The range of the values of the term independent of $x$ in the expansion of ${\left( {x{{\sin }^{ - 1}}{\rm{\alpha }} + \frac{{{{\cos }^{ - 1}}{\rm{\alpha }}}}{x}} \right)^{10}},{\rm{\alpha }} \in \left[ { - 1,1} \right]$ is

$\left[ { - \frac{{{{\rm{\;}}^{10}}{{\rm{C}}_5}{\rm{.}}{{\rm{\pi }}^{10}}}}{{{2^5}}},{\rm{\;}}\frac{{{{\rm{\;}}^{10}}{{\rm{C}}_5}{\rm{.}}{{\rm{\pi }}^{10}}}}{{{2^{20}}}}} \right]$

$\left[ {\frac{{{{\rm{\;}}^{10}}{{\rm{C}}_5}{\rm{.}}{{\rm{\pi }}^{10}}}}{{{2^5}}},{\rm{\;}} - \frac{{{{\rm{\;}}^{10}}{{\rm{C}}_5}{{\rm{\pi }}^{10}}}}{{{2^{20}}}}} \right]$

$\left[ {\frac{{{{\rm{\;}}^{10}}{{\rm{C}}_5}{\rm{.}}{{\rm{\pi }}^5}}}{{{2^5}}},{\rm{\;}}\frac{{{{\rm{\;}}^{10}}{{\rm{C}}_5}{\rm{.}}{{\rm{\pi }}^5}}}{{{2^{20}}}}} \right]$

$\left[ { - \frac{{{{\rm{\;}}^{10}}{{\rm{C}}_5}{\rm{.}}{{\rm{\pi }}^5}}}{{{2^5}}},{\rm{\;}}\frac{{{{\rm{\;}}^{10}}{{\rm{C}}_5}{\rm{.}}{{\rm{\pi }}^5}}}{{{2^{20}}}}} \right]$

$\left( {\begin{array}{*{20}{c}} n \\ 0 \end{array}} \right)\left( {\begin{array}{*{20}{c}} n \\ r \end{array}} \right) + \left( {\begin{array}{*{20}{c}} n \\ 1 \end{array}} \right)\left( {\begin{array}{*{20}{c}} n \\ {r + 1} \end{array}} \right) + ... + \left( {\begin{array}{*{20}{c}} n \\ {n - r} \end{array}} \right)\left( {\begin{array}{*{20}{c}} n \\ n \end{array}} \right) = $

$\left( {\begin{array}{*{20}{c}} {2n} \\ {n - r} \end{array}} \right)$

$\left( {\begin{array}{*{20}{c}} {2n} \\ {r - 1} \end{array}} \right)$

$\left( {\begin{array}{*{20}{c}} {2n} \\ r \end{array}} \right)$

$\left( {\begin{array}{*{20}{c}} {2n} \\ {n - r + 1} \end{array}} \right)$

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JEE • MATHEMATICS • BINOMIAL THEOREM

Binomial theorem practice · Mixed difficulty

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271 Easy

271 of 951

Analyzing +4 | -1

If the second, third and fourth term in the expansion of

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

are 240,720 and 1080 respectively, then the value of

n

View answer and explanation

Correct option: D

The correct answer is D (5) because by setting up ratios of consecutive terms in the binomial expansion, we can solve for n.

272 Easy

272 of 951

Application +4 | -1

The two values of the rational exponent

m

such that in the binomial expansion of

(1-x)^m

, the coefficient of

x^2

is 3 , are

View answer and explanation

Correct option: C

3,-2

Option C (3, -2) is correct because these are the two values of m that satisfy the given condition for the coefficient of x² in the binomial expansion.

273 Hard

273 of 951

Analyzing +4 | -1

{C_\mathrm{r}}

stands for

{\;^n}{C_r}

, the sum of the given series

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

where

n

is an even positive integer, is

View answer and explanation

Correct option: D

{\left( { - 1} \right)^{n/2}}\left( {n + 2} \right)

Option D (

{{\left( { - 1} \right)}^{n/2}}\left( {n + 2} \right)

) is correct because simplifying the given series using properties of binomial coefficients leads to the result

{{\left( { - 1} \right)}^{n/2}}\left( {n + 2} \right)

274 Easy

274 of 951

Application +4 | -1

The coefficient of

{a^3}{b^4}c

in the expansion of

{\left( {1 + a - b + c} \right)^9}

is equal to

View answer and explanation

Correct option: D

\frac{{9!}}{{3!4!}}

Option D (9!/(3!4!)) is correct because the coefficient of a³b⁴c in the multinomial expansion of (1+a-b+c)⁹ is calculated using the multinomial coefficient formula.

275 Medium

275 of 951

Analyzing +4 | -1

The sum of

\sum\limits_{{r=0}}^n(-1)^r \frac{{ }^n C_r}{{ }^{r+2} C_r}

View answer and explanation

Correct option: D

\frac{2}{n+2}

Option D (

\frac{{2}}{{(n+2)}}

) is correct because after simplifying the summation, the expression reduces to this form.

276 Easy

276 of 951

Analyzing +4 | -1

n

is an integer greater than 1, then

a - {\;^n}{C_1}\left( {a - 1} \right) + {\;^n}{C_2}\left( {a - 2} \right) + ... + {( - 1)^n}\left( {a - n} \right)

is equal to

View answer and explanation

Correct option: B

Option B (0) is correct because the given expression simplifies to 0 using binomial theorem properties.

277 Easy

277 of 951

Application +4 | -1

Last four digits of the number

N=7^{100}-3^{100}

View answer and explanation

Correct option: D

8000

Option D (8000) is correct because the last four digits of the expression

7^{100}-3^{100}

can be found using binomial expansion and modular arithmetic.

278 Easy

278 of 951

Application +4 | -1

If the ratio of the coefficient of third and fourth term in the expansion of

{\left( {x - \frac{1}{{2x}}} \right)^n}

1:2

, then the value of

n

will be

View answer and explanation

Correct option: D

- 10

The correct answer is D (-10). The value of n is found by equating the ratio of the coefficients of the third and fourth terms to 1:2 and solving for n.

279 Easy

279 of 951

Understanding +4 | -1

The value of

r

for which

{}^{20}{C_r}\,\,{}^{20}{C_0} + {}^{20}{C_{r - 1}}\,\,{}^{20}{C_1} + {}^{20}{C_{r - 2}}\,\,{}^{20}{C_2} + .... + {}^{20}{C_0}{}^{20}{C_r}

is maximum, is

View answer and explanation

Correct option: D

Option D (20) is correct because the expression represents the coefficient of

x^r

(1+x)^{40}

, which is maximum when r = 20.

280 Easy

280 of 951

Application +4 | -1

The value of

{ }^{15} C_0^2-{ }^{15} C_1^2+{ }^{15} C_2^2-\ldots-{ }^{15} C_{15}^2

View answer and explanation

Correct option: C

Option C (0) is correct because the alternating sum of squares of binomial coefficients is zero when n is odd.

281 Easy

281 of 951

Application +4 | -1

The last digit of the number

(32)^{32}

View answer and explanation

Correct option: B

Option B (6) is correct because the last digit of (32)³² is determined by the last digit of 2³², which repeats in a cycle of 4, and 32 is a multiple of 4, so the last digit is 6.

282 Medium

282 of 951

Application +4 | -1

If the coefficients of

x^2

and

x^3

in the expansion of

(3+k x)^9

are equal, then the value of

k

View answer and explanation

Correct option: B

9 / 7

Option B (9 / 7) is correct because equating the coefficients of x² and x³ leads to the equation

{ }^9 C_2 3^7 k^2 = { }^9 C_3 3^6 k^3

, which simplifies to k = 9/7.

283 Easy

283 of 951

Application +4 | -1

The coefficient of

x^7

in the expansion o

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

, is

View answer and explanation

Correct option: B

-144

Option B (-144) is correct because the coefficient of

x^7

in the expansion of

(1-x-x^2+x^3)^6

can be found by factoring and then applying the binomial theorem.

284 Medium

284 of 951

Application +4 | -1

The value of

\left( {\frac{{{\;^{50}}{C_0}}}{1} + \frac{{{\;^{50}}{C_2}}}{3} + \frac{{{\;^{50}}{C_4}}}{5} + ... + \frac{{{\;^{50}}{C_{50}}}}{{51}}} \right)

View answer and explanation

Correct option: A

\frac{{{2^{50}}}}{{51}}

Option A is correct. The given sum can be simplified using binomial coefficient identities and properties of series.

285 Hard

285 of 951

Application +4 | -1

The term independent of

x

\left[\sqrt{\frac{x}{3}}+\sqrt{\left(\frac{3}{2 x^2}\right)}\right]^{10}

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the term independent of

x

requires the power of

x

to be zero, but the resulting value of

r

is not an integer, meaning there is no constant term.

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286 Medium

286 of 951

Analyzing +4 | -1

\frac{1}{n+1}{ }^{n} C_{n}+\frac{1}{n}{ }^{n} C_{n-1}+\ldots+\frac{1}{2}{ }^{n} C_{1}+{ }^{n} C_{0}=\frac{1023}{10}

then

n

is equal to

View answer and explanation

Correct option: A

Option A (9) is correct because solving the given summation involving binomial coefficients leads to n = 9.

287 Easy

287 of 951

Application +4 | -1

For all integers

n \ge 1,

which of the following is divisible by 9?

View answer and explanation

Correct option: B

{4^n} - 3n - 1

Option B (

{4^n} - 3n - 1

) is correct because when expanding

4^n

using the binomial theorem, the expression becomes divisible by 9.

288 Easy

288 of 951

Application +4 | -1

{\;^{20}}{C_4} + 2.{\;^{20}}{C_3} + {\;^{20}}{C_2} - {\;^{22}}{C_{18}}

is equal to

View answer and explanation

Correct option: A

Option A (0) is correct. Using the properties of binomial coefficients, the expression simplifies to zero.

289 Medium

289 of 951

Application +4 | -1

For

n > 3, a, b \in \mathbf{R}

, let

S(n, a, b)

=\sum\limits_{{r=0}}^n(-1)^r \frac{n(n-1) \cdots(n-r+1)}{r !}(a-r)(b-r)

then

S\left(5, \frac{1}{2}, \frac{1}{7}\right)

View answer and explanation

Correct option: A

Option A (0) is correct because the given summation simplifies to zero due to the properties of binomial coefficients and alternating sums.

290 Easy

290 of 951

Application +4 | -1

The term independent of

x

in the expansion of

\left(\sqrt{\left(\frac{x}{3}\right)}+\sqrt{\left(\frac{3}{2 x^2}\right)}\right)^{10}

View answer and explanation

Correct option: D

none of these

Option D (none of these) is correct because the value of r is not a whole number, indicating there's no term independent of x.

291 Medium

291 of 951

Analyzing +4 | -1

Suppose

[x]

denote the greatest integer

\leq x

, and

n \in {N}

, then

\lim \limits_{{n \rightarrow \infty}} \frac{\left[{ }^n C_0 x^2\right]+\left[{ }^n C_1 x^2\right]+\cdots+\left[{ }^n C_n x^2\right]}{2^n-2}

is equal to

View answer and explanation

Correct option: B

x^2

Option B (

x^2

) is correct because using the sandwich theorem, the limit of the given expression as n approaches infinity is

x^2

292 Hard

292 of 951

Analyzing +4 | -1

The range of the values of the term independent of

x

in the expansion of

{\left( {x{{\sin }^{ - 1}}\mathrm{\alpha } + \frac{{{{\cos }^{ - 1}}\mathrm{\alpha }}}{x}} \right)^{10}},\mathrm{\alpha } \in \left[ { - 1,1} \right]

View answer and explanation

Correct option: B

\left[ { - \frac{{{\mathrm{\;}^{10}}{\mathrm{C}_5}\mathrm{.}{\mathrm{\pi }^{10}}}}{{{2^5}}},\mathrm{\;}\frac{{{\mathrm{\;}^{10}}{\mathrm{C}_5}\mathrm{.}{\mathrm{\pi }^{10}}}}{{{2^{20}}}}} \right]

Option B is correct because the range of the term independent of x is found by analyzing the term and using properties of inverse trigonometric functions.

293 Easy

293 of 951

Application +4 | -1

If the binomial expansion of

{\left( {a + b\;x} \right)^{ - 2}}

\frac{1}{4} - 3\;x + \ldots ,

then

\left( {a,b} \right) =

View answer and explanation

Correct option: A

\left( {2,\;12} \right)

Option A

\left( {2,\;12} \right)

is correct because it satisfies the given binomial expansion condition by comparing coefficients.

294 Medium

294 of 951

Application +4 | -1

The coefficient of

x^5

\left(1+2 x+3 x^2+\ldots\right)^{\frac{-3}{2}}

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the given expression simplifies to (1-x)³, and the coefficient of x⁵ in this polynomial is 0.

295 Easy

295 of 951

Application +4 | -1

The coefficient of

{x^5}

{\left( {1 + {x^2}} \right)^5}{\left( {1 + x} \right)^4}

View answer and explanation

Correct option: C

Option C (60) is correct because the coefficient of x⁵ is obtained by multiplying the appropriate terms in each expansion and summing the results.

296 Medium

296 of 951

Application +4 | -1

The value of

{ }^{n-1} C_0 \cdot{ }^n C_2+{ }^{n-1} C_1 \cdot{ }^n C_3+{ }^{n-1} C_2 \cdot{ }^n C_4 +\ldots+{ }^{n-1} C_{n-2} \cdot{ }^n C_n \text { is }

View answer and explanation

Correct option: C

{ }^{2 n-1} C_{n-2}

Option C (

{ }^{2 n-1} C_{n-2}

) is correct because it represents the coefficient of

x^{n-2}

in the expansion of

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

297 Easy

297 of 951

Application +4 | -1

The last two digits of the number

19^{9^4}

View answer and explanation

Correct option: A

Option A (19) is correct because by using the binomial theorem and focusing on the last two digits, we find the remainder when divided by 100 is 19.

298 Easy

298 of 951

Analyzing +4 | -1

The largest integer

\lambda

such that

2^\lambda

divides

3^{2^n}-1, n \in N

View answer and explanation

Correct option: D

n+2

Option D (

n+2

) is correct because the largest integer

\lambda

such that

2^\lambda

divides

3^{2^n}-1

n+2

299 Medium

299 of 951

Application +4 | -1

\left( {\begin{array}{cccccccccccccccccccc} n \\ 0 \end{array}} \right)\left( {\begin{array}{cccccccccccccccccccc} n \\ r \end{array}} \right) + \left( {\begin{array}{cccccccccccccccccccc} n \\ 1 \end{array}} \right)\left( {\begin{array}{cccccccccccccccccccc} n \\ {r + 1} \end{array}} \right) + ... + \left( {\begin{array}{cccccccccccccccccccc} n \\ {n - r} \end{array}} \right)\left( {\begin{array}{cccccccccccccccccccc} n \\ n \end{array}} \right) =

View answer and explanation

Correct option: C

\left( {\begin{array}{cccccccccccccccccccc} {2n} \\ {n - r} \end{array}} \right)

Option C

\left( {\begin{array}{cccccccccccccccccccc} {2n} \\ {n - r} \end{array}} \right)

is correct because it represents the coefficient of

x^{n-r}

in the expansion of

(1+x)^n(1+x)^n = (1+x)^{2n}

300 Easy

300 of 951

Application +4 | -1

The ratio of the coefficient of

{x^{15}}

to the term independent of

x

{\left( {{x^2} + \frac{2}{x}} \right)^{15}},

View answer and explanation

Correct option: C

\frac{1}{32}

Option C (1/32) is correct because it represents the ratio of the coefficient of x¹⁵ to the term independent of x after expanding the given binomial expression.

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