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Test: Binomial Theorem- 1 - JEE MCQ


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25 Questions MCQ Test Physics for JEE Main & Advanced - Test: Binomial Theorem- 1

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Test: Binomial Theorem- 1 - Question 1

If n is a rational number, which is not a whole number, then the number of terms in the expansion of (1+x)n, |x| < 1, is

Test: Binomial Theorem- 1 - Question 2

If n is a +ve integer, then the binomial coefficients equidistant from the beginning and the end in the expansion of (x+a)n are

Detailed Solution for Test: Binomial Theorem- 1 - Question 2

(x+a)n = nC0 xn + nC1 x(n-1) a1 + nC2 x(n-2) a2 + ..........+ nC(n-1) xa(n-1) + nCn  an
Now, nC0 = nCn, nC1 = nCn-1,    nC2 = nCn-2,........
therefore, nCr = nCn-r
The binomial coefficients equidistant from the beginning and the end in the expansion of (x+a)n are equal.

Test: Binomial Theorem- 1 - Question 3

If the rth term in the expansion of  contains x4, then r =

Detailed Solution for Test: Binomial Theorem- 1 - Question 3

Test: Binomial Theorem- 1 - Question 4

If the coefficients of x−7 and x−8 in the expansion of are equal then n =

Detailed Solution for Test: Binomial Theorem- 1 - Question 4

coefficient of x-7 in [2+1/3x]n is nC7(2)n-7 (1/3)7
coefficient of x-8 in [2+1/3x]n is nC8(2)n-8 (1/3)8
nC7 × (2n-7) × 1/37 = nC8 × (2n-8) × 1/38
nC8/nC7 = 6
⟹ (n-7)/8 = 6
⟹ n = 55

Test: Binomial Theorem- 1 - Question 5

The largest term in the expansion of (1+x)19 when x = 1/2 is

Detailed Solution for Test: Binomial Theorem- 1 - Question 5

 Let Tr and Tr+1 denote the rthand(r+1)th terms in
the expansion of (1+x)19
 Tr = 19Cr-1 xr-1 and Tr+1 = 19Cr xr .
∴Tr+1/Tr = 19Cr xr/(19Cr-1 xr-1)
⇒Tr+1/Tr = 19Cr  19Cr-1 x
⇒Tr+1/Tr = 19!/(19−r)!r! × x[(19−r+1)!(r−1)]/10!
⇒Tr+1/Tr = x(20−r)/r
⇒Tr+1/Tr = (20−r)/r × 1/2   [∵x not equal to 1/2]
Now
Tr+1/Tr > 1
⇒ (20−r)/r × 1/2 > 1
⇒ 20 > 3r
r > 20/3
∴ (6+1)th i.e. 7th term is the greatest term.

Test: Binomial Theorem- 1 - Question 6

If coefficients of three successive terms in the expansion of (x+1)n are in the ratio 1 : 3 : 5, then n is equal to

Test: Binomial Theorem- 1 - Question 7

The exponent of power of x occurring in the 7th term of expansion of 

Detailed Solution for Test: Binomial Theorem- 1 - Question 7

Test: Binomial Theorem- 1 - Question 8

The term independent of x in the expansion of 

Test: Binomial Theorem- 1 - Question 9

The number of dissimilar terms in the expansion of (a+b)n is n + 1, therefore number of dissimilar terms in the expansion of (a+b+c)12 is

Test: Binomial Theorem- 1 - Question 10

The term containing x3 in the expansion of (x−2y)7 is

Test: Binomial Theorem- 1 - Question 11

The coefficients of xp and xq (p, q are + ve integers) in the binomial expansion of (1+x)p+q are

Detailed Solution for Test: Binomial Theorem- 1 - Question 11

The ratio of the coefficients of xp and xq in the expansion of (1+x)p+q is

Test: Binomial Theorem- 1 - Question 12

If 2nd, 3rd and 4th terms in the expansion of (x+a)n are 240, 720 and 1080 respectively, then the value of n is

Detailed Solution for Test: Binomial Theorem- 1 - Question 12

General term Tr+1 of (x+y)n is given by 
Tr+1 = nCr xn-r yr
T2 = nC2 xn-2 y = 240
T3 = nC3 xn-3 y2 = 720
T4 = nC4 xn-4 y3 = 1080
T3/T2 = [(n-1)/2] * [y/x] = 3......(1)
T4/T2 = {[(n-1)(n-2)]/(3*2)} * x2/y2 = 9/2
T4/T3 = [(n-2)/3] * [y/x] = 3/2...(2)
Dividing 1 by 2
[(n-1)/2] * [3/(n-2)] = 2
⇒ 3n−3 = 4n−8
⇒ 5 = n

Test: Binomial Theorem- 1 - Question 13

If the first three terms in the expansion of (x+a)n are 729, 7290 and 30375 respectively, then the value of n is

Detailed Solution for Test: Binomial Theorem- 1 - Question 13




Now, dividing (1) by (2) we get,





Now, Dividing (2) by (3) we get,


Test: Binomial Theorem- 1 - Question 14

The coefficient of xn in expansion of (1+x)(1−x)n is

Detailed Solution for Test: Binomial Theorem- 1 - Question 14

We can expand (1+x)(1−x)n as 
= (1+x)(nC0​− nC1​.x + nC2​.x+ ........ +(−1)n nCn​xn)


Coefficient of xn is 
(−1)n nC​+ (−1)n−1 nCn−1
​=(−1)n(1−n)

Test: Binomial Theorem- 1 - Question 15

The two consecutive terms in the expansion of (3+2x)74, which have equal coefficients, are

Detailed Solution for Test: Binomial Theorem- 1 - Question 15

Tr+1 = 74Cr3 373-r (2x)r
Tr+2 = 74Cr+1 373-r (2x)r+1
 as co−efficients are equal 
74Cr3 373-r 2r = 74Cr+1 373-r 2r+1
 ((74!/74−r)!r!)×=(74!/(73−r)!(r+1)!) × 2
3(73−r)!(r+1)! = 2(74−r)!r!
= 3(73−r)!(r+1)! = 2(74−r)!(73−r)!r!
= 3(r+1) = 2(74−r)
= 3r+3 = 148−2r
∴ 5r = 148−3
r = (145/5) 
r = 29
Hence terms are (29+1)th and(29+2)nd or 30th and 31th term

Test: Binomial Theorem- 1 - Question 16

The coefficient of y in the expansion of (y² + c/y)5 is 

Detailed Solution for Test: Binomial Theorem- 1 - Question 16

Given, binomial expression is (y² + c / y)5 
Now, Tr+1 = 5Cr × (y²)5-r × (c / y)r 
= 5Cr × y10-3r × Cr 
Now, 10 – 3r = 1 
⇒ 3r = 9 
⇒ r = 3 
So, the coefficient of y = 5C3 × c³ = 10c³

Test: Binomial Theorem- 1 - Question 17

The term independent of x in the expansion of 

Test: Binomial Theorem- 1 - Question 18

The coefficient of y in the expansion of 

Detailed Solution for Test: Binomial Theorem- 1 - Question 18

5C0 (c/y)0 (y2)5-0 + 5C1 (c/y)1 (y2)5-1 + 5C2 (c/y)2 (y2)5-2 +....... + 5C5 (c/y)5 (y2)5-5
∑(r = 0 to 5) 5Cr (c/y)r (y2)5-r
We need coefficient of y ⇒2(5−r)−r=1
⇒ 10 − 3r = 1
⇒ r = 3
So, cofficient of y = 5C3.C3
= 10C3

Test: Binomial Theorem- 1 - Question 19

The number o subsets of a set containing n distinct elements is

Test: Binomial Theorem- 1 - Question 20

The number of terms in the expansion of [(x+4y)3(x−4y)3]2 is

Test: Binomial Theorem- 1 - Question 21

Test: Binomial Theorem- 1 - Question 22

Coefficient of x5 in the expansion of (1+x2)5(1+x)4 is  

Detailed Solution for Test: Binomial Theorem- 1 - Question 22

Co-efficient of x5 in (1+x2)5 (1+x)4
= {5C0 (1)0 + 5C1 (1)x2 + 5C2 (1)(x2)2 + 5C3 (1)(x2)3 + 5C4 (1)(x2)4 + 5C5 (1)(x2)5} * {4C0 (1) + 4C1 (x) + 4C2 (x)2 + 4C3 (x)3 + 4C4 (x)4}
= {5C2(x)4 * 4C1(x) + 5C1(x)4 + 5C1(x)2 * 4C3(x)3
∴ coefficient of x5
=> (5×4)/(2×1) × 4)x5 + (5×4)x5
⇒(10×4+20)x5
⇒ 60x5

Test: Binomial Theorem- 1 - Question 23

If three successive terms in the expansion of (1+x)a have their coefficients in the ratio 6 : 33 : 110, then n is equal to

Detailed Solution for Test: Binomial Theorem- 1 - Question 23

(1 + x)ⁿ  = 1 + ⁿC₁x¹  + ⁿC₂x²+..............................+ⁿCnxⁿ
Three consecutive terms coefficients
ⁿCₐ  : ⁿCₐ₊₁  :  ⁿCₐ₊₂  :::  6 : 33 :  110
⇒  ⁿCₐ  = 6K  =>  n!/(a!)(n-a)!  = 6K  => n! = 6K (a!)(n-a)!
ⁿCₐ₊₁   = 33K   => n!/(a+1)!(n-a-1)! = 33K  => n!  = 33K (a+1)!(n-a-1)!
ⁿCₐ₊₂ = 110K  => n!/(a + 2)!(n-a-2)! = 110K  => n! = 110K (a + 2)!(n-a-2)!
6K (a!)(n-a)!  = 33K (a+1)!(n-a-1)!
⇒ 2  (a!)(n-a)(n-a - 1)! = 11 (a + 1)a! (n-a-1)!
⇒ 2(n-a) = 11(a + 1)
⇒ 2n - 2a = 11a + 11
⇒ 2n = 13a + 11
⇒ 13a = 2n - 11
33K (a+1)!(n-a-1)!  = 110K (a + 2)!(n-a-2)!
⇒ 3 (a+1)!(n-a-1)(n-a-2)!  = 10 (a + 2)(a + 1)!(n-a-2)!
⇒ 3 (n - a - 1) = 10(a + 2)
⇒ 3n - 3a - 3 = 10a + 20
⇒  3n = 13a + 23
⇒ 13a = 3n - 23
2n - 11 = 3n - 23
⇒ n = 12

Test: Binomial Theorem- 1 - Question 24

(1.003)is nearly equal to

Test: Binomial Theorem- 1 - Question 25

5th term from the end in the expansion of 

Detailed Solution for Test: Binomial Theorem- 1 - Question 25

5th term in the expansion of (x2/2 − 2/x2)12 is
Tr+ 1= nCr xr* y(n−r)
T5 = 12C4[(x2/2)4 (-2/x2)8]
= (12! * x8 * 28)/ (4! * 8! *24 * x16)
= 7920x−4

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