Practice Start Test

Binomial Theorem

JEE Mathematics · 151 questions · Page 3 of 16 · Click an option or "Show Solution" to reveal answer

Q21
If 20C1 + (22) 20C2 + (32) 20C3 + ..... + (202 ) 20C20 = A(2β\beta ), then the ordered pair (A, β\beta ) is equal to :
A (420, 19)
B (420, 18)
C (380, 18)
D (380, 19)
Correct Answer
Option B
Solution

S =

12 20C1+22 20C2+32 20C3+..........+202 20C20{1^2}\,{}^{20}{C_1} + {2^2}\,{}^{20}{C_2} + {3^2}\,{}^{20}{C_3} + .......... + {20^2}\,{}^{20}{C_{20}}

⇒\Rightarrow

∑r=120r2 20Cr\sum\limits_{r = 1}^{20} {{r^2}\,{}^{20}{C_r}}

⇒\Rightarrow

∑r=120r .(r.20Cr)\sum\limits_{r = 1}^{20} {r\,.\left( {r.{}^{20}{C_r}} \right)}

⇒\Rightarrow

20∑r=120r .19Cr−120\sum\limits_{r = 1}^{20} {r\,.} {}^{19}{C_{r - 1}}

⇒\Rightarrow

20∑r=120(r−1+1) .19Cr−120\sum\limits_{r = 1}^{20} {(r - 1 + 1)\,.} {}^{19}{C_{r - 1}}

⇒\Rightarrow

20∑r=120(r−1) .19Cr−1+20∑r=12019Cr−120\sum\limits_{r = 1}^{20} {(r - 1)\,.} {}^{19}{C_{r - 1}} + 20\sum\limits_{r = 1}^{20} {{}^{19}{C_{r - 1}}}

⇒\Rightarrow

20×19∑r=22018Cr−2+20×21920 \times 19\sum\limits_{r = 2}^{20} {{}^{18}{C_{r - 2}}} + 20 \times {2^{19}}

⇒\Rightarrow 20 ×\times 19 ×\times 218 + 20 ×\times 219 ⇒\Rightarrow

20×218(19+2)=20×21×21820 \times {2^{18}}\left( {19 + 2} \right) = 20 \times 21 \times {2^{18}}

⇒\Rightarrow

420×218420 \times {2^{18}}
Q22
The sum of the co-efficients of all even degree terms in x in the expansion of (x+x3−1)6{\left( {x + \sqrt {{x^3} - 1} } \right)^6} + (x−x3−1)6{\left( {x - \sqrt {{x^3} - 1} } \right)^6}, (x > 1) is equal to:
A 32
B 26
C 29
D 24
Correct Answer
Option D
Solution

Let

(a+x)n{\left( {a + x} \right)^n}

= Odd trems(A) + Even terms(B) So

(a−x)n{\left( {a - x} \right)^n}

= Odd terms(A) - Even terms(B) ∴\therefore

(a+x)n−(a−x)n{\left( {a + x} \right)^n} - {\left( {a - x} \right)^n}

= (A + B) + (A - B) = 2A = 2[odd terms] = 2[ T1 + T3 + T5 + ....... ] So in case of

(x+x3−1)6{\left( {x + \sqrt {{x^3} - 1} } \right)^6}

+

(x−x3−1)6{\left( {x - \sqrt {{x^3} - 1} } \right)^6}

= 2[ T1 + T3 + T5 + T5 ] = 2[

6C0x6+6C2x4(x3−1){}^6{C_0}{x^6} + {}^6{C_2}{x^4}\left( {{x^3} - 1} \right)
+6C4x2(x3−1)2+6C6(x3−1)3+ {}^6{C_4}{x^2}{\left( {{x^3} - 1} \right)^2} + {}^6{C_6}{\left( {{x^3} - 1} \right)^3}

] = 2[

x6+15x4(x3−1)+15x2(x3−1)2+(x3−1)3{x^6} + 15{x^4}\left( {{x^3} - 1} \right) + 15{x^2}{\left( {{x^3} - 1} \right)^2} + {\left( {{x^3} - 1} \right)^3}

] = 2[

x6+15(x7−x4){x^6} + 15\left( {{x^7} - {x^4}} \right)
+15x2(x6−2x3+1)+(x9−3x6+3x3−1)+ 15{x^2}\left( {{x^6} - 2{x^3} + 1} \right) + \left( {{x^9} - 3{x^6} + 3{x^3} - 1} \right)

] ∴\therefore Sum of coefficient of all even degree terms = 2[ 1 - 15 + 15 + 15 - 3 - 1 ] = 24

Q23
The coefficient of t4 in the expansion of (1−t61−t)3{\left( {{{1 - {t^6}} \over {1 - t}}} \right)^3} is :
A 14
B 15
C 10
D 12
Correct Answer
Option B
Solution
(1−t61−t)3{\left( {{{1 - {t^6}} \over {1 - t}}} \right)^3}

= (1 −- t6)3 (1 −- t)−-3 = (1 −- 3C1t6 + 3C2t12 −- 3C3t18) ×\times (1 −- t)−-3 coefficient of t4 is 1 ×\times coefficient of t4 in (1 −- t)−-3 = 1 ×\times 3+4−-1C4 (By multinomial theorem) = 6C4 = 15

Q24
If ∑i=120(20Ci−120Ci+20Ci−1)3=k21{\sum\limits_{i = 1}^{20} {\left( {{{{}^{20}{C_{i - 1}}} \over {{}^{20}{C_i} + {}^{20}{C_{i - 1}}}}} \right)} ^3} = {k \over {21}} then k is equal to
A 100
B 200
C 50
D 400
Correct Answer
Option A
Solution
∑i=120(20Ci−120Ci+20Ci−1)3=k21{\sum\limits_{i = 1}^{20} {\left( {{{^{20}{C_{i - 1}}} \over {^{20}{C_i}{ + ^{20}}{C_{i - 1}}}}} \right)} ^3} = {k \over {21}}
⇒  ∑i=120(20Ci−121Ci)3=k21\Rightarrow \,\,\sum\limits_{i = 1}^{20} {{{\left( {{{{}^{20}{C_{i - 1}}} \over {{}^{21}{C_i}}}} \right)}^3}} = {k \over {21}}
⇒  ∑i=120(i21)3=k21\Rightarrow \,\,\sum\limits_{i = 1}^{20} {{{\left( {{i \over {21}}} \right)}^3}} = {k \over {21}}
⇒  1(21)3[20(21)2]2=k21\Rightarrow \,\,{1 \over {{{\left( {21} \right)}^3}}}{\left[ {{{20\left( {21} \right)} \over 2}} \right]^2} = {k \over {21}}
⇒100=k\Rightarrow 100 = k
Q25
If the term independent of xx in the expansion of (ax2+12x3)10\left(\sqrt{\mathrm{a}} x^2+\dfrac{1}{2 x^3}\right)^{10} is 105 , then a2\mathrm{a}^2 is equal to :
A 6
B 4
C 2
D 9
Correct Answer
Option B
Solution
(ax2+12x3)10Tr+1=10Cr(ax2)10−r(12x3)r\begin{aligned} & \left(\sqrt{a} x^2+\frac{1}{2 x^3}\right)^{10} \\ & T_{r+1}={ }^{10} C_r\left(\sqrt{a} x^2\right)^{10-r}\left(\frac{1}{2 x^3}\right)^r \end{aligned}

Independent of

x⇒20−2r−3r=0x \Rightarrow 20-2 r-3 r=0
r=4r=4

Independent of

xx

is

10C4(a)6(12)4=105{ }^{10} C_4(\sqrt{a})^6\left(\frac{1}{2}\right)^4=105
2102×8a3=105⇒a=2a2=4\begin{gathered} \frac{210}{2 \times 8} a^3=105 \\ \Rightarrow \quad a=2 \\ a^2=4 \end{gathered}
Q26
Let (x + 10)50 + (x −- 10)50 = a0 + a1x + a2x2 + . . . . + a50x50, for all x ∈ \in R; then a2a0{{{a_2}} \over {{a_0}}} is equal to
A 12.25
B 12.75
C 12.00
D 12.50
Correct Answer
Option A
Solution

(10 + x)50 + (10 −- x)50 ⇒\Rightarrow a2 = 2.50C2 1048, a0 = 2.1050

a2a0=50C2102=12.25{{{a_2}} \over {{a_0}}} = {{^{50}{C_2}} \over {{{10}^2}}} = 12.25
Q27
If the constant term in the binomial expansion of (x−kx2)10{\left( {\sqrt x - {k \over {{x^2}}}} \right)^{10}} is 405, then |k| equals :
A 3
B 9
C 1
D 2
Correct Answer
Option A
Solution
(x−kx2)10{\left( {\sqrt x - {k \over {{x^2}}}} \right)^{10}}

rth term of the expansion, Tr+1 = 10Cr

(x)10−r(−kx2)r{\left( {\sqrt x } \right)^{10 - r}}{\left( {{{ - k} \over {{x^2}}}} \right)^r}

= 10Cr.

x10−r2.(−k)r.x−2r{x^{{{10 - r} \over 2}}}.{\left( { - k} \right)^r}.{x^{ - 2r}}

= 10Cr.

x10−5r2.(−k)r{x^{{{10 - 5r} \over 2}}}.{\left( { - k} \right)^r}

If it is constant term then

10−5r2{{{10 - 5r} \over 2}}

= 0 ⇒\Rightarrow r = 2 T3 = 10C2.(-k)2 = 405 ⇒\Rightarrow k2 =

40545{{405} \over {45}}

= 9 ⇒\Rightarrow k = ±\pm 3 ⇒\Rightarrow |k| = 3

Q28
If for some positive integer n, the coefficients of three consecutive terms in the binomial expansion of (1 + x)n + 5 are in the ratio 5 : 10 : 14, then the largest coefficient in this expansion is :
A 330
B 792
C 252
D 462
Correct Answer
Option D
Solution

Consider the three consecutive coefficients as

n+5Cr, n+5Cr+1, n+5Cr+2^{n + 5}{C_r},{\,^{n + 5}}{C_{r + 1}},{\,^{n + 5}}{C_{r + 2}}

∵\because

n+5Crn+5Cr+1=12{{^{n + 5}{C_r}} \over {^{n + 5}{C_{r + 1}}}} = {1 \over 2}
⇒r+1n+5−r=12⇒3r=n+3\Rightarrow {{r + 1} \over {n + 5 - r}} = {1 \over 2} \Rightarrow 3r = n + 3

...(i) and

n+5Cr+1n+5Cr+2=57{{^{n + 5}{C_{r + 1}}} \over {^{n + 5}{C_{r + 2}}}} = {5 \over 7}

⇒\Rightarrow

⇒r+2n+4−r=57⇒12r=5n+6\Rightarrow {{r + 2} \over {n + 4 - r}} = {5 \over 7} \Rightarrow 12r = 5n + 6

...(ii) From (i) and (ii) n = 6 Largest coefficient in the expansion is

11C6{^{11}{C_6}}

= 462

Q29
If {p} denotes the fractional part of the number p, then {32008}\left\{ {{{{3^{200}}} \over 8}} \right\}, is equal to :
A 58{5 \over 8}
B 78{7 \over 8}
C 18{1 \over 8}
D 38{3 \over 8}
Correct Answer
Option C
Solution
{32008}\left\{ {{{{3^{200}}} \over 8}} \right\}

=

{(32)1008}\left\{ {{{{{\left( {{3^2}} \right)}^{100}}} \over 8}} \right\}

=

{(1+8)1008}\left\{ {{{{{\left( {1 + 8} \right)}^{100}}} \over 8}} \right\}

=

{1+100C1.8+100C2.82+....+100C100.81008}\left\{ {{{1 + {}^{100}{C_1}.8 + {}^{100}{C_2}{{.8}^2} + .... + {}^{100}{C_{100}}{{.8}^{100}}} \over 8}} \right\}

=

{1+8K8}\left\{ {{{1 + 8K} \over 8}} \right\}

=

{18+K}\left\{ {{1 \over 8} + K} \right\}

where K

∈\in

Integer ∴\therefore Fractional part =

18{{1 \over 8}}
Q30
Let α\alpha > 0, β\beta > 0 be such that α\alpha 3 + β\beta 2 = 4. If the maximum value of the term independent of x in the binomial expansion of (αx19+βx−16)10{\left( {\alpha {x^{{1 \over 9}}} + \beta {x^{ - {1 \over 6}}}} \right)^{10}} is 10k, then k is equal to :
A 176
B 336
C 352
D 84
Correct Answer
Option B
Solution

General term Tr + 1 = 10Cr

α10−r.(x)10−r9.βr(x)−r6{\alpha ^{10 - r}}.{\left( x \right)^{{{10 - r} \over 9}}}.{\beta ^r}{\left( x \right)^{ - {r \over 6}}}

= 10Cr

α10−rβr.(x)10−r9−r6{\alpha ^{10 - r}}{\beta ^r}.{\left( x \right)^{{{10 - r} \over 9} - {r \over 6}}}

If Tr + 1 is independent of x ∴\therefore

10−r9−r6{{{10 - r} \over 9} - {r \over 6}}

= 0 ⇒\Rightarrow r = 4 ∴\therefore T5 = 10C4

α6β4{\alpha ^6}{\beta ^4}

Also given, α\alpha3 + β\beta2 = 4 By AM-GM inequality

α3+β22≥(α3β2)12{{{\alpha ^3} + {\beta ^2}} \over 2} \ge {\left( {{\alpha ^3}{\beta ^2}} \right)^{{1 \over 2}}}

⇒\Rightarrow (2)2 ≥\ge

α3β2{{\alpha ^3}{\beta ^2}}

⇒\Rightarrow

α6β4{\alpha ^6}{\beta ^4}

≤\le 16 ∴\therefore 10k = 10C4 (16) ⇒\Rightarrow k = 336

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