Practice Start Test

Binomial Theorem

JEE Mathematics · 151 questions · Page 2 of 16 · Click an option or "Show Solution" to reveal answer

Q11
If xx is so small that x3{x^3} and higher powers of xx may be neglected, then (1+x)32−(1+12x)3(1−x)12{{{{\left( {1 + x} \right)}^{{3 \over 2}}} - {{\left( {1 + {1 \over 2}x} \right)}^3}} \over {{{\left( {1 - x} \right)}^{{1 \over 2}}}}} may be approximated as
A 1−38x21 - {3 \over 8}{x^2}
B 3x+38x23x + {3 \over 8}{x^2}
C −38x2 - {3 \over 8}{x^2}
D x2−38x2{x \over 2} - {3 \over 8}{x^2}
Correct Answer
Option C
Solution
(1+x)32{\left( {1 + x} \right)^{{3 \over 2}}}

= 1 +

32x+32.121.2x2+...{3 \over 2}x + {{{3 \over 2}.{1 \over 2}} \over {1.2}}{x^2} + ...

= 1 +

32x+38x2{3 \over 2}x + {3 \over 8}{x^2}

(As

xx

is so small, so

x3{x^3}

and higher powers of

xx

neglected)

(1+x)32−(1+12x)3(1−x)12{{{{\left( {1 + x} \right)}^{{3 \over 2}}} - {{\left( {1 + {1 \over 2}x} \right)}^3}} \over {{{\left( {1 - x} \right)}^{{1 \over 2}}}}}

=

(1+32x+38x2)−(1+3C0.x2+3C1.(x2)2)(1−x)2{{\left( {1 + {3 \over 2}x + {3 \over 8}{x^2}} \right) - \left( {1 + {}^3{C_0}.{x \over 2} + {}^3{C_1}.{{\left( {{x \over 2}} \right)}^2}} \right)} \over {{{\left( {1 - x} \right)}^2}}}

=

38x2−34x2(1−x)2{{{3 \over 8}{x^2} - {3 \over 4}{x^2}} \over {{{\left( {1 - x} \right)}^2}}}

=

x2(38−34)(1−x)−2{{x^2}\left( {{3 \over 8} - {3 \over 4}} \right){{\left( {1 - x} \right)}^{ - 2}}}

=

−38x2(1−12(−x)+....){ - {3 \over 8}{x^2}\left( {1 - {1 \over 2}\left( { - x} \right) + ....} \right)}

=

−38x2−316x3- {3 \over 8}{x^2} - {3 \over {16}}{x^3}

[ As x3 is so small we can ignore

−316x3-{3 \over {16}}{x^3}

] =

−38x2- {3 \over 8}{x^2}
Q12
The coefficient of xn{x^n} in expansion of (1+x)(1−x)n\left( {1 + x} \right){\left( {1 - x} \right)^n} is
A (−1)n−1n{\left( { - 1} \right)^{n - 1}}n
B (−1)n(1−n){\left( { - 1} \right)^n}\left( {1 - n} \right)
C (−1)n−1(n−1)2{\left( { - 1} \right)^{n - 1}}{\left( {n - 1} \right)^2}
D (n−1)\left( {n - 1} \right)
Correct Answer
Option B
Solution

Given

(1+x)(1−x)n\left( {1 + x} \right){\left( {1 - x} \right)^n}

=

(1−x)n{\left( {1 - x} \right)^n}

+

x(1−x)nx{\left( {1 - x} \right)^n}

General term of

(1−x)n{\left( {1 - x} \right)^n}

=

nCr.(−1)r.xr{}^n{C_r}.{\left( { - 1} \right)^r}.{x^r}

∴\therefore Term containing

xn{x^n}

in

(1−x)n{\left( {1 - x} \right)^n}

=

nCn.(−1)n.xn{}^n{C_n}.{\left( { - 1} \right)^n}.{x^n}

So coefficient of

xn{x^n}

in

(1−x)n{\left( {1 - x} \right)^n}

=

nCn.(−1)n{}^n{C_n}.{\left( { - 1} \right)^n}

General term of

x(1−x)nx{\left( {1 - x} \right)^n}

=

nCr.(−1)r.xr+1{}^n{C_r}.{\left( { - 1} \right)^r}.{x^{r + 1}}

∴\therefore Term containing

xn{x^n}

in

x(1−x)nx{\left( {1 - x} \right)^n}

=

nCn−1.(−1)n−1.xn{}^n{C_{n - 1}}.{\left( { - 1} \right)^{n - 1}}.{x^n}

So coefficient of

xn{x^n}

in

x(1−x)nx{\left( {1 - x} \right)^n}

=

nCn−1.(−1)n−1{}^n{C_{n - 1}}.{\left( { - 1} \right)^{n - 1}}

∴\therefore coefficient of

xn{x^n}

in

(1−x)n{\left( {1 - x} \right)^n}

+

x(1−x)nx{\left( {1 - x} \right)^n}

=

nCn.(−1)n{}^n{C_n}.{\left( { - 1} \right)^n}

+

nCn−1.(−1)n−1{}^n{C_{n - 1}}.{\left( { - 1} \right)^{n - 1}}

=

(−1)n−1[nCn−1−nCn]{\left( { - 1} \right)^{n - 1}}\left[ {{}^n{C_{n - 1}} - {}^n{C_n}} \right]

=

(−1)n−1[n−1]{\left( { - 1} \right)^{n - 1}}\left[ {n - 1} \right]

=

(−1)n[1−n]{\left( { - 1} \right)^n}\left[ {1 - n} \right]
Q13
If the coefficints of x3{x^3} and x4{x^4} in the expansion of (1+ax+bx2)(1−2x)18\left( {1 + ax + b{x^2}} \right){\left( {1 - 2x} \right)^{18}} in powers of xx are both zero, then (a, b)\left( {a,\,b} \right) is equal to:
A (14,2723)\left( {14,{{272} \over 3}} \right)
B (16,2723)\left( {16,{{272} \over 3}} \right)
C (16,2513)\left( {16,{{251} \over 3}} \right)
D (14,2513)\left( {14,{{251} \over 3}} \right)
Correct Answer
Option B
Solution
(1+ax+bx2)(1−2x)18\left( {1 + ax + b{x^2}} \right){\left( {1 - 2x} \right)^{18}}

=

(1−2x)18+ax(1−2x)18+bx2(1−2x)18{\left( {1 - 2x} \right)^{18}} + ax{\left( {1 - 2x} \right)^{18}} + b{x^2}{\left( {1 - 2x} \right)^{18}}

=

(1+ax+bx2)[18C0−18C1(2x)+18C2(2x)2−18C3(2x)3+....]\left( {1 + ax + b{x^2}} \right)\left[ {{}^{18}{C_0} - {}^{18}{C_1}\left( {2x} \right) + {}^{18}{C_2}{{\left( {2x} \right)}^2} - {}^{18}{C_3}{{\left( {2x} \right)}^3} + ....} \right]

Coefficient of x3 is

(−2)3.18C3+a(−2)2.18C2+b(−2).18C1{\left( { - 2} \right)^3}.{}^{18}{C_3} + a{\left( { - 2} \right)^2}.{}^{18}{C_2} + b\left( { - 2} \right).{}^{18}{C_1}

= 0

⇒153a−9b=1632\Rightarrow 153a - 9b = 1632

....... (1) Coefficient of x4 is

(−2)4.18C4+a(−2)3.18C3+b(−2)2.18C2{\left( { - 2} \right)^4}.{}^{18}{C_4} + a{\left( { - 2} \right)^3}.{}^{18}{C_3} + b{\left( { - 2} \right)^2}.{}^{18}{C_2}

= 0

⇒3b−32a=−240\Rightarrow 3b - 32a = -240

....... (2) Solving (1) and (2), we get

aa

= 16, b =

1723{{172} \over 3}
Q14
The sum of coefficients of integral power of xx in the binomial expansion (1−2x)50{\left( {1 - 2\sqrt x } \right)^{50}} is :
A 12(350−1){1 \over 2}\left( {{3^{50}} - 1} \right)
B 12(250+1){1 \over 2}\left( {{2^{50}} + 1} \right)
C 12(350+1){1 \over 2}\left( {{3^{50}} + 1} \right)
D 12(350){1 \over 2}\left( {{3^{50}}} \right)
Correct Answer
Option C
Solution
(1−2x)50{\left( {1 - 2\sqrt x } \right)^{50}}

=

50C0+50C1.(−2x)+50C2.(−2x)2+....{}^{50}{C_0} + {}^{50}{C_1}.\left( { - 2\sqrt x } \right) + {}^{50}{C_2}.{\left( { - 2\sqrt x } \right)^2} + ....

Now we need to find out those coefficient where degree of x is integer and you can see at odd terms power of x is integer.

Let

(1−2x)50{\left( {1 - 2\sqrt x } \right)^{50}}

= Odd(A) - Even(B) So

(1+2x)50{\left( {1 + 2\sqrt x } \right)^{50}}

= A + B ∴\therefore 2A =

(1+2x)50{\left( {1 + 2\sqrt x } \right)^{50}}

+

(1−2x)50{\left( {1 - 2\sqrt x } \right)^{50}}
⇒A=12[(1+2x)50+(1−2x)50]\Rightarrow A = {1 \over 2}\left[ {{{\left( {1 + 2\sqrt x } \right)}^{50}} + {{\left( {1 - 2\sqrt x } \right)}^{50}}} \right]

Now to find sum of coefficient of A, put x = 1. ∴\therefore Sum of coefficient of A =

12[(1+2)50+(1−2)50]{1 \over 2}\left[ {{{\left( {1 + 2} \right)}^{50}} + {{\left( {1 - 2} \right)}^{50}}} \right]

=

12[(3)50+1]{1 \over 2}\left[ {{{\left( 3 \right)}^{50}} + 1} \right]
Q15
If nn is a positive integer, then (3+1)2n−(3−1)2n{\left( {\sqrt 3 + 1} \right)^{2n}} - {\left( {\sqrt 3 - 1} \right)^{2n}} is :
A an irrational number
B an odd positive integer
C an even positive integer
D a rational number other than positive integers
Correct Answer
Option A
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) = 2B = 2[even terms] = 2[ T2 + T4 + T6 + ....... ] So in case of

(3+1)2n−(3−1)2n{\left( {\sqrt 3 + 1} \right)^{2n}} - {\left( {\sqrt 3 - 1} \right)^{2n}}

= 2[ T2 + T4 + T6 + ....... ] = 2[

2nC1.(3)2n−1+2nC3.(3)2n−3+...{}^{2n}{C_1}.{\left( {\sqrt 3 } \right)^{2n - 1}} + {}^{2n}{C_3}.{\left( {\sqrt 3 } \right)^{2n - 3}} + ...

Here in

(3)2n−1{\left( {\sqrt 3 } \right)^{2n - 1}}

, 2n - 1 is odd number. So there will be always

3{\sqrt 3 }

in

(3)2n−1{\left( {\sqrt 3 } \right)^{2n - 1}}

. So

(3+1)2n−(3−1)2n{\left( {\sqrt 3 + 1} \right)^{2n}} - {\left( {\sqrt 3 - 1} \right)^{2n}}

will be always irrational number.

Q16
The coefficient of x−5 in the binomial expansion of (x+1x23−x13+1−x−1x−x12)10,{\left( {{{x + 1} \over {{x^{{2 \over 3}}} - {x^{{1 \over 3}}} + 1}} - {{x - 1} \over {x - {x^{{1 \over 2}}}}}} \right)^{10}}, where x ≠ \ne 0, 1, is :
A 1
B 4
C −- 4
D −- 1
Correct Answer
Option A
Solution
(x+1x2/3−x1/3+1−x−1x−x1/2)10{\left( {{{x + 1} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{x - 1} \over {x - {x^{1/2}}}}} \right)^{10}}

=

((x1/3)3+(11/3)3x2/3−x1/3+1−(x)2−(1)2x−x1/2)10{\left( {{{{{\left( {{x^{1/3}}} \right)}^3} + {{\left( {{1^{1/3}}} \right)}^3}} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{{{\left( {\sqrt x } \right)}^2} - {{\left( 1 \right)}^2}} \over {x - {x^{1/2}}}}} \right)^{10}}

=

((x1/3+1)(x2/3−x1/3+1)x2/3−x1/3+1−(x+1)(x−1)x(x−1))10{\left( {{{\left( {{x^{1/3}} + 1} \right)\left( {{x^{2/3}} - {x^{1/3}} + 1} \right)} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)} \over {\sqrt x \left( {\sqrt x - 1} \right)}}} \right)^{10}}

=

((x1/3+1)−(x+1)x)10{\left( {\left( {{x^{1/3}} + 1} \right) - {{\left( {\sqrt x + 1} \right)} \over {\sqrt x }}} \right)^{10}}

=

((x1/3+1)−(1+1x))10{\left( {\left( {{x^{1/3}} + 1} \right) - \left( {1 + {1 \over {\sqrt x }}} \right)} \right)^{10}}

=

(x1/3−1x1/2)10{\left( {{x^{1/3}} - {1 \over {{x^{1/2}}}}} \right)^{10}}

[Note: For

(xα±1xβ)n{\left( {{x^\alpha } \pm {1 \over {{x^\beta }}}} \right)^n}

the

(r+1)\left( {r + 1} \right)

th term with power m of x is

r=nα−mα+βr = {{n\alpha - m} \over {\alpha + \beta }}

] Here

α=13\alpha = {1 \over 3}

,

β=12\beta = {1 \over 2}

and m = -5 then

r=10×13−(−5)13+12r = {{10 \times {1 \over 3} - (-5)} \over {{1 \over 3} + {1 \over 2}}}

=

253×65{{25} \over 3} \times {6 \over 5}

= 10 ∴\therefore T11 is the term with x-5. ∴\therefore T11 =

10C10{}^{10}{C_{10}}

= 1

Q17
If (27)999 is divided by 7, then the remainder is :
A 1
B 2
C 3
D 6
Correct Answer
Option D
Solution

We have,

(27)9997{{{{\left( {27} \right)}^{999}}} \over 7}

=

(28−1)9997{{{{\left( {28 - 1} \right)}^{999}}} \over 7}

=

28 λ−17{{28\,\lambda - 1} \over 7}

=

28 λ−7+7−1λ{{28\,\lambda - 7 + 7 - 1} \over \lambda }

=

7(4λ−1)+67{{7\left( {4\lambda - 1} \right) + 6} \over 7}
∴   \therefore\,\,\,

Remainder = 6

Q18
The smallest natural number n, such that the coefficient of x in the expansion of (x2+1x3)n{\left( {{x^2} + {1 \over {{x^3}}}} \right)^n} is nC23, is :
A 23
B 58
C 38
D 35
Correct Answer
Option C
Solution

General term

Tr+1=nCrx2n−2r.x−3r{T_{r + 1}} = {}^n{C_r}{x^{2n - 2r}}.{x^{ - 3r}}

∴\therefore

2n−5r=1⇒2n=5r+12n - 5r = 1 \Rightarrow 2n = 5r + 1

∴\therefore

r=2n−15r = {{2n - 1} \over 5}

⇒\Rightarrow Coefficient of x =

nC(2n−15)=nC23{}^n{C_{\left( {{{2n - 1} \over 5}} \right)}} = {}^n{C_{23}}

⇒\Rightarrow

2n−15=23  or  n−(2n−15)=23{{2n - 1} \over 5} = 23\,\,or\,\,n - \left( {{{2n - 1} \over 5}} \right) = 23

⇒\Rightarrow 2n - 1 = 115 ⇒\Rightarrow n = 58 and n = 38 ∴\therefore smallest n = 38

Q19
If the coefficients of x2 and x3 are both zero, in the expansion of the expression (1 + ax + bx2 ) (1 – 3x)15 in powers of x, then the ordered pair (a,b) is equal to :
A (28, 861)
B (28, 315)
C (–21, 714)
D (–54, 315)
Correct Answer
Option B
Solution

(1 + ax + bx2)(1 – 3x)15 Co-eff. of x2 = 1.15C2(–3)2 + a.15C1(–3) + b.15C0

=15×142×9−15×3a+b=0= {{15 \times 14} \over 2} \times 9 - 15 \times 3a + b = 0

(given) ⇒\Rightarrow 945 – 45a + b = 0 ...(i) Now co-eff. of x3 = 0 ⇒\Rightarrow 15C3(–3)3 + a.15C2(–3)2 + b.15C1(–3) = 0

⇒15×14×133×2×(−3×3×3)+a×15×14×92–b×3×15=0\Rightarrow {{15 \times 14 \times 13} \over {3 \times 2}} \times ( - 3 \times 3 \times 3) + a \times {{15 \times 14 \times 9} \over 2 } – b × 3 × 15 = 0

⇒\Rightarrow 15 × 3[–3 × 7 × 13 + a × 7 × 3 – b] = 0 ⇒\Rightarrow 21a – b = 273 ...(ii) From (i) and (ii) a = +28, b = 315 ≡\equiv (a, b) ≡\equiv (28, 315)

Q20
The term independent of x in the expansion of (160−x881).(2x2−3x2)6\left( {{1 \over {60}} - {{{x^8}} \over {81}}} \right).{\left( {2{x^2} - {3 \over {{x^2}}}} \right)^6} is equal to :
A 36
B - 108
C - 36
D - 72
Correct Answer
Option C
Solution

Given expression =

(160−x881).(2x2−3x2)6\left( {{1 \over {60}} - {{{x^8}} \over {81}}} \right).{\left( {2{x^2} - {3 \over {{x^2}}}} \right)^6}

=

160(2x3−3x2)6−x881(2x2−3x2)6{1 \over {60}}{\left( {2{x^3} - {3 \over {{x^2}}}} \right)^6} - {{{x^8}} \over {81}}{\left( {2{x^2} - {3 \over {{x^2}}}} \right)^6}

So its general term is Tr + 1 =

1606Cr(2x2)6−r(−3x2)r−x8816Cr(2x2)6−r(−3x2)r{1 \over {60}}{}^6{C_r}{\left( {2{x^2}} \right)^{6 - r}}{\left( { - {3 \over {{x^2}}}} \right)^r} - {{{x^8}} \over {81}}{}^6{C_r}{\left( {2{x^2}} \right)^{6 - r}}{\left( { - {3 \over {{x^2}}}} \right)^r}

=

1606Cr(2)6−r(−3)rx12−4r−1816Cr(2)6−r(−3)rx20−4r{1 \over {60}}{}^6{C_r}{\left( 2 \right)^{6 - r}}{\left( { - 3} \right)^r}{x^{12 - 4r}} - {1 \over {81}}{}^6{C_r}{\left( 2 \right)^{6 - r}}{\left( { - 3} \right)^r}{x^{20 - 4r}}

.....(i) For this term to be independent of x, put r = 3 in 1st part and r = 5 in 2nd part.

So from (i) the term independent of x =

160×23×(−3)3×6C3+(−181)(2)(−3)5×6C5{1 \over {60}} \times {2^3} \times {\left( { - 3} \right)^3} \times {}^6{C_3} + \left( { - {1 \over {81}}} \right)(2){( - 3)^5} \times {}^6{C_5}

= -72 + 36 = -36

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