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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.
| 201. |
The coefficient of `a^8b^4c^9d^9`in `(a b c+a b d+a c d d+b c d)^(10)`is`10 !`b. `(10 !)/(8!4!9!9!)`c. `2520`d. none of theseA. `10!`B. `(10!)/(8!4!9!9!)`C. `2520`D. none of these |
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Answer» Correct Answer - C `a^(10)b^(10)c^(10)d^(10)(1/a+1/b+1/c+1/d)^(10)` Therefore the required coefficient is equal to the coefficient of `a^(-2)b^(-6)c^(-1)d^(-1)` in `(1/a+1/b+1/c+1/d)^(10)`, which is given by `= (10!)/(2!6!1!1!) = (10 xx 9 xx 8 xx 7)/(2) = 2520` |
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| 202. |
In the expansion of `(7^(1/3)+ 11^(1/9))^6561`, the number of terms free from radicals is:A. there are exactly `730` rational termsB. there are exactly 5832 irrational termsC. the term which involves greatest binomial coefficients is irrationalD. the term which involves greatest binomial coefficients is rational |
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Answer» Correct Answer - A::B::C General term is `.^(6561)C_(r) 7^((6561-r)/(3)) 11^((r)/(9))` To make the term free of radical sign, r should be a multiple of 9. `:. r=0, 9, 18,27,"…..",6561`. Hence, there are 730 rational terms. The greatest binomial coefficients are `.^(6561)C_((6561-1)/(2))` and `.^(6561)C_((6561+11)/(2))` or `.^(6561)C_(3280)` and `.^(6561)C_(3281)`. Now `3280` are `3281` are not a multiple of 3, hence, both terms involving greatest binomial coefficients are irrational . |
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| 203. |
If ` y = (1)/(3) + (1*3)/(3 *6) + (1 * 3*5)/(3*6*9) +…` then the value of `y^(2) + 2y ` isA. 2B. -2C. 0D. none of these |
| Answer» Correct Answer - A | |
| 204. |
If` y=3 x + 6 x^(2) + 10 x^(3) +…` then x =A. `(4)/(3) - (1*4)/(3^(2)*2) y^(2) + (1*4*7)/(3^(2)*3) y^(3)...`B. `(4)/(3) + (1*4)/(3^(2)*2) y^(2) - (1*4*7)/(3^(2)*3) y^(3)...`C. `(4)/(3) + (1*4)/(3^(2)*2) y^(2) + (1*4*7)/(3^(2)*3) y^(3)...`D. none of these |
| Answer» Correct Answer - D | |
| 205. |
If p is nearly equal to q and n `gt` 1 , such that `((n+1) p+(n-1)q)/((n-1)p+(n +1)q) = ((p)/(q))^(k)` , then the value of k, isA. n = 2r is a positive integral mulitple of 3B. `(1)/(n)`C. n+1D. `(1)/(n+1)` |
| Answer» Correct Answer - B | |
| 206. |
If the middle term in the binomial expansion of `(1/x+xsinx^(10))`is equal to `(63)/8,`find the value of `xdot` |
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Answer» `(1/x+xsinx)^10` Middle term in the above expansion `= 10/2 +1 = 6` Then, `T_6 = C(10,5)(1/x)^5(xsinx)^5` `=> 63/8 = (10*9*8*7*6)/(5*4*3*2*1)sin^5x` `=>1/8 = 4sin^5x` `=>sin^5x = 1/32` `=>sinx = 1/2` `=>x = pi/6`General value for `x = npi+(-1)^npi/6, n in Z.` |
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| 207. |
Given that the 4th term in the expansion of `[2+(3//8x)]^(10)`has the maximum numerical value. Then find the range of value of `xdot` |
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Answer» If `T_(r+1)` is the greatest term in `(a+bx)^(n)`, `(r/(n-r+1)) a/b le |x|le ((r+1)/(n-r)) a/b` Given that the fourth terms is in the expansion of `(2+3/(8)x)^(10)`. `rArr (3/(10-3+1)) (2)/((3//8)) le |x|le ((3+1)/(10-3))(2)/((3//8))` `rArr 2 le |x| le 64/21` |
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| 208. |
Find the sum `.^(n)C_(0) + 2 xx .^(n)C_(1) + xx .^(n)C_(2) + "….." + (n+1) xx .^(n)C_(n)`. |
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Answer» Method I : `.^(n)C_(0)+2xx.^(n)C_(1)+3xx.^(n)C_(2)+"...."+(n+1)xx .^(n)C_(n)` `= underset(r=0)overset(n)sum(r+1).^(n)C_(r)` `=underset(r=0)overset(n)sum[r.^(n)C_(r)+.^(n)C_(r)]` `=n underset(r=0)overset(n)sum.^(n-1)C_(r-1)+underset(r=0)overset(n)sum.^(n)C_(r)` `= n(.^(n-1)C_(0) + .^(n-1)C_(1) + .^(n-1)C_(2)+"..."+.^(n-1)C_(n-1)) + (.^(n)C_(0)+.^(n)C_(1)+.^(n)C_(2)+"....."+.^(n)C_(n))` `= n2^(n-1) + 2^(n)` `= (n+2)2^(n-1)` Method II : We have `(1+x)^(n) = .^(n)C_(0)+.^(n)C_(1)x+.^(n)C_(2)x^(2)+"....."+.^(n)C_(n)x^(n)` `:. x(1+x)^(n) = .^(n)C_(0)x+.^(n)C_(1)x^(2)+.^(n)C_(2)x^(3) + "....." + .^(n)C_(n)x^(n+1)` Differentiating w.r.t. x, we get `n(n1+x)^(n-1)x+(1+x)^(n)=.^(n)C_(0)+2xx.^(n)C_(1)x+3xx.^(n)C_(2)x^(2)+"..."+(n+1)xx.^(n)C_(n)x^(n)` Putting `x = 1`, we get `n2^(n-1)+2^(n)=.^(n)C_(0) +2xx.^(n)C_(1)+3xx.^(n)C_(2)+"....."+(n+1)xx.^(n)C_(n)` |
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| 209. |
If the third term in the expansion of `(1+x)^mi s-1/8x^2,`then find the value of `mdot`A. 2B. 43467C. 3D. 4 |
| Answer» Correct Answer - B | |
| 210. |
If the third term in the expansion of `(1+x)^mi s-1/8x^2,`then find the value of `mdot` |
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Answer» Correct Answer - `m = 1/2` We have, `(1+x)^(m) = 1+mx+(m(m+1))/(2!) x^(2) + "……"` Given that the third therm is `-(1//8)^(2)`, hence `(m(m-1))/(2)x^(2) = =1/8x^(2)` or `4m^(2) - 4m = -1` or `(2m-1)^(2) = 0` or `m =1/2` |
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| 211. |
Find the coefficient of `x^(-25)` in the expansion of `((x^(2))/(2)-(3)/(x^(3)))^(15)`A. `(-1365)/(16)xx3^(11)`B. `(1365)/(16)xx3^(11)`C. `(-16)/(1365)xx3^(11)`D. None of these |
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Answer» Correct Answer - A N/a |
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| 212. |
No. of terms in the expansion of `(1+3x+3x^(2)+x^(3))^(10)` is:A. 31B. 32C. 10D. 11 |
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Answer» Correct Answer - A N/a |
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| 213. |
Find the value of `sum_(p=1)^(n) (underset(m=p)overset(n)sum""^(n)C_(m)""^(m)C_(p))`. And hence, find the value of `lim_(nrarroo) (1)/(3^(n)) sum_(m=p)^(n) ""^(n)C_(m)""^(m)C_(p))`. |
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Answer» Correct Answer - `(3^(n) - 2^(n))` and `1` `S = underset(p=1)overset(n)sum(underset(m=p)overset(n)sum.^(n)C_(m).^(m)C_(p))` `= underset(p=1)overset(n)sum(.^(n)C_(p).^(p)C_(p)+.^(n)C_(p+1).^(p)+1)C_(p) + "....."+.^(n)C_(n).^(n)C_(p))` `= underset(p=1)overset(n)sum` [coefficient of `x^(p)` in `{.^(n)C_(p)(1+x)^(p)+.^(n)C_(p+1)(1+x)^(p+1)+"....."+.^(n)C_(n)(1+x)^(n)}]` `= underset(p=1)overset(n)sum` [Coefficeint of `x^(p)` in `{.^(n)C_(0)+.^(n)C_(1)(1+x)+.^(n)C_(2)(1+x)^(2)+"....."+.^(n)C_(p-1)(1+x)^(p-1)+.^(n)C_(p)(1+x)^(p)+"....."+.^(n)C_(n)(1+x)^(n)}]` `= underset(p=1)oversetg(n)sum` [coefficient of `x^(p)` in `{1+(1+x)}^(n)`] `= underset(p=1)overset(n)sum` [coefficient of `x^(p)` in `(2+x)^(n)`] `= underset(p=1)overset(n)sum[.^(n)C_(p)2^(n-p)]` ` = .^(n)C_(1)2^(n-1)+.^(n)C_(2)2^(n-2)+"...".^(n)C_(n)` `= .^(n)C_(0)2^(n)+.^(n)C_(1)2^(n-1)+.^(n)C_(2)2^(n-2)+"...."+.^(n)C_(n)-.^(n)C_(0)2^(n)` `= (1+2)^(n)-2^(n)` `= 3^(n) - 2^(n)` Alternate solution : `underset(p=1)overset(n)sum(underset(m=p)overset(n)sum.^(n)C_(m).^(m)C_(p))= underset(p=1)overset(n)sum(underset(m=p)overset(n)sum(n!)/((n-m)!m!)(m!)/((m-p)!p!))` `= underset(p=1)overset(n)sum(underset(m=p)overset(n)sum(n!)/((n-m)!)(1)/((m-p)!p!))` `=underset(p=1)overset(n)sum(n!)/((n-p)!p!)(underset(m=p)overset(n)sum((n-p)!)/((n-m)!(m-p)!))` `= underset(p=1)overset(n)sum.^(n)C_(P)(underset(m=p)overset(n)sum.^(n-p)C_(m-p))` `= underset(p=1)overset(n)sum .^(n)C_(p)2^(n-p)` `= .^(n)C_(1)2^(n-1)+.^(n)C_(2)2^(n-2)+"....."+.^(n)C_(n)2^(0)` `= (2+1)^(n)-.^(n)C_(0)2^(n)` `= 3^(n) - 2^(n)` Also, `underset(nrarroo)"lim"1/(3^(n)) underset(p=1)overset(n)sum(overset(n)underset(m=p)sum.^(n)C_(m).^(m)C_(p))= underset(nrarroo)"lim"(3^(n)-2^(n))/(3^(n)) = 1` |
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| 214. |
15th term in the expansion of `(sqrt(2)-sqrt(y)^(17)` is :A. `860x^(3//2)y^(7)`B. `680x^(7)y^(3//2)`C. `680x^(3//2)y^(7)`D. `860x^(3)y^(7//2)` |
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Answer» Correct Answer - C N/a |
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| 215. |
Prove that `.^(n)C_(1) + 2 xx .^(n)C_(2) + 3 xx .^(n)C_(3) + "…." + n xx .^(n)C_(n) = n2^(n-1)`. Hence, prove that `.^(n)C_(1).(.^(n)C_(2))^(2).(.^(n)C_(3))^(3)"......."(.^(n)C_(n))^(n) le ((2^(n))/(n+1))^(.^(n+1)C_(2)) AA n in N`. |
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Answer» Method I : `.^(n)C_(1) + 2.^(n)C_(2)+3.^(n)C_(3)+"….."+n.^(n)C_(n)` `=underset(r=1)overset(n)sumr.^(n)C_(r)` `= n underset(r=1)overset(n)sum.^(n-1)C_(r-1)` `= n(.^(n-1)C_(0) + .^(n-1)C_(1) + .^(n-1)C_(2) + .^(n-1)C_(3)+"…."+.^(n-1)C_(n-1))` `= n2^(n-1)` Method II : We have `(1+x)^(n) = .^(n)C_(0) + .^(n)C_(1)x+.^(n)C_(2)x^(2)+"...."+.^(n)C_(n)x^(n)` Differentiating w.r.t x, we get `n(1+x)^(n-1)= .^(n)C_(1)+2 xx .^(n)C_(2)x + 3 xx .^(n)C_(3)x^(2) +"......" + n xx .^(n)C_(n)x^(n-1)` Putting `x = 1`, we get `n2^( n-1) = .^(n)C_(1)+2xx .^(n)C_(2) + "....." + n xx .^(n)C_(n)` Using `A.M. ge G.M.`, we get `(.^(n)C_(1)+2.^(n)C_(2)+3.^(n)C_(3)+".....+n.^(n)C_(n))/((n(n+1))/(2))` `[(.^(n)C_(1))(.^(n)C_(2))^(2)(.^(n)C_(3))^(3)"....."(.^(n)C_(n))^(n)]^((1)/((pi(n-1))/2))` `rArr (n2^(n-1))/((n(n+1))/(2))ge [(.^(n)C_(1))(.^(n)C_(2))^(2)(.^(n)C_(3))^(3)"....."(.^(n)C_(n))^(n)]^((2)/(n(n+1)))` `(.^(n)C_(1))(.^(n)C_(2))^(2)(.^(n)C_(3))^(3)"....."(.^(n)C_(n))^(n)le((2^(n))/(n+1))^((n(n+1))/(2))` or `(.^(n)C_(1))(.^(n)C_(2))^(2)(.^(n)C_(3))^(3)"....."(.^(n)C_(n))^(n)le((2^(n))/(n+1))^(.^(n+1)C_(2))` |
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| 216. |
If the constant term in the binomial expansion of `(x^2-1/x)^n ,n in N`is 15, then the value of `n`is equal to.A. `6`B. `9`C. `12`D. `15` |
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Answer» Correct Answer - A `(a)` For `(x^(2)-(1)/(x))^(n)` `T_(r+1)=^(n)C_(r )(x^(2))^(n-r)(-1)^(r )x^(-r)=^(n)C_(r )x^(2n-3r)(-1)^(r )` Constant term `=^(n)C_(r )(-1)^(r )` if `2n=3r` i.e. coefficient of `x=0` Hence `.^(n)C_(2n//3)(-1)^(2n//3)=15=^(6)C_(4)impliesn=6` |
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| 217. |
If `p^(4)+q^(3)=2(p gt 0, q gt 0)`, then the maximum value of term independent of `x` in the expansion of `(px^((1)/(12))+qx^(-(1)/(9)))^(14)` isA. `"^(14)C_(4)`B. `"^(14)C_(6)`C. `"^(14)C_(7)`D. `"^(14)C_(12)` |
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Answer» Correct Answer - B `(b)` `(px^((1)/(12))+qx^(-(1)/(9)))^(14)` General term `T_(r+1)=14C_(r )(px^((1)/(12)))^(14-r)(qx^((-1)/(9)))^(r )` `=^(14)C_(r )p^(14-r)q^(r )x^((14-r)/(12)-(r )/(9))` Term is independent of `r`, then `(14-r)/(12)-(r )/(9)=0` `:.r=6` `:.` Term independent of `x` is `"^(14)C_(5)p^(8)q^(6)=^(14)C_(6)(p^(4)q^(3))^(2)` Now `p^(4)`, `q^(3)` are positive Using `AM ge GM` `(p^(4)+q^(3))/(2) ge (p^(4)q^(3))^(1//2)implies(p^(4)q^(3))^(2) le 1` `implies` Maximum value of term independent of `x` is `"^(14)C_(6)`. |
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| 218. |
Find the coefficient of `t^8` in the expansion of `(1+2t^2-t^3)^9`.A. `1680`B. `2140`C. `2520`D. `2730` |
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Answer» Correct Answer - C `(c )` `((1+2t^(2))-t^(3))^(9)` `="^(9)C_(0)(1+2t^(2))^(9)-^(9)C_(1)(1+2t^(2))^(8)*t^(3)+^(9)C_(2)(1+2t^(2))^(7)*t^(6)-^(9)C_(3)(1+2t^(2))^(6)*t^(9)+.....-^(9)C_(9)(t^(3))^(9)` `:.` Coefficient of `t^(8)` in the expansion of `(1+2t^(2)-t^(3))^(9)` `=^(9)C_(0)("coefficient of" t^(8) "in" (1+2t^(2))^(9)) -^(9)C_(1)("coefficient of" t^(5) "in" (1+2t^(2))^(8))+^(9)C_(2)("coefficient of"t^(2) "in" (1+2t^(2))^(7))` `=^(9)C_(9)*^(9)C_(4)2^(4)-0+^(9)C_(2)*^(7)C_(1)*2` `=2520` |
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| 219. |
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. `25`B. `20`C. `15`D. None of these |
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Answer» Correct Answer - C `(c )` `((1)/(x^(2))-x^(3))^(n)` `T_(r+1)=(n!)/(r!(n-r)!)(-1)^(n-r)x^(5r-2n)` If `5r-2n=5`, then `5r=2n+5` `r=(2n)/(5)+1` If `5r-2n=10`, then `5r=2n+10` `r=(2n)/(5)+2` Let `n=5k` According to question `(5k!)/((2k+1)!(3k-1)!)-(5k!)/((2k+2)!(3k-2)!)=0` `(1)/(3k-1)-(1)/(2k+2)=0` `k=3`, `n=15` |
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| 220. |
If in the expansion of `(1 + ax)^(n),n in `N, the coefficient of x and `x^(2) ` are 8 and 24 respectively, thenA. a = 2, n=4B. a = 4, n=2C. a = 2, n=6D. a = -2, n=4 |
| Answer» Correct Answer - A | |
| 221. |
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. 5B. 10C. 15D. 20 |
| Answer» Correct Answer - C | |
| 222. |
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. 25B. 20C. 15D. none of these |
| Answer» Correct Answer - C | |
| 223. |
The sum of the magnitudes of the coefficients in the expansion of `(1 - x + x^(2) - x^(3))^(n)` ,is |
| Answer» Correct Answer - D | |
| 224. |
The sum of the coefficients in `(1+x+3x^2)^2143` is (A) `2^2143` (B) 0 (C) 1 (D) -1A. -1B. 1C. 0D. none of these |
| Answer» Correct Answer - a | |
| 225. |
Find the term independent of `x`in the expansion of `(1+x+2x^3)[(3x^2//2)-(1//3)]^9` |
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Answer» Correct Answer - `17/54` In the expansion of `E = ( 3/2x^(2) - 1/ (3x))^(9)` , we have `T_(r+1) = (-1)^(r)*^(9)C_(r)*(3/2x^(2))(9-r) (1/(3x))^(r)` `rArr T_(r+1) = (-1)^(r)*^(9)C_(r)*(3(9-2r))/(2^((9-r)))* x^((18-3r)).` `(1+ x + 2x^(3)) [(a_(0) xx 1/x^(3)+a_(1) xx 1/x + a^(2))" from E"]` `=(1+x+2x^(3)) [ {(-1)^(7)*^(9)C_(7)*3^(-5)/2^(2) xx1/(x^(3))}+ {(-1)^(6)*^(9)C_(6) * 3^(-3)/2^(3)xxx^(0)}]` `[(x=-1 rArr18-3r = -1 rArr " r is faction"),(18-3r=0 rArr r=6 and 18-3r = -3 rArr r=7)]` `=(1+ x+ 2x^(3))[(-1)/(27x^(3))+7/18]` `:. " required term " ((-2)/27+7/18) = 17/54.` |
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| 226. |
If the sum of the coefficient in the expansion of `(alpha^2x^2-2alphax+1)^(51)`vanishes, then find the value of`alpha`A. 2B. -1C. 1D. -2 |
| Answer» Correct Answer - C | |
| 227. |
The ratio of the coefficient of `x^(15)` to the term independent of x in the expansion of `(X^(2)+(2)/(x))^(15)` is |
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Answer» Correct Answer - `1:32` `T_(r+1) = .^(15) C (x^(2)) ^(15-r) xx (2/x)^(r) = .^(15) C _(r) xx 2 ^(r) xx x^(30-3r)` `[30 - 3r = 15 rArr 3r = 15 rArr r = 5 ] and [ 30 - 3r = 0 rArr 3r = 30 rArr r = 10].` Find ` (T_(6) : T _(11) ) = ( T _ ( 5+ 1) : T _ ( 10+1)).` |
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| 228. |
The number of terms in the expansion of `(x + y + z)^(10)`, isA. 11B. 33C. 66D. 1000 |
| Answer» Correct Answer - C | |
| 229. |
If the sum of the coefficient in the expansion of` (alpha x^(2) - 2x + 1)^(35)` is equal to the sum of the coefficient of the expansion of `(x - alpha y)^(35)`, then `alpha` = |
| Answer» Correct Answer - B | |
| 230. |
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//4`B. `1//16`C. `1//32`D. `1//32` |
| Answer» Correct Answer - C | |
| 231. |
The range of values of the term independent of x in the expansion of `(x sin^(-1) alpha + (cos^(-1)alpha)/(x))^(10), a in [-1,1]`isA. `[-(""^(10)C_(5) pi^(10))/(2^(5)),(""^(10)C_(5) pi^(10))/(2^(20))]`B. `[(""^(10)C_(5) pi^(2))/(2^(20)),(""^(10)C_(5) pi^(2))/(2^(5))]`C. `[1,2]`D. `(1,2)` |
| Answer» Correct Answer - A | |
| 232. |
The term independent of x in the expansion of `(1 - x)^(2) (x + (1)/(x))^10`, isA. `""^(11)C_(5)`B. `""^(10)C_(5)`C. `""^(10)C_(4)`D. none of these |
| Answer» Correct Answer - A | |
| 233. |
The number of terms with integral coefficients in the expansion of `(17^(1/3)+35^(1/2)x)^(600)` is (A) `100` (B) `50` (C) `150` (D) `101`A. 100B. 50C. 101D. none of these |
| Answer» Correct Answer - C | |
| 234. |
Find the coefficient of `x^(50)`in the expansion of `(1+x)^(101)xx(1-x+x^2)^(100)dot` |
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Answer» `(1+x)^(101)(1-x+x^(2))^(100)` `= (1+x)[(1+x)^(100) (1-x+x^(2))^(100)]` `= (1+x)(1-x^(3))^(100)` `= (1-x^(3))^(100) + x(1-x^(3))^(100)` `= (1-x^(3))^(100) + x(1-x^(3))^(100)` Now, coefficient of `x^(50)` in `[(1-x^(3))^(100) + x(1-x^(3))^(100)]` `=` Coefficient of `x^(50)` in `(1-x^(3))^(100)` + Coefficient of `x^(49)` in `(1-x^(3))^(100)` `= 0` (as 49 and 50 are not a multiple of 3) |
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| 235. |
Find the coefficient of `x^k` in `1 +(1 +x) +(1 +x)^2+.. +(1+x)^n (0 |
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Answer» The expansion being in G.P., we have `E = 1 + (1+x) + (1+x)^(2)+"…….."+(1+x)^(n)` `= ((1+x)^(n+1)-1)/((1+x)-1) = x^(-1)[(1+x)^(n+1)-1]` Therefore, the coefficient of `x^(k)` in E is equal to the coefficient of `x^(k+1)` in`[(1+x)^(n+1)-1]`, which is given by ` .^(n+1)C_(k+1)`. |
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| 236. |
Find the coefficient of `x^(20)`in `(x^2+2+1/(x^2))^(-5)(1+x^2)^(40)dot` |
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Answer» Given expansion is `E = ((x+(1)/(x))^(2))^(-5) (1+x^(2))^(40)` `= ((x^(2) + 1)/(x))^(-10) (1+x^(2))^(40) = x^(10)(1+x^(2))^(30)` Coefficient of `x^(20)` in E = Coefficient of `x^(10)` in `(1+x^(2))^(30)` `= .^(30)C_(5)` |
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| 237. |
Find the coefficient of `x^(13)`in the expansion of `(1-x)^5xx(1+x+x^2+x^2)^4dot` |
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Answer» `E = (1-x)^(5)(1-+x)^(4) (1+x^(2))^(4)` `= (1-x)(1-x^(2))^(4)(1+x^(2))^(4)` ` = (1-x)(1-x^(4))^(4)` ` = (1-x)[1-4(x^(4))+6(x^(4))^(2) - 4(x^(4))^(3)+(x^(4))^(4)]` Coefficient of `x^(13)` is `(-1)(-4) = 4` |
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| 238. |
The coefficients of three consecutive terms of `(1+x)^(n+5)`are in the ratio 5:10:14. Then `n=`___________. |
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Answer» Let the three consecutive terms in `(1+x)^(n+5)` be `t_(r), t_(r+1),t_(r+2)` having coefficient `""^(n+5)C_(r-1) , ""^(n+5)C_(r), ""^(n+5)C_(r+1)` Given , `""^(n+5)C_(r-1) ,:""^(n+5)C_(r) : ""^(n+5)C_(r+1) =5:10:14` `:. (""^(n+5)C_(r))/(""^(n+5)C_(r-1))=(10)/(5) and (""^(n+5)C_(r+1))/(""^(n+5)C_(r))=(14)/(10) ` `implies (n+5-(r-1))/(r)=2 and (n-r+5)/(r+1)=(7)/(5)` `implies n-r+6 =2r` ` and 5n-5r+25=7r+7` ` implies n+6 =3r and 5n+18=12r ` `:. (n+6)/(3)=(5n+18)/(12)` `implies 4n+24 =5n+ 18implies n=6 ` |
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| 239. |
If `a_1,a_2, a_3, a_4`be the coefficient of four consecutive terms in the expansion of `(1+x)^n ,`then prove that: `(a_1)/(a_1+a_2)+(a_3)/(a_3+a_4)=(2a_2)/(a_2+a_3)dot` |
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Answer» First four coefficient terms of `(1+x)^n` are, `1, C(n,1), C(n,2),C(n,3)` `:. a_1 = 1` `a_2 = n` `a_3 = (n(n-1))/2` `a_4 = (n(n-1)(n-2))/6` Now, `L.H.S. = a_1/(a_1+a_2) + a_3/(a_3+a_4)` `=1/(n+1) +( (n(n-1))/2)/( (n(n-1))/2+ (n(n-1)(n-2))/6)` `=1/(n+1) +(1/(1+((n-2))/3))` `=1/(n+1)+3/(n+1) = 4/(n+1)` Now, `R.H.S. = (2a_2)/(a_2+a_3)` `=(2n)/(n+(n(n-1))/2)` `=(2*2)/(2+n-1)` `=4/(n+1)` `:. L.H.S. = R.H.S.` |
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| 240. |
If `a_1,a_2, a_3, a_4`be the coefficient of four consecutive terms in the expansion of `(1+x)^n ,`then prove that: `(a_1)/(a_1+a_2)+(a_3)/(a_3+a_4)=(2a_2)/(a_2+a_3)dot`A. `P + q = r`B. `(1 )/(2) (a_(2))/(a_(2) + a_(3))`C. `(2 a_(2))/(a_(2) +a_(3))`D. `(2 a_(3))/(a_(2) +a_(3))` |
| Answer» Correct Answer - c | |
| 241. |
Find the last three digits in `11^(50)` |
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Answer» Expansion of `(10 +1)^(50)=""^(50)C_010^(50)+""^(50)C_(1)10^(49)+…..+""^(50)C_(48)10^2+""^(50)C_(49)10+""^(50)C_(50)` `= ubrace(""^(50)C_(0)10^(50)+""^(50)C_(1)10^(49)+.......+""^(50)C_(47)10^(3))_("1000k")+49xx25xx100+500+1` `rArr 1000k +123001` `rArr` Last 3 digits are 001 |
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| 242. |
If the expansion of `(x - (1)/(x^(2)))^(2n)` contains a term independent of x, then `n` is a multiple of 2. |
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Answer» Given Binomial expansion is `(x - (1)/(x^(2)))^(2n)` Let `T_(r + 1)` term is independent of `x` Then, `T_(r + 1) = .^(2n)C_(r) (x)^(2n - r) (-(1)/(x^(2)))^(r)` `= .^(2n)C_(r) x^(2n - r) (-1)^(r) x^(-2r) = .^(2n)C_(r) x^(2n - 3r) (-1)^(r)` For independent of `x` `2n - 3 r = 0` `:. r = (2n)/(3)` Which is not a integer So, the given expansion is not possible |
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| 243. |
Find the last three digits of the number `27^(27)dot` |
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Answer» Correct Answer - Last three digits are 803 `27^(27) = 3(10-1)^(40)` `= 3(10^(40) - .^(40)C_(1) .10^(39) +"….." + .^(40)C_(38).10^(2)-.^(40)C_(39).10+1)` `= (1000 lambda - 400 + 1)` `= 3(1000 lambda - 399)` Therefore, last 3 digits of this number of `803`. |
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| 244. |
Write last two digits of the number `3^(400)dot` |
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Answer» Correct Answer - 1 Given that, `3^(400) = g^(200) = (10 - 1)^(200)` `rArr (10 - 1)^(200) = .^(200)C_(n) 10^(200) - .^(200)C_(1) 10^(199) + .... - .^(200)C_(199) 10^(1) + .^(200)C_(200) 1^(200)` `rArr (10 - 1)^(200) = 10^(200) - 200 xx 10^(199) + ...... - 10 xx 200 + 1` So, it is clear that the last two digits are 01. |
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| 245. |
Find (i) the last digit, (ii) the last two digits, and (iii) the lastthree digits of `17^(256)dot` |
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Answer» We have `17^(256) = (17^(2))^(128) = (289)^(128) = (290-1)^(128)` `:. 17^(256) = .^(128)C_(0)(290)^(128)-.^(128)C_(1)(290)^(127)+.^(128)C_(2)(290)^(126)-"....."` `- .^(128)C_(125)(290)^(3)+.^(128)C_(126)(290)^(2)-.^(128)C_(127)(290)+1` `=[.^(128)C_(0)(290)^(128) - .^(128)C_(1)(290)^(127)+.^(128)C_(2)(290)^(126)-"....."` `-.^(128)C_(125)(290)^(3)]+.^(128)C_(126)(290)^(2) -.^(128)C_(127)(290)+1` `=1000m+((128)(127))/(2)(290)^(2)-128xx290+1` `= 1000 m + (128)(127)(290)(145)-128xx290-1` `= 1000m+(128)(290)(127xx145-1)+1` `=1000m+(128)(290)(18414)+1` `=1000m+683527680+1` `= 1000m+683527000+680+1` `= 1000(m+683527)+681` Hence, the last three digits of `17^(256)` are `6,8,1`. As a result, the last two digits of `17^(256)` are `8, 1` and the last digit of `17^(256)` is 1. |
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| 246. |
`" if" C_(0)C_(1)C_(2),……C_(n)` are the binomial coefficients in the expansion of `(1+x)^(n)` then prove that: `C_(0)C_(2)+C_(1)C_(3)+C_(2)C_(4)+……+C_(n-2)C_(n)=(|ul2n)/(|uln-2|uln+2)`A. `((2n)!)/((n-2)!(n+2)!)`B. `((2n)!)/(((n-2)!)^(2))`C. `((2n)!)/(((n+2)!)^(2))`D. none of these |
| Answer» Correct Answer - A | |
| 247. |
In the binomial expansion of `(1+x)^n`, coefficients of the fifth, sixth and seventh terms are in A.P. find all the values of `n`for which this can happen. |
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Answer» Coefficients of fifth, sixth and seventh terms in `(1+x)^n` are `C(n,4),C(n,5) and C(n,6).` As, they are in `A.P.`, `:. 2**C(n,5) = C(n,4)+C(n,6)` `=>2*(n!)/(5!(n-5!)) = (n!)/(4!(n-4!)) + (n!)/(6!(n-6!))` `=>2*1/(5(n-5)) = 1/((n-4)(n-5)) + 1/(6*5)` `=>2/(5(n-5)) = 1/((n-4)(n-5)) + 1/30` `=>2/(5(n-5)) = (30+(n-4)(n-5))/(30(n-4)(n-5))` `=>12(n-4) = 30+(n^2+20-9n)` `=>12n-48 = n^2+50 -9n` `=>n^2-21n+98 =0` `=>n^2-14n-7n+98 =0` `=>(n-14)(n-7) = 0` `=> n = 14 or n = 7` So, `n` can be `14` and `7` for the given expansion. |
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| 248. |
If n be a positive integer and `P_n` denotes the product of the binomial coefficients in the expansion of `(1 +x)^n`, prove that `(P_(n+1))/P_n=(n+1)^n/(n!)`. |
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Answer» `P_(n+1)/P_n=((n+1)C_0*(n+1)C_1*...(n+1)C_(n+1))/(nC_0*nC_1*...nC_n` `P_n=n/(1!)*(n(n-1))/(2!)*...(n!)/(n!)` `=(n^n(n-1)^(n-1)(n-1)^(n-2)...(33)^1)/(1!*2!*3!...n!` `P(n+1)=(n+1)/1*(n(n+1))/(2!)*(n(n+1)(n-1))/(3!)...(n+1)/(n+1)` `=((n+1)^(n+1)*(n)^n*(n-1)^(n-1)...1)/(1!*2!*3!...(n+1)!)` `P(n+1)/P_n=((n+1)^(n+1)/((n+1)!))` `=(n+1)^n/(n!)`. |
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| 249. |
If the seventh term from the beginning and end in the binomialexpansion of `(2 3+1/(3 3))^n ,""`are equal, find `ndot` |
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Answer» Given expansions is `(root(3)(2) + (1)/(root(3)(3)))^(n)` `:. T_(7) = T_(6 + 1) = .^(n)C_(6) (root(3)(2))^(n - 6) ((1)/(root(3)(3)))^(6)` Since, `T_(7)` from end is same as the `T_(7)` from beginning of `((1)/(root(3)(3))) + ((1)/(root(3)(3)) + root(3)(2))^(n)` Then, `T_(7) = .^(n)C_(6) ((1)/(root(3)(3)))^(n - 6) (root(3)(2))^(6)` Given, that, `.^(n)C_(6)(2)^((n -6)/(3)) (3)^(-6//3) = .^(n)C_(6) (3)^(-((n - 6))/(3)) 2^(6//3)` `rArr (2)^((n - 12)/(3)) = ((1)/(3^(1//3)))^(n - 12)` Which is true, when `(n - 12)/(3) = 0` `rArr n - 12 0 rArr n = 12` |
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| 250. |
Given positive integers `r>1,n> 2, n` being even and the coefficient of `(3r)th` term and `(r+ 2)th` term in the expansion of `(1 +x)^(2n)` are equal; find rA. 3rB. 3r+1C. 2rD. 2r+1 |
| Answer» Correct Answer - C | |