Practice Start Test

Binomial Theorem

JEE Mathematics · 151 questions · Page 6 of 16 · Click an option or "Show Solution" to reveal answer

Q51
If 12 . 310+122 . 39+  .....  +  1210 . 3=K210 . 310{1 \over {2\,.\,{3^{10}}}} + {1 \over {{2^2}\,.\,{3^9}}} + \,\,.....\,\, + \,\,{1 \over {{2^{10}}\,.\,3}} = {K \over {{2^{10}}\,.\,{3^{10}}}}, then the remainder when K is divided by 6 is :
A 1
B 2
C 3
D 5
Correct Answer
Option D
Solution
12 . 310+122 . 39+  ....  +  1210 . 3=K210 . 310{1 \over {2\,.\,{3^{10}}}} + {1 \over {{2^2}\,.\,{3^9}}} + \,\,....\,\, + \,\,{1 \over {{2^{10}}\,.\,3}} = {K \over {{2^{10}}\,.\,{3^{10}}}}
⇒12 . 310[(32)10−132−1]=K210 . 310\Rightarrow {1 \over {2\,.\,{3^{10}}}}\left[ {{{{{\left( {{3 \over 2}} \right)}^{10}} - 1} \over {{3 \over 2} - 1}}} \right] = {K \over {{2^{10}}\,.\,{3^{10}}}}
=310−210210 . 310=K210 . 310⇒K=310−210= {{{3^{10}} - {2^{10}}} \over {{2^{10}}\,.\,{3^{10}}}} = {K \over {{2^{10}}\,.\,{3^{10}}}} \Rightarrow K = {3^{10}} - {2^{10}}

Now

K=(1+2)10−210K = {(1 + 2)^{10}} - {2^{10}}
=10C0+10C12+10C223+  ....  +  10C10210−210= {}^{10}{C_0} + {}^{10}{C_1}2 + {}^{10}{C_2}{2^3} + \,\,....\,\, + \,\,{}^{10}{C_{10}}{2^{10}} - {2^{10}}
=10C0+10C12+6λ+10C9 . 29= {}^{10}{C_0} + {}^{10}{C_1}2 + 6\lambda + {}^{10}{C_9}\,.\,{2^9}
=1+20+5120+6λ= 1 + 20 + 5120 + 6\lambda
=5136+6λ+5= 5136 + 6\lambda + 5
=6μ+5= 6\mu + 5
λ, μ∈N\lambda ,\,\mu \in N

∴\therefore remainder = 5

Q52
The remainder when 32022 is divided by 5 is :
A 1
B 2
C 3
D 4
Correct Answer
Option D
Solution
32022{3^{2022}}
=(32)1011= {({3^2})^{1011}}
=(9)1011= {(9)^{1011}}
=(10−1)1011= {(10 - 1)^{1011}}
=1011C0(10)1011+  .....  +  1011C1010 . (10)1−1011C1011= {}^{1011}{C_0}{(10)^{1011}} + \,\,.....\,\, + \,\,{}^{1011}{C_{1010}}\,.\,{(10)^1} - {}^{1011}{C_{1011}}
=10[1011C0(10)1010+  ......  +  1011C1010]−1= 10\left[ {{}^{1011}{C_0}{{(10)}^{1010}} + \,\,......\,\, + \,\,{}^{1011}{C_{1010}}} \right] - 1
=10 K−1= 10\,K - 1

[As

10[1011C0 . (10)1010+  ......  +  1011C1010]10\left[ {{}^{1011}{C_0}\,.\,{{(10)}^{1010}} + \,\,......\,\, + \,\,{}^{1011}{C_{1010}}} \right]

is multiple of 10]

=10K+5−5−1= 10K + 5 - 5 - 1
=10K−5+5−1= 10K - 5 + 5 - 1
=5(2K−1)+4= 5(2K - 1) + 4

∴\therefore Unit digit = 4 when divided by 5.

Q53
For two positive real numbers a and b such that 1a2+1b3=4{1 \over {{a^2}}} + {1 \over {{b^3}}} = 4, then minimum value of the constant term in the expansion of (ax18+bx−112)10{\left( {a{x^{{1 \over 8}}} + b{x^{ - {1 \over {12}}}}} \right)^{10}} is :
A 1052{{105} \over 2}
B 1054{{105} \over 4}
C 1058{{105} \over 8}
D 10516{{105} \over 16}
Correct Answer
Option C
Solution

Given, Binomial expansion,

(ax18+bx−112)10{\left( {a{x^{{1 \over 8}}} + b{x^{ - {1 \over {12}}}}} \right)^{10}}

General term,

Tr+1=10Cr . (ax18)10−r . (bx−112)r{T_{r + 1}} = {}^{10}{C_r}\,.\,{\left( {a{x^{{1 \over 8}}}} \right)^{10 - r}}\,.\,{\left( {b{x^{ - {1 \over {12}}}}} \right)^r}
=10Cr . a10−r . br . x10−r8 . x−r12= {}^{10}{C_r}\,.\,{a^{10 - r}}\,.\,{b^r}\,.\,{x^{{{10 - r} \over 8}}}\,.\,{x^{ - {r \over {12}}}}
=10Cr . a10−r . br . x(10−r8−r12)= {}^{10}{C_r}\,.\,{a^{10 - r}}\,.\,{b^r}\,.\,{x^{\left( {{{10 - r} \over 8} - {r \over {12}}} \right)}}
=10Cr . a10−r . br . x30−3r−2r24= {}^{10}{C_r}\,.\,{a^{10 - r}}\,.\,{b^r}\,.\,{x^{{{30 - 3r - 2r} \over {24}}}}
=10Cr . a10−r . br . x30−5r24= {}^{10}{C_r}\,.\,{a^{10 - r}}\,.\,{b^r}\,.\,{x^{{{30 - 5r} \over {24}}}}

For constant term,

30−5r24=0{{30 - 5r} \over {24}} = 0
⇒r=6\Rightarrow r = 6

∴\therefore Constant term,

Tr+1=T6+1=10C6 . a4 . b6{T_{r + 1}} = {T_{6 + 1}} = {}^{10}{C_6}\,.\,{a^4}\,.\,{b^6}
=10!6! 4!a4 . b6= {{10!} \over {6!\,4!}}{a^4}\,.\,{b^6}
=10×9×8×74×3×2×1 . a4 . b6= {{10 \times 9 \times 8 \times 7} \over {4 \times 3 \times 2 \times 1}}\,.\,{a^4}\,.\,{b^6}
=210a4b6= 210{a^4}{b^6}

We know,

GM≥HMGM \ge HM

For terms a2 and b3,

a2b3≥21a2+1b3\sqrt {{a^2}{b^3}} \ge {2 \over {{1 \over {{a^2}}} + {1 \over {{b^3}}}}}
⇒a2b3≥24\Rightarrow \sqrt {{a^2}{b^3}} \ge {2 \over 4}
⇒a2b3≥14\Rightarrow {a^2}{b^3} \ge {1 \over 4}
⇒(a2b3)2≥116\Rightarrow {({a^2}{b^3})^2} \ge {1 \over {16}}

∴\therefore

a4b6≥116{a^4}{b^6} \ge {1 \over {16}}

∴\therefore Minimum value of

a4b6=116{a^4}{b^6} = {1 \over {16}}

∴\therefore Minimum value of constant term

T7=210×a4b6{T_7} = 210 \times {a^4}{b^6}
=210×116= 210 \times {1 \over {16}}
=1058= {{105} \over 8}
Q54
The remainder when (11)1011+(1011)11(11)^{1011}+(1011)^{11} is divided by 9 is
A 1
B 4
C 6
D 8
Correct Answer
Option D
Solution
Re((11)1011+(1011)119)=Re(21011+3119){\mathop{\rm Re}\nolimits} \left( {{{{{(11)}^{1011}} + {{(1011)}^{11}}} \over 9}} \right) = {\mathop{\rm Re}\nolimits} \left( {{{{2^{1011}} + {3^{11}}} \over 9}} \right)

For

Re(210119){\mathop{\rm Re}\nolimits} \left( {{{{2^{1011}}} \over 9}} \right)
21011=(9−1)337=337C09337(−1)0+337C19336(−1)1+337C29335(−1)2+  .....  +  337C33790(−1)337{2^{1011}} = {(9 - 1)^{337}} = {}^{337}{C_0}{9^{337}}{( - 1)^0} + {}^{337}{C_1}{9^{336}}{( - 1)^1} + {}^{337}{C_2}{9^{335}}{( - 1)^2} + \,\,.....\,\, + \,\,{}^{337}{C_{337}}{9^0}{( - 1)^{337}}

So, remainder is 8 and

Re(3119)=0{\mathop{\rm Re}\nolimits} \left( {{{{3^{11}}} \over 9}} \right) = 0

So, remainder is 8

Q55
∑i,j=0i≠jnnCi nCj\sum\limits_{\begin{array}{ll}{i,j = 0} \\ {i \ne j} \end{array}} ^n {{}^n{C_i}\,{}^n{C_j}} is equal to
A 22n−2nCn2^{2 n}-{ }^{2 n} C_{n}
B 22n−1−2n−1Cn−1{2^{2n - 1}} - {}^{2n - 1}{C_{n - 1}}
C 22n−122nCn2^{2 n}-\dfrac{1}{2}{ }^{2 n} C_{n}
D 22n−1+2n−1Cn{2^{2n - 1}} + {}^{2n - 1}{C_n}
Correct Answer
Option A
Solution
∑i, j=0  i≠jnnCi nCj=∑i, j=0nnCi nCj−∑i=jnnCi nCj\sum\limits_{i,\,j = 0\,\,i \ne j}^n {{}^n{C_i}\,{}^n{C_j} = \sum\limits_{i,\,j = 0}^n {{}^n{C_i}\,{}^n{C_j} - \sum\limits_{i = j}^n {{}^n{C_i}\,{}^n{C_j}} } }
=∑j=0nnCi ∑j=0nnCj−∑i=0nnCi Ci= \sum\limits_{j = 0}^n {{}^n{C_i}\,\sum\limits_{j = 0}^n {{}^n{C_j} - \sum\limits_{i = 0}^n {{}^n{C_i}\,{C_i}} } }
=2n . 2n−2nCn= {2^n}\,.\,{2^n} - {}^{2n}{C_n}
=22n−2nCn= {2^{2n}} - {}^{2n}{C_n}
Q56
The remainder when (2021)2022+(2022)2021(2021)^{2022}+(2022)^{2021} is divided by 7 is
A 0
B 1
C 2
D 6
Correct Answer
Option A
Solution
(2021)2022+(2022)2021{(2021)^{2022}} + {(2022)^{2021}}
=(7k−2)2022+(7k1−1)2021= {(7k - 2)^{2022}} + {(7{k_1} - 1)^{2021}}
=[(7k−2)3]674+(7k1)2021−2021(7k1)2020 + .... −1= {\left[ {{{(7k - 2)}^3}} \right]^{674}} + {(7{k_1})^{2021}} - 2021{(7{k_1})^{2020}}\, + \,....\, - 1
=(7k2−1)674+(7m−1)= {(7{k_2} - 1)^{674}} + (7m - 1)
=(7n+1)+(7m−1)=7(m+n)= (7n + 1) + (7m - 1) = 7(m + n)

(multiple of 7) ∴\therefore Remainder = 0

Q57
The remainder when 72022+320227^{2022}+3^{2022} is divided by 5 is :
A 0
B 2
C 3
D 4
Correct Answer
Option C
Solution
72022+32022=(72)1011+(32)1011=(50−1)1011+(10−1)1011=(501011−1011.501010+…−1)+(101011−1011.101010+…..−1)=5m−1+5n−1=5(m+n)−2=5(m+n)−5+3=5(m+n−1)+3=5k+3∴ Remainder =3\begin{aligned} & 7^{2022}+3^{2022} \\\\ & =\left(7^2\right)^{1011}+\left(3^2\right)^{1011} \\\\ &=(50-1)^{1011}+(10-1)^{1011} \\\\ &= (50^{1011}-1011.50^{1010}+\ldots-1) \\\\ & + (10^{1011}-1011.10^{1010}+\ldots . .-1) \\\\ &= 5 m-1+5 n-1=5(m+n)-2 \\\\ &= 5(m+n)-5+3=5(m+n-1)+3 \\\\ &= 5 k+3 \\\\ & \therefore \text{ Remainder }=3 \end{aligned}
Q58
∑r=120(r2+1)(r!)\sum\limits_{r=1}^{20}\left(r^{2}+1\right)(r !) is equal to
A 22!−21!22 !-21 !
B 22!−2(21!)22 !-2(21 !)
C 21!−2(20!)21 !-2(20 !)
D 21!−2021 !-20 !
Correct Answer
Option B
Solution

Given,

∑r=120(r2+1)(r!)\sum\limits_{r = 1}^{20} {({r^2} + 1)(r!)}

Let,

f(r)=(r2+1)(r!)f(r) = ({r^2} + 1)(r!)
=(r2)(r!)+r!= ({r^2})(r!) + r!
=r(r r!)+r!= r(r\,r!) + r!
=r[(r+1−1)r!]+r!= r[(r + 1 - 1)r!] + r!
=r[(r+1)r!−r!]+r!= r[(r + 1)r! - r!] + r!
=r[(r+1)!−(r!)]+r!= r[(r + 1)! - (r!)] + r!
=r(r+1)!−r(r!)+r!= r(r + 1)! - r(r!) + r!
=(r+2−2)(r+1)!−r(r!)+r!= (r + 2 - 2)(r + 1)! - r(r!) + r!
=(r+2)(r+1)!−2(r+1)!−[(r+1−1)(r!)]+r!= (r + 2)(r + 1)! - 2(r + 1)! - [(r + 1 - 1)(r!)] + r!
=(r+2)!−2(r+1)!−(r+1)!+r!+r!= (r + 2)! - 2(r + 1)! - (r + 1)! + r! + r!
=(r+2)!−3(r+1)!+2r!= (r + 2)! - 3(r + 1)! + 2r!
=[(r+2)!−(r+1)!]−2[(r+1)!−r!]= [(r + 2)! - (r + 1)!] - 2[(r + 1)! - r!]

∴\therefore

∑r=120f(r)\sum\limits_{r = 1}^{20} {f(r)}
=∑r=120[(r+2)!−(r+1)!]−2∑r=120[(r+1)!−r!]= \sum\limits_{r = 1}^{20} {[(r + 2)! - (r + 1)!] - 2\sum\limits_{r = 1}^{20} {[(r + 1)! - r!]} }
=[(22!+21!+20! + ..... + 4!+3!)−(21!+20!+19! + .... + 3!+2!]−2[(21!+20! + ..... + 3!+2!)−(20!+19! + ..... 1!)]= [(22! + 21! + 20!\, + \,.....\, + \,4! + 3!) - (21! + 20! + 19!\, + \,....\, + \,3! + 2!] - 2[(21! + 20!\, + \,.....\, + \,3! + 2!) - (20! + 19!\, + \,.....\,1!)]
=[(22!)−(2!)]−2[(21)!−(1!)]= [(22!) - (2!)] - 2[(21)! - (1!)]
=22!−2!−2 . (21)!+2 . 1!= 22! - 2! - 2\,.\,(21)! + 2\,.\,1!
=22!−2 . (21)!= 22! - 2\,.\,(21)!
Q59
If the coefficient of x15x^{15} in the expansion of (ax3+1 bx1/3)15\left(\mathrm{a} x^{3}+\dfrac{1}{\mathrm{~b} x^{1 / 3}}\right)^{15} is equal to the coefficient of x−15x^{-15} in the expansion of (ax1/3−1bx3)15\left(a x^{1 / 3}-\dfrac{1}{b x^{3}}\right)^{15}, where aa and bb are positive real numbers, then for each such ordered pair (a,b)(\mathrm{a}, \mathrm{b}) :
A a = 3b
B ab = 1
C ab = 3
D a = b
Correct Answer
Option B
Solution

For

(ax3+1bx13)\left( {a{x^3} + {1 \over {b{x^{{1 \over 3}}}}}} \right)
Tr+1=15Cr(ax3)15−r(1bx13)1{T_{r + 1}} = {}^{15}{C_r}{(a{x^3})^{15 - r}}{\left( {{1 \over {b{x^{{1 \over 3}}}}}} \right)^1}

∴\therefore

x15→3(15−r)−r3=15{x^{15}} \to 3(15 - r) - {r \over 3} = 15
⇒30=10r3⇒r=9\Rightarrow 30 = {{10r} \over 3} \Rightarrow r = 9

Similarly, for

(ax13−1bx3)15{\left( {a{x^{{1 \over 3}}} - {1 \over {b{x^3}}}} \right)^{15}}
Tr+1=15Cr(ax13)15−r(−1bx3)2{T_{r + 1}} = {}^{15}{C_r}{\left( {a{x^{{1 \over 3}}}} \right)^{15 - r}}{\left( { - {1 \over {b{x^3}}}} \right)^2}

∴\therefore For

x−15→15−r3−3r=−15⇒r=6{x^{ - 15}} \to {{15 - r} \over 3} - 3r = - 15 \Rightarrow r = 6

∴\therefore

15C9a6b9=15C6a9b6⇒ab=1{}^{15}{C_9}{{{a^6}} \over {{b^9}}} = {}^{15}{C_6}{{{a^9}} \over {{b^6}}} \Rightarrow ab = 1
Q60
The coefficient of x301{x^{301}} in (1+x)500+x(1+x)499+x2(1+x)498 + ... + x500{(1 + x)^{500}} + x{(1 + x)^{499}} + {x^2}{(1 + x)^{498}}\, + \,...\, + \,{x^{500}} is :
A 500C300{}^{500}{C_{300}}
B 501C200{}^{501}{C_{200}}
C 500C301{}^{500}{C_{301}}
D 501C302{}^{501}{C_{302}}
Correct Answer
Option B
Solution

The coefficient of

x301{x^{301}}

in

(1+x)500+x(1+x)499+x2(1+x)498 + ... + x500{(1 + x)^{500}} + x{(1 + x)^{499}} + {x^2}{(1 + x)^{498}}\, + \,...\, + \,{x^{500}}
500C301+499C300+498C299 + ... + 199C0{}^{500}{C_{301}} + {}^{499}{C_{300}} + {}^{498}{C_{299}}\, + \,...\, + \,{}^{199}{C_0}
=500C199+499C199+498C199 + ... + 199C199= {}^{500}{C_{199}} + {}^{499}{C_{199}} + {}^{498}{C_{199}}\, + \,...\, + \,{}^{199}{C_{199}}
=501C200= {}^{501}{C_{200}}
Ready for a full JEE mock test? Timed · full syllabus · instant results
Take a Mock Test →