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JEE • MATHEMATICS • BINOMIAL THEOREM

Binomial theorem practice · Mixed difficulty

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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181 Easy

181 of 951

Analyzing +4 | -1

If the coefficients of

{x^2}

and

{x^3}

are both zero, in the expansion of the expression

\left( {1 + ax + b{x^2}} \right){(1 - 3x)^{15}}

in powers of

x,

then the ordered pair

(a,b)

is equal to

View answer and explanation

Correct option: A

\left( {28,315} \right)

Option A (

\left( {28,315} \right)

) is correct because by setting the coefficients of

x^2

and

x^3

to zero, we can solve the resulting system of equations to find

a = 28

and

b = 315

182 Medium

182 of 951

Application +4 | -1

Let

x

be the

7^{\text {th }}

term from the beginning and

y

be the

7^{\text {th }}

term from the end in the expansion of

\left(3^{1 / 3}+\frac{1}{4^{1 / 3}}\right)^n

. If

y=12 x

then the value of

n

is :

View answer and explanation

Correct option: A

Option A (9) is correct because by equating the expressions for the 7th term from the beginning and end, and using the given relation y=12x, we can solve for n.

183 Easy

183 of 951

Application +4 | -1

a, b, c

are odd positive integers, then number of integral solutions of

a+b+c=13

, is

View answer and explanation

Correct option: A

Option A (21) is correct because it represents the number of non-negative integer solutions to the transformed equation.

184 Medium

184 of 951

Application +4 | -1

The value of

{ }^{47} C_4+\sum\limits_{{j=0}}^3{ }^{50-j} C_3+\sum\limits_{{k=0}}^5{ }^{56-k} C_{53-k} \text { is }

View answer and explanation

Correct option: A

{ }^{57} C_4

Option A (

{ }^{57} C_4

) is correct because the given expression simplifies to

{ }^{57} C_4

using combinatorial identities.

185 Easy

185 of 951

Application +4 | -1

\cot 7 \frac{1^{\circ}}{2}=\sqrt{n_1}+\sqrt{n_2}+\sqrt{n_3}+\sqrt{n_4}

(where

n_1, n_2

n_3, n_4

are natural numbers) then the value of

\sum_{i=1}^4 n_i

View answer and explanation

Correct option: A

Option D (8) is correct because the area of the quadrilateral formed by the foci of the two ellipses is 8 square units.

186 Easy

186 of 951

Application +4 | -1

If in the expansion of

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

, the coefficient of

r\,th

and

\left( {r + 2} \right)

th term be equal, then

r

is equal to

View answer and explanation

Correct option: C

\frac{n}{2}

Option C (

\frac{n}{2}

) is correct because if the coefficients of the r-th and (r+2)-th terms are equal, then

{\;^n}{C_{r - 1}} = {\;^n}{C_{r + 1}}

, leading to

r = \frac{n}{2}

187 Medium

187 of 951

Analyzing +4 | -1

The coefficient of

{x^{{2^{m + 1}}}}

in the expansion of

E = \frac{1}{{\left( {1 + x} \right)\left( {1 + {x^2}} \right)\left( {1 + {x^4}} \right)\left( {1 + {x^8}} \right) \ldots \left( {1 + {x^{{2^m}}}} \right)}}\;,\;\left| x \right| < 1

View answer and explanation

Correct option: C

Option C (1) is correct because after simplifying the expression, the coefficient of x^(2^(m+1)) in the resulting series is 1.

188 Medium

188 of 951

Application +4 | -1

{ }^{14} C_7+\sum\limits_{{i=1}}^3{ }^{17-i} C_6=

View answer and explanation

Correct option: B

{ }^{17} {C}_7

Option B (

{ }^{17} {C}_7

) is correct because the summation can be simplified using the identity

{n \choose k} + {n \choose k+1} = {n+1 \choose k+1}

repeatedly.

189 Medium

189 of 951

Application +4 | -1

{\;^n}{C_0},\;{\;^n}{C_1},\;{\;^n}{C_2} \ldots .{\;^n}{C_n}\;

denote the coefficient of the binomial expansion

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

, then the value of

{C_1} + 3{C_3} + 5{C_5} + ...

View answer and explanation

Correct option: A

n{2^{n - 2}}

Option A (n2^(n-1)) is correct because the value of the series C1 + 3C3 + 5C5 + ... is equal to n2^(n-2).

190 Easy

190 of 951

Analyzing +4 | -1

If the coefficients of three consecutive terms in the expansion of

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

are in the ratio

1:7:42

, then the value of

n

View answer and explanation

Correct option: C

Option D (None of these) is correct because the value of n is 55, which is not among the provided options.

191 Easy

191 of 951

Application +4 | -1

If the sum of the coefficients in the expansion of

\left(\alpha x^2-2 x+1\right)^{35}

is equal to the sum of the coefficients in the expansion of

(x-\alpha y)^{35}

, then

\alpha

is equal to

View answer and explanation

Correct option: C

Option C (1) is correct because equating the sum of coefficients and solving the resulting equation leads to alpha = 1.

192 Medium

192 of 951

Application +4 | -1

(1+\mathrm{x})^n=\mathrm{C}_0+\mathrm{C}_1 \mathrm{x}+\mathrm{C}_2 \mathrm{x}^2+\ldots+\mathrm{C}_{\mathrm{n}} \mathrm{X}^n

, then

\mathrm{C}_0+2 \mathrm{C}_1+3 \mathrm{C}_2+\ldots+(\mathrm{n}+1) \mathrm{C}_{\mathrm{n}}

is equal to-

View answer and explanation

Correct option: A

2^{n-1}(n+2)

Option A (

2^{n-1}(n+2)

) is correct because differentiating the binomial expansion and adding it to the original expansion gives the desired sum.

193 Medium

193 of 951

Application +4 | -1

The greatest value of the term independent of

x

in the expansion of

{\left( {x\sin \mathrm{\alpha } + {x^{ - 1}}\cos \mathrm{\alpha }} \right)^{10}},\;\mathrm{\alpha } \in R

View answer and explanation

Correct option: C

\frac{1}{{{2^5}}}\;{(^{10}}{C_5})

Option C (

\frac{1}{{{{2^5}}}}\;{(^{10}}{C_5})

) is correct because the greatest value of the term independent of x is obtained when sin(2α) = 1.

194 Medium

194 of 951

Analyzing +4 | -1

The coefficient of

{x^{256}}

in the expansion of

{\left( {1 - \mathrm{ }x} \right)^{101}}{\left( {\mathrm{ }{x^2} + \mathrm{ }x + 1} \right)^{100}}

View answer and explanation

Correct option: B

^{100}{C_{15}}

The correct answer is B (

^{100}{C_{15}}

) because it is the coefficient of

{x^{256}}

obtained after simplifying the expression using the identity

{a^3} - {b^3} = (a - b)({a^2} + ab + {b^2})

195 Easy

195 of 951

Application +4 | -1

The sum of coefficient in the expansion of

(1-x)^{10}

View answer and explanation

Correct option: B

Option B (0) is correct because when x=1, (1-x)¹⁰ becomes (1-1)¹⁰ = 0¹⁰ = 0, which is the sum of the coefficients.

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196 Medium

196 of 951

Application +4 | -1

I

is integral part of

(2+\sqrt{3})^n

and

f

is its fractional part. Then

(I+f)(1-f)

View answer and explanation

Correct option: B

Option B (1) is correct because

(I+f)(1-f) = (2+\sqrt{3})^n(2-\sqrt{3})^n = 1

, based on the properties of the integral and fractional parts.

197 Medium

197 of 951

Application +4 | -1

\left(1+x-2 x^2\right)^6=\sum\limits_{{r=0}}^{12} A_r x^r

, then value of

A_2+A_4

+A_6+\ldots+A_{12}

, is

View answer and explanation

Correct option: B

Option B (31) is correct because by substituting x=1 and x=-1 into the given expression and adding the resulting equations, we can isolate the sum of even-powered coefficients.

198 Easy

198 of 951

Application +4 | -1

If the sum of the coefficient in the expansion of

{\left( {x - 2y + 3z} \right)^n}

is 128, then the greatest coefficient in the expansion of

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

View answer and explanation

Correct option: A

Option A (35) is correct because the sum of coefficients gives n=7, and the greatest coefficient in (1+x)⁷ is 7C3 or 7C4, which equals 35.

199 Medium

199 of 951

Application +4 | -1

\sum\limits_{{r=1}}^n(-1)^{r+1} \cdot \frac{{ }^n C_r}{r+1}

is equal to

View answer and explanation

Correct option: D

\frac{n}{n+1}

Option D (

\frac{n}{n+1}

) is correct because by manipulating the summation and using binomial coefficient identities, we can simplify the expression to

\frac{n}{n+1}

200 Easy

200 of 951

Analyzing +4 | -1

Let

f(x)=|x-a|, a \neq 0

then

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because none of the given equations hold true for a general

a \neq 0

when

f(x) = |x-a|

201 Medium

201 of 951

Application +4 | -1

The value of

\begin{array}{l} \left({ }^{21} \mathrm{C}_{1}-{ }^{10} \mathrm{C}_{1}\right)+\left({ }^{21} \mathrm{C}_{2}-{ }^{10} \mathrm{C}_{2}\right)+\left({ }^{21} \mathrm{C}_{3}-{ }^{10} \mathrm{C}_{3}\right)+\left({ }^{21} \mathrm{C}_{4}-{ }^{10} \mathrm{C}_{4}\right) \\ +\ldots .+\left({ }^{21} \mathrm{C}_{10}-{ }^{10} \mathrm{C}_{10}\right) \text { is : } \end{array}

View answer and explanation

Correct option: A

2^{20}-2^{10}

Option A (

2^{20}-2^{10}

) is correct because it involves summing binomial coefficients and using properties of binomial expansions.

202 Medium

202 of 951

Application +4 | -1

Middle term in the expansion of

\left(1+3 x+3 x^2+x^3\right)^6

View answer and explanation

Correct option: C

10 \mathrm{th}

Option C (10th) is correct because the given expression simplifies to (1+x)¹⁸, and the middle term of this expansion is the 10th term.

203 Easy

203 of 951

Application +4 | -1

5^{81}+\lambda

is divisible by 26, then minimum positive value of

\frac{\lambda}{3}

must be

View answer and explanation

Correct option: A

Option A (7) is correct because by using the binomial theorem and divisibility rules, the minimum positive value of lambda/3 is found to be 7.

204 Medium

204 of 951

Analyzing +4 | -1

The greatest value of the term independent of

x

, as

\alpha

varies over

R

, in the expansion of

{\left( {x\cos \mathrm{\alpha } + \frac{{\sin \mathrm{\alpha }}}{x}} \right)^{20}}

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the greatest value of the term independent of x is actually

{\;^{20}}{C_{10}}{\left( {\frac{1}{2}} \right)^{10}}

, which is not among the given options.

205 Easy

205 of 951

Application +4 | -1

The coefficient of

{x^6}\;{a^{ - 2}}

in the expansion of

{\left( {\frac{{{x^2}}}{a} - \frac{a}{x}} \right)^{12}},

View answer and explanation

Correct option: C

Option C (0) is correct because the powers of x and a in the general term of the expansion cannot simultaneously satisfy the conditions for x⁶ and a^-2.

206 Medium

206 of 951

Analyzing +4 | -1

C_r

stands for

{ }^n C_r

, then the sum of the series

\frac{2\left(\frac{n}{2}\right) !\left(\frac{n}{2}\right) !}{n !}\left[C_0^2-2 C_1^2+3 C_2^2 \ldots \ldots+(-1)^n(n+1) C_n^2\right]

where '

{n}

' is an even positive integer is equal to

View answer and explanation

Correct option: C

(-1)^{n / 2}(n+2)

Option C (

(-1)^{n / 2}(n+2)

) is correct because after simplifying the series, the sum equals this expression.

207 Easy

207 of 951

Understanding +4 | -1

The expansion of

\frac{1}{(4-3 x)^{1 / 2}}

by binomial theorem will be valid, if

View answer and explanation

Correct option: D

none of these

Option D (none of these) is correct because the expansion is valid when

|\frac{{3}}{{4}}x| < 1

, which means

-\frac{{4}}{{3}} < x < \frac{{4}}{{3}}

. None of the provided options match this condition.

208 Easy

208 of 951

Application +4 | -1

If the number of integral terms in the expansion of

{\left( {{3^{\frac{1}{2}}} + {5^{\frac{1}{8}}}} \right)^n}

is exactly 33, then the least value of

n

View answer and explanation

Correct option: C

256

Option C (256) is correct because it is the least value of n that results in exactly 33 integral terms in the binomial expansion.

209 Easy

209 of 951

Analyzing +4 | -1

\alpha \,\,\mathrm{and}\,\,\beta

be the coefficients of

{x^4}\,\,\mathrm{and}\,\,{x^2}

respectively in the expansion of

{(x + \sqrt {{x^2} - 1} )^6} + {(x - \sqrt {{x^2} - 1} )^6},

then

View answer and explanation

Correct option: B

\alpha - \beta = - 132

Option B (

\alpha - \beta = - 132

) is correct because it is the result of calculating the coefficients of x⁴ and x² in the given expansion and then subtracting them.

210 Easy

210 of 951

Understanding +4 | -1

The expansion of

\frac{1}{(4-3 x)^{1 / 2}}

, binomial theorem will be valid, if

View answer and explanation

Correct option: D

None of these

None of these is correct because the binomial expansion is valid when

|x| < \frac{4}{3}

Revision notes and key takeaways

Review BINOMIAL THEOREM in your textbook or class notes, then use this set to check your understanding.

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