x + sqrt(x^(3) - 1))^(5) + (x - sqrt(x^(3) - 1))^(5), x gt 1` `= 2 (.^(5)C_(0) x^(5) + .^(5)C_(2) x_(3) (sqrt(x^(3) - 1))^(2) .^(5)C_(4) x (sqrt(x^(3) - 1))^(4))` `= 2(x^(5) + 10 x^(3) (x^(3) - 1) + 5x(x^(3) - 1)^(2))` `= 2 (x^(5) + 10x^(6) - 10 x^(3) + 5x^(7) - 10x^(4) + 5x)` Sum of coefficients of all degree terms is `2 (1 - 10 + 5 + 5) = 2` |
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| 180. |
The positive value of `lambda` for which the coefficient of `x^(2)` in the expression `x^(2) (sqrt(x) + (lambda)/(x^(2)))^(10)` is 720 isA. 3B. `sqrt(5)`C. `2 sqrt(2)`D. 4 |
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Answer» The general term in the expansion of binomial expression `(a + b)^(n)` is `T_(r + 1) = .^(n)C_(r ) a^(n - r) b^(r )`, so the general term in the expansio of binomial expression `x^(2) (sqrt(x) + (lambda)/(x^(2)))^(10)` is `T_(r + 1) = x^(2) (.^(10)C_(r ) (sqrt(x))^(10-r) ((lambda)/(x^(2)))^(r )) = .^(10)C_(r ) x^(2). x^((10 - r)/(2) lambda_(r ) x^(-2r)` `= .^(10)C_(r ) lambda^(r ) x^(2 + (10 - 2)/(2) - 2r)` No, for the coefficeint of `x^(2)`, put `2 + (10 - r)/(2) - 2r = 2` `implies (10 - r)/(2) - 2r = 0` `implies 10 - r = 4r implies r = 2` So, the coefficient of `x^(2)` is `.^(10)C_(2) lambda^(2) = 720` [given] `implies (10 !)/(2!8) lambda_(2) = 720 implies (10.9.8)/(2.8) lambda^(2) = 720` `implies 45 lambda^(2) = 720` `implies lambda^(2) = 16 implies lambda = +- 4` `:. lambda = 4" "[lambda gt 0]` |
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| 181. |
If the third term in expansion of `(1+x^(log_2x))^5` is `2560` then `x` is equal to (a) `2sqrt2` (b) `1/8` (c) `1/4` (d) `4sqrt2`A. `4 sqrt(2)`B. `(1)/(4)`C. `(1)/(8)`D. `2 sqrt(2)` |
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Answer» The `(r + 1)` th term in the expansion of `(a + x)^(n)` is given by `T_(r + 1) = .^(n)C_(r ) a^(n - r) x^(r )` `:. 3^(rd)` term in the expansion of `(1 + x^(log_(2)x))^(5)` is `.^(5)C_(2) (1)^(5 - 2) (x^(log_(2)x))^(2)` `implies .^(5)C_(2) (1)^(5 - 2) (x^(log_(2)x))^(2) = 2560` given `implies 10 (x^(log_(2)x))^(2) = 2560` `implies x^((2log_(2)x)) = 256` `implies log_(2) x^(2log_(2)x) = log_(2) = 256` (taking `log_(2)` on both sides ) `implies 2(log_(2)x)(log_(2)x) = 8` `(log_(2)x) = 4` `implies log_(2) = +- 2` `implies log_(2) x = 2` or `log_(2) x = - 2` `implies x = 4` or `x = 2^(-2) = (1)/(4)` |
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| 182. |
If the coefficient of `x^(7)` in `(ax - b^(-1) x^(-2))^(11)` , then ab=A. 1B. 2C. 3D. 4 |
| Answer» Correct Answer - a | |
| 183. |
If the value of `"^(n)C_(0)+2*^(n)C_(1)+3*^(n)C_(2)+...+(n+1)*^(n)C_(n)=576`, then `n` isA. `7`B. `5`C. `6`D. `9` |
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Answer» Correct Answer - A `(a)` `S=^(n)C_(0)+2.^(n)C_(1)+3.(n)C_(2)+...+(n+1).^(n)C_(n)` Here `T_(r )=(r+1)^(n)C_(r )=n^(n-1)C_(r-1)+^(n)C_(r )` `:.S=n^(2n-1)+2^(n)=576` (given) `:.n=7` |
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| 184. |
Given `"^(8)C_(1)x(1-x)^(7)+2*^(8)C_(2)x^(2)(1-x)^(6)+3*^(8)C_(3)x^(3)(1-x)^(5)+...+8*x^(8)=ax+b`, then `a+b` isA. `4`B. `6`C. `8`D. `10` |
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Answer» Correct Answer - C `(c )` `.^(n)C_(1)x(1-x)^(n-1)+2*^(n)C_(2)x^(2)(1-x)^(n-2)+3*^(n)C_(3)x^(3)(1-x)^(n-3)+....+n*^(n)C_(n)x^(n)` `=sum_(r=1)^(n)r*^(n)C_(r )x^(r )(1-x)^(n-r)` `=sum_(r=1)^(n)n^(n-1)C_(r-1)x^(r)(1-x)^(n-r)` `=nsum_(r=1)^(n).^(n-1)C_(r-1)x*x^(r-1)(1-x)^((n-1)-(r-1))` `=nxsum_(r=1)^(n).^(n-1)C_(r-1)x^(r-1)(1-x)^((n-1)-(r-1))` `=nx[x+(1-x)]^(n-1)` `=nx` `.^(8)C_(1)x(1-x)^(7)+2*^(8)C_(2)x^(2)(1-x)^(6)+3*^(8)C_(3)x^(3)(1-x)^(5)+....+8.x^(8)` `=8x` |
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| 185. |
The value of `((100),(0))((200),(150))+((100),(1))((200),(151))+......+((100),(50))((200),(200))` equals (where `((n),(r ))="^(n)C_(r)`)A. `((300),(50))`B. `((100),(50))((200),(150))`C. `((100),(50))^(2)`D. `((300),(50))^(2)` |
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Answer» Correct Answer - A `(a)` Writing `((200),(r ))=((200),(200-r))`, we have `((100),(0))((200),(50))+((100),(1))((200),(49))+((100),(2))((200),(48))+......+((100),(50))((200),(0))` `=` Coefficient of `x^(50)` in the expansion of `(1+x)^(100)(x+1)^(200)` `=` Coefficient of `x^(50)` in the binomial expansion of `(1+x)^(300)` `=((300),(50))` |
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| 186. |
Let `f(n)= (k=1)Σ^n k^2(n C_k)^ 2` then the value of f(5) equalsA. `1000`B. `1250`C. `1750`D. `2500` |
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Answer» Correct Answer - C `(c )` `k^(n)C_(k)=n^(n-1)C_(k-1)` `:.sum_(k=1)^(n)k^(2)("^(n)C_(k))^(2)` `=n^(2)sum_(k=1)^(n)("^(n-1)C_(k-1))^(2)` `=n^(2)("^(2n-2)C_(n-1))` `:. F(5)=25("^(8)C_(4))=1750` |
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| 187. |
Using binomial theorem, expand each of the following: `(1-2x)^(5)` |
| Answer» Correct Answer - `1-10x+40x^(2)-80x^(3)+80x^(4)-32x^(5)` | |
| 188. |
Using binomial theorem, expand each of the following: `(2x-3)^(6)` |
| Answer» Correct Answer - `64x^(6)-576x^(5)+2160x^(4)-4320x^(3) +4860x^(2)-2916x+729` | |
| 189. |
Using binomial theorem, expand each of the following:`(3x+2y)^(5)` |
| Answer» Correct Answer - `243x^(5)+ 810x^(4)y+1080x^(3)y^(3) + 720x^(2) y^(3)+240xy^(4)+32y^(5)` | |
| 190. |
Find the coefficient of :` x `in the expansion of `(1-3x+7x^2)(1-x)^(16)dot` |
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Answer» Correct Answer - -19 In the expansion of `E=(1-x)^(16),` we have `T_(r+1) = (-1)^(r)*^(16) C_(r)x^(r).` Product of given expressions `=( 1- 3x + 7x^(2))(1-16x + …)`. Terms containing `x= [1 xx (-16x)+ (-3x) xx 1] = ( -19x).` `:. " coefficient of " x= -19.` |
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| 191. |
Sum of the coefficients of terms of degree 13 in the expansion of`(1+x)^(11) (1+y^(2)-z)^(10)` isA. `.^(10)C_(3)`B. `.^(10)C_(4)`C. `.^(11)C_(3)`D. `.^(11)C_(4)` |
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Answer» Correct Answer - B Required sum of coefficient `=` coefficient of `t^(15)` in `(1+t)^(11) (1+t^(2) - t)^(100)` `=` coefficeint of `t^(13)` in `(1+t)(1+t^(3))^(10)` `=` coefficient of `t^(13)` in `((1+t^(3))^(10) + t(1+t^(3))^(10))` `=` coefficient of `t^(12)` in `(1+t^(3))^(10)` `= .^(10)C_(4)` |
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| 192. |
Expand `(x^(2) + 2y)^(5)` using binomial expansion. |
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Answer» Using the binomial expansion, we have `(x^(2) + 2y)^(5)= ^(5).C_(0)(x^(2))^(5)+^(5).C_(1)(x^(2))^(4) (2y)+^(5).C_(2)(x^(2))^(3)(2y)^(2)+^(5).C_(3)(x^(2))^(2) (2y)^(3)+^(5).C_(4)x^(2)(2y)^(4)+^(5).C_(5)(2y)^(5)` `x^(10)+10x^(8)y + 40 x^(6)y^(2)+80x^(4)y^(3)+ 80x^(2)y^(2)+32y^(5)`. |
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| 193. |
If the coefficient of `x^(n)` in `(1+x)^(101) (1+x^(3)-x^(6))^(30)` isA. `3r+1`B. `3r`C. `3r+2`D. none of these |
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Answer» Correct Answer - C We havec `(1+x)^(101)(1-x+x^(2))^(100)` `= (1+x)((1+x)(1-x+x^(2)))^(100)` `= (1+x)(1+x^(3))^(100)` `= (1+x){C_(0) + C_(1)x^(3)+C_(2)x^(6)+"…."+C_(100)x^(300)}`, where `C_(r) = .^(100)C_(r)` `= (1+x)underset(r=0)overset(n)sum.^(n)C_(r)x^(3r)=underset(r=0)overset(n)sum.^(n)C_(r)x^(3r)+underset(r=0)overset(n)sum.^(n)C_(r)x^(3r+1)` Hence, there will be no term containing `3r+2`. |
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| 194. |
Expand `x(x^(3)- 2/(x^(2)))^(6)` using binomial expansion. |
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Answer» Using the binomial expansion, we get `(x^(3)-2/x^(2))^(6)=.^(6)C_(0)(x^(3))^(6)-.^(6)C_(1)(x^(3))^(5)(2/x^(2))+.^(6)C_(2)(x^(3))^(4)(2/x^(2))^(2)-.^(6)C_(3)(x^(3))^(3)(2/x^(2))^(3) +.^(6)C_(4)(x^(3))^(2)(2/x^(2))^(4)-.^(6)C_(5)x^(3)(2/x^(2))^(5)+.^(6)C_(6)(2/x^(2))^(6)` `=x^(18)-12x^(13)+60 x^(8) 160x^(3) +240/x^(2)_192/x^(7)+64/x^(12)`. |
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| 195. |
Show that the middle term in the expansion of `(1+x)^(2n)i s((1. 3. 5 (2n-1)))/(n !)2^n x^n ,w h e r en`is a positive integer. |
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Answer» Clearly, the number of terms in the expansion of `(1+x)^(2n)" is "(2n+1).` `:. " middle term " =((2n)/2+1)" th term "= (n+1)" th term "= T _(n+1).` Now, `T_(n+1) = .^(2n) C _(n)* x^(n) = ((2n)!)/((n!) xx (n !))* x ^(n) ` `=(1* 2* 3 * 4 ... ( 2n - 2) (2n-1) (2n))/((n!) xx ( n !)) * x^(n)` `=( [ 1 * 3 * 5 ... ( 2n - 3) ( 2n-1)] xx [2* 4* 6 ... ( 2n - 2) (2n) ])/((n!) (n!)) * x^(n) ` `=([1 * 3* 5 ... ( 2n-1)] xx 2 ^(n) xx [ 1* 2 * 3 ... ( n-1) *n])/((n!) xx (n!))* x^(n)` `= ([ 1* 3* 5 ... ( 2n-1)] xx 2^(n) xx x^(n))/((n!))*` Hence , the middle term in the given expansion is `(1* 3* 5 ... ( 2n-1) xx 2^(n) xx x^(n))/((n!))*` |
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| 196. |
find the term independent of `x` in the expansion of `(sqrtx/sqrt3+sqrt3/(2x^2))^10` |
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Answer» Let `T_(r+1)` be independent of x. Here ,`T_(r+1) = .^(10) C_(r) xx (sqrt(x/3))^((10-r)) xx ((sqrt(3))/(2x^(2)))^(r)` `T_(r+1) = .^(10) C _(r) xx (x/3)^(((10-r))/2) xx 3^(r/2) xx 1/(2^(r) x^(2r))` `=.^(10)C_(r) xx x^((5-r/2 -2r)) xx1/(3^((5-r/2))) xx 3^(r/2) xx 2^(-r)` `=.^(10) C_(r) xx x^((5-(5r)/2)) xx3^((r-5)) xx 2 ^(-r)." "`...(i) Now, `T_(r+1)` will be independent of x only when the power of x in it is 0. `:. 5- (5r)/2 = 0 rArr (5r)/2 rArr =5 rArr r = 2 rArr r + 1 = 3 .` Thus, `T_(3)` is independent of x. Putting `r=2` in (i), we get `T_(3) = T _(2+1)` `=.^(10) C _(2) xx 3^((2-5)) xx 2 ^(-2) xx x^(0) = ((10 xx9)/(2 xx 1) xx 3^(-3) xx 2 ^(-2)).` `= (45/(3^(3) xx 2^(2))) = ( 45/(27 xx 4)) = 5 /12.` Hence, the term independent of x in the given expansion is `5/12`. |
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| 197. |
The coefficient of `x^(28)`in the expansion of `(1+x^3-x^6)^(30)`is`1`b. `0`c. `^30 C_6`d. `^30 C_3`A. 1B. 0C. `.^(30)C_(6)`D. `.^(30)C_(3)` |
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Answer» Correct Answer - B `(1+x^(3)-x^(6))^(30)` `= {1+x^(3)(1-x^(3))}^(30)` `=.^(30)C_(0) + .^(30)C_(1)x^(3)(1-x^(3))+.^(30)C_(2)x^(6)(1-x^(3))^(2) + "…."` Obviously, each term will contain `x^(3m), m in N`. But 28 is not divisible by 3. Therefore ther will be no term containing `x^(28)`. |
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| 198. |
If the term free form x in the expansion of `(sqrt(x)-m/(x^(2)))^(10)` is 405, find the value of m. |
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Answer» The general term in the given expansion is given by `T_(r+1) = (-1)^(r) xx .^(10) C_(r) xx (sqrt(x))^((10-r)) xx (m/(x^(2)))^(r)` `=(-1)^(r) xx .^(10) C _(r) xx x^((5-r/2)) xx (m^(r))/(x^(2r))` `= (-1)^(r) xx . ^ (10) C _(r) xx x ^((5-r/2-2r)) xx m ^(r)` `= (-1)^(r) xx.^(10)C_(r) xx x^((5-(5r)/2)) xx m^(r)." "`...(i) Let `T_(r+1)` be free from x. Then, the power of x in `T_(r+1)` must be 0. `:. 5 - (5r)/2 = 0 rArr (5r)/2 = 5 rArr r=2 rArr r+1 = 3 .` So, `T_(3)` will be free from x. Now, `T_(3) = T_(2+1)` `=(-1)^(20 )xx .^(10) C _(2) xx x^(0) xx m ^(2) " "` [putting r=2 in (i) ] `= ( (10 xx 9)/2 xx m^(2)) = 45m^(2).` But, it is given that the term free from x is 405. `:. 45m^(2) = 405 rArr m^(2) = 9 rArrm = pm3.` Hence, `m=pm 3.` |
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| 199. |
Find the middle term in the expansion of `(ax-b/x^(2))^(12)` |
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Answer» The general term in the given expansion is given by `T_(r+1)=(-1)^(r) xx ^12C_(r) xx (ax)^((12-r)) xx (b/x^(2))^(r)` ` rArr T _(r+1)= (-1)^(r) xx ^12C_(r) xxa^((12-r)) xx b^(r) xx x^((12-3r))" " `. Since the given binomial contains an even power, so it has only one middle term. Middle term `=(12/2+1)`th term =7th term. Now, `T_(7)=T_((6+1))` `(-1)^(6) xx ^(12)C_(6) xx a^((12-6)) xx b^(6)xx x^((12-3xx6))" "` [putting r = 6(i) ]. `.= ^(12)C_(6) xxa^(6)b^(6) xxx^(-6)` ` = ((12xx11xx10xx9xx8xx7)/(6xx 5xx4xx 3xx2xx1) xx (a^(6)b^(6))/(x^(6))) xx(924a^(6)b^(6))/x^(6).` Hence, the middle term in the given expansion is `(924a^(6)b^(6))/x^(6)`. |
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| 200. |
The coefficient of `x^(7)` in the expansion of `(1-x-x^(3)+x^(4))^(8)` is equal toA. `-648`B. `792`C. `-792`D. `648` |
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Answer» Correct Answer - C `(1-x-x^(2)+x^(4))^(8)=(1-x)^(8)(1-x^(3))^(8)` `= (1-.^(8)C_(1)x+.^(8)C_(1)x^(2)-.^(8)C_(3)x^(3)+"….")xx(1-.^(8)C_(1)x^(3)+.^(8)C_(2)x^(6)-.^(8)C_(3)x^(9)+"…..")` `:.` Coefficient of `x^(7)= -.^(8)C_(7)-.^(8)C_(1)xx.^(8)C_(4)-.^(8)C_(1) xx.^(8)C_(2)` `= - 792` |
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