Saturday, August 30, 2008

Binomial Theorem - Model Problems - Past JEE

1. The coefficient of x4 in [(x/2)- (3/x²)]10 is

a. 405/256
b. 504/259
c. 450/263
d. none of these

(JEE 1983)

2. Let n be a positive integer. If the coefficient of 2nd, 3rd, and 4th terms in the expansion of (1 + x)n are in A.P. then the value of n is ________________ .
(JEE 1994)


3. If the binomial expansion of (a-b)n, n≥5, the sum of the 5th and 6th terms is zero. The a/b equals

a. (n-5)/6
b. (n-4)/5
c. 5/(n-4)
d. 6/(n-5)

(JEE 2001)


4. The coefficient of t24 in (1+t²)12(1+t12)(1+24) is

a. 12C6 + 3
b. 12C6 + 1
c. 12C6
d. 12C6 + 2

(JEE 2003)


Answers and Solutions

1. The coefficient of x4 in [(x/2)- (3/x²)]10 is

a. 405/256
b. 504/259
c. 450/263
d. none of these

(JEE 1983)

Answer (a)

Solution:
(r+1) th term of the expansion is given by
10Cr (x/2) 10-r(-3/x²) r

= 10Cr (x) 10-r(1/ x²)r(1/2) 10-r(-3) r

= 10Cr(-3) r(1/2) 10-r(x) 10-3r

In this term (x) 10-3r is equal to x4
 10-3r = 4 or r = 2

So coefficient =
10Cr(-3) r(1/2) 10-r

10C2(-3) 2(1/2) 10-2

= [10*9/2]*9*(1/28)

= (45*9)/256
= 405/256




2. Let n be a positive integer. If the coefficients of 2nd, 3rd, and 4th terms in the expansion of (1 + x)n are in A.P. then the value of n is ________________ .
(JEE 1994)


Answer: 7

Solution:

The question requires concepts from binomial theorem and concept from arithmetic progression.

Coefficients of 2nd, 3rd, and 4th terms in the expansion of (1 + x)n are
nC1, nC2, nC3

According to arithmetic progression properties, three terms can be taken as a-d, a, and a+d. Hence sum of first and third terms = 2*second term

=> 2(nC2) = nC1 + nC3

=> 2 n(n-1)/2 = n + [n(n-1)(n-2)/6]

=> n-1 = 1 +[(n² – 3n+2)/6]

=> n² – 3n+2 = 6n-12
=> n² – 9n+14 = 0
=> (n-2)(n-7) = 0

Since 4th term will be there only when n>2, n is equal to 7.





3. If the binomial expansion of (a-b)n, n≥5, the sum of the 5th and 6th terms is zero. The a/b equals

a. (n-5)/6
b. (n-4)/5
c. 5/(n-4)
d. 6/(n-5)

(JEE 2001)

Answer: (b)

Solution

5th term = 4+1th term = nC4 a n-4(-b) 4

6th term = 5+1th term = nC5 a n-5(-b) 5

As sum of 5th terms and 6th terms is zero

nC4 a n-4(-b) 4 + nC5 a n-5(-b) 5 = 0

nC4 a n-4(b) 4 = nC5 a n-5(b) 5

a n-4(b) 4/ a n-5(b) 5 = nC5/nC4

a/b = [n!/(n-5)!(5!)]/[n!/(n-4)!(4!)

= (n-4)!/(5(n-5)!)
= (n-4)/5



4. The coefficient of t24 in (1+t²)12(1+t12)(1+24) is

a. 12C6 + 3
b. 12C6 + 1
c. 12C6
d. 12C6 + 2

(JEE 2003)

Answer: (d)

Solution:
Multiplying the last two expressions we have to find coefficient of t24 in (1+t²)12(1+t12+24)

t24 term will come as a sum of three terms in the expanded expression

1st term is t24 term in 1st expression multiplied by 1 in the second expression.
2nd term is t12 term in 1st expression multiplied by t12 term in the second expression.
3rd term is 1 in the first expression multiplied by t24 term in the second expression.

Coefficients of 1 and 3 terms are 1 and 1.

Coefficient of 2 term = coefficient t12 term in 1 st expression which is (1+t²)12

7th term or 6+1th term in the expansion will have t12 term

Hence coefficient of 7th term = 12C6

Therefore coefficient of t24 term will be

12C6 + 2

1 comment:

truthfinder said...

thanks for the questions!