Test: Binomial Theorem (Competition Level)- 1


30 Questions MCQ Test Mathematics For JEE | Test: Binomial Theorem (Competition Level)- 1


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QUESTION: 1

The sum of coefficients of (1 + x - 3x2)2134 is

Solution:

For the sum of coefficient, put x = 1, to obtain the sum 
which implies (1 + 1 - 3)2134 = 1

QUESTION: 2

The sum rCr + r+1Cr+2Cr + ....+nCr (n > r) equals 

Solution:

C(n, r)  + c(n -1, r)  +  C(n - 2, r) +  . . .   + C(r, r)
= r+1Cr+1  + r+1Cr + r+2Cr + . . .  .  +  n-1Cr + nCr
= n+1Cr+1 (applying same rule again and again)        
(∴ nCr + nCr-1 = n+1Cr)

QUESTION: 3

The expansion [x2 + (x6 - 1)1/2]5 + [x2 - (x6 - 1)1/2]5 is a polynomial of degree 

Solution:

(x2 + (x6 - 1)1/2)5 + (x2 - (x6 - 1)1/2)5
= 2(5C0(x2)5 + 5C2 (x2)3 (x6 - 1) + 5C4 (x6 - 1)2)
Here last term is of 14 degree.

QUESTION: 4

The term independent of x in 

Solution:

The  general term 


The term independent of x, (or the constant term) corresponds to x18-3r being x0 or 18 - 3r = 0 ⇒ r = 6 .

QUESTION: 5

The value of the greatest term in the expansion of 

Solution:





Hence t1 < t2 < t3 < t4 < t5 <t6 <t7 < t8 < t9 < t10
Hence, t8 is  the greatest   term and its   value is

QUESTION: 6

9n+1 - 8n- 9 is divisible by

Solution:

(1+ 8)n+1 = n+1C0 + n+1C1(8)1+ n+1C2(8)2 + n+1C3 (8)3 + .... + n+1Cn+1(8)n+1     (1)
⇒ 9n+1 = 1+(n+1)x8+n+1C2(8)3 + ....+n+1Cn-1(8)n+1
⇒ 9n+1 - 8n - 9 = (8)2 [n+1C+ n+1C3(8)+n+1C4(8)2+...+n+1Cn+1(8)n-1]
⇒ 9n+1 - 8n - 9 = 64 x an integer ⇒ 9n+1 - 8n - 9 is divisible by 64.

QUESTION: 7

The first integral term in the expansion of 

Solution:


∴ Tr+1 = 9Cr (31/2)9-r (21/3)r = 9Cr 3(9-r)/2.2r/3
For first integral term for r = 3;
T3+1 = 9C333.21 i.e., T3+1 = T4 (4th term)

QUESTION: 8

The number of irrational terms in the expansion of (21/5 + 31/10)55 is

Solution:

(21/5 + 31/10)55 
Total term = 55 + 1 = 56
Tr+1 = 55C2(55-r)/5 3r/10 
Here r = 0, 10, 20, 30, 40, 50
Number of rational terms = 6;    
Number of rational terms = 56 - 6 = 50

QUESTION: 9

The number of terms in the expansion of (2x + 3y- 4z)n is 

Solution:

We have, (2 + 3 - 4z)n = {2 + (3 - 4)}n
= nC0 (2x)n (3y - 4z)° + nC1 (2x)n-1(3y-4z)1 + nC2 (2x)n-2 (3y - 4z)2 +.. + nCn-1(2x)1 (3y - 4z)n-1 + nC2(3y-4z)n
Clearly, the first term in the above   expansion gives one term, second term gives two terms, third term gives three terms and so on.
So, Total number of term = 1 +2+3+...+n+(n+1) = 

QUESTION: 10

In the expansion of the coefficient of x-10 will be 

Solution:

Given expansion is 
∴ General term = Tr+1 = 12Cr (a/x)12-r (bx)r = 12Cr (a)12-r br x-12+2r
Since, we have to find coefficient of x-10
∴ -12 + 2r = -10 ⇒ r = 1
Now, then  coefficient of x-10 is 12C1 (a)11 (b)1 = 12a11b

QUESTION: 11

If (1 + ax)n = 1 + 8x + 24x2 + ….., then the values of a and n are equal to

Solution:




∴ n = 4,a = 2

QUESTION: 12

The product of middle terms in the expansion of is equal to

Solution:

Correct Answer :- A

Explanation : (x + 1/x)11, it has 12 terms

Therefore, there are two middle terms (6th and 7th)

T6T7 = [11C5(x)6(1/x)5] * [11C6(x)5(1/x)6]

11C5 11C6

QUESTION: 13

The middle term in the expansion of (1 – 2x + x2)n is 

Solution:


Here 2n is even   integer, therefore,

term  will be the middle term.
Now, ( n + 1)th term in (1-x)2n = 2nCn(1)2n-n(-x)n = 2nCn(-x)n

QUESTION: 14

The sum of the  binomial coefficients in the expansion of (x-3/4 + ax 5/4)n lies between 200 and 400 and the term independent of x equals 448. The value of a is

Solution:





  

QUESTION: 15

23C0 + 23C2 + 23C4 +...+23C22 equals

Solution:

Given  sum = sum of odd terms 

QUESTION: 16

Solution:

(1- x+2x2)10 = a0+a1x+a2x2 +..a19x19 + a20x20 
Replacing x by-x, (1+x+ 2x2)10 = a- 1x+a2x2...a19x19 +a20x20
Subtracting and putting x = 1, 210 - 410 = 2(a1 +a3 +...+a19)
⇒ 

QUESTION: 17

The greatest coefficient in the expansion of (1 + x)2n + 2 is 

Solution:

Here 2n + 2 is even  
Greatest  coefficient

QUESTION: 18

(nC0)+ (nC1)+ (nC2)2 +...+ (nCn)2 equals

Solution:

(1+x)n = nC0 + nC1x + nC2x2 - ..+ nCnxn....       (i)
(x+1)n = nC0xn + nC1xn-1 + nC2xn-2 + ..... +nCn ......(ii)

Multiply (i) and (ii)  and consider  the coefficient xn of both sides, we have 
 coefficient of 

QUESTION: 19

The value of C1 + 3C3 + 5C5 + 7C7 + ...., where C0 , C3 , C5 , C7 ,..... are binomial coefficients is 

Solution:

We know (1+x)n = C0 +C1x+C2x2 +C3x3 + ...+ Cnxnn(1+ x)r-1 = C1+ 2C2x + 3C3x2 +4C4x3 +...+nCnxn-1 
Putting x =1 and —1, then add
n2n-1 = C1 + 2C2 +3C3 +4C4 +..+nCn
0= C- 2C2 + 3C3 - 4C4 +... ⇒ n2n-2
C1 + 3C3 +5C5 +...

QUESTION: 20

Fractional part of 

Solution:


= 8(1+31)15 = 8{15C15C131+..+15C15(31)15}
278 = 8 + an integer multiple of 31;

QUESTION: 21

(103)86 - (86)103  is divisible by

Solution:

Given expression




QUESTION: 22

23n - bn- a is divisible by 49 then (a, b) is

Solution:




QUESTION: 23

The number of dissimilar terms in the expansion of (a  + b + c)2n+1 – (a + b – c)2n+1 is

Solution:

Given expansion  



Number of dissimilar  term = (2n + 1) + (2n -1) + ....(2n - 3) + .... + 5 + 3 + 1

QUESTION: 24

The numbers of  terms in the expansion of 

Solution:





Where ao = sum of all absolute terms   = 1 +100C2 .2 + ....
Similarly  a1 , a2 , ...a100 and  b1 , b2 ...b100 are coefficients  obtained   after simplification. 
∴ Total  number  of terms  = 1 + 100 + 100 = 201

QUESTION: 25

The coefficient of x50 in the expansion of (1+x)1000 + 2x(1+x)999 + 3x2(1+x)998 + ....+1001x1000 is

Solution:




Subtract  above  equations, 

 

⇒ 


Coefficient of x50 in S = coefficient of x50 in

QUESTION: 26

The coefficient of term independent of x in the expansion of 

Solution:

Given expression 




⇒ 




which is independent of x if 

Hence  required  coefficient

QUESTION: 27

The value of  where [x] represents integral part of ‘x’ is 

Solution:


where I is integer and 0 < f <1.

Now, 


∴ I + f + F is integer.


QUESTION: 28

If in the expansion of (1 + x)m(1 – x)n, the coefficient of x and x2 are 3 and -6 respectively, then m is 

Solution:





Given m - n = 3


⇒ 
⇒ (m - n)2 - m + n = -12 ⇒ m + n = 9 + 12 = 21   (2) using (1)
Solving  (1) and (2), we get m = 12.

QUESTION: 29

 is equal to

Solution:




QUESTION: 30

Coefficient of t24 in (1+t2)12(1+t12)(1+t24) is

Solution:

t24 in (1+t2)12(1+t12)(1+t24) is
Here, coefficient of t24 in 

⇒ coefficient of t24 in

⇒ coefficient of t24 in 


coefficient of 
 coefficient of 

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