π Binomial Theorem Solved Problems for JEE Exam β Class 11 Math
π Author: Neeraj Anand
π Published by: ANAND TECHNICAL PUBLISHERS
π Available at: ANAND CLASSES
π For CBSE Board & JEE Mains/Advanced Aspirants
π’ Mastering Binomial Theorem with Solved Problems
The Binomial Theorem is an essential topic in Class 11 Mathematics, particularly for students preparing for CBSE Board Exams, JEE Mains, and JEE Advanced. It provides a powerful method for expanding algebraic expressions of the form (a + b)βΏ and is widely used in problem-solving.
In this article, we will cover:
βοΈ Basic & Advanced Problems on Binomial Expansion
βοΈ Application-Based Problems for JEE
βοΈ Shortcuts & Tricks for Fast Calculation
βοΈ Important Questions with Solutions
π₯ Download the PDF by Neeraj Anand (ANAND CLASSES) for detailed explanations and extra practice questions!
Solved Examples
Example.1 :: Find the positive value of Ξ» for which the coefficient of x2Β in the expression x2[βxΒ + (Ξ»/x2)]10Β is 720.
Solution:
β x2 [10Cr . (βx)10-r . (Ξ»/x2)r] = x2 [10Cr . Ξ»r . x(10-r)/2 . x-2r]
= x2 [10Cr . Ξ»r . x(10-5r)/2]
Therefore, r = 2
Hence, 10C2 . Ξ»2 = 720
β Ξ»2 = 16
β Ξ» = Β±4.
Example.2: Let (x + 10)50Β + (x β 10)50Β = a0Β +a1x + a2Β x2Β + . . . . . + a50Β x50Β for all x βR, then a2/a0Β is equal to?
Solution:
β (x + 10)50 + (x β 10)50:
a2 = 2 Γ 50C2 Γ 1048
a0 = 2 Γ 1050
β a2/a0 = 50C2/102 = 12.25.
Example.3 : Find the coefficient of x9Β in the expansion ofΒ (1 + x) (1 + x2Β ) (1 + x3) . . . . . . (1 + x100).
Solution:
x9 can be formed in 8 ways.
i.e., x9 x1+8 x2+7 x3+6 x4+5, x1+3+5, x2+3+4
β΄ The coefficient of x9 = 1 + 1 + 1 + . . . . + 8 times = 8.
Example.4 : Find the total number of terms in the expansion ofΒ (x + a)100Β + (x β a)100.
Solution:
β (x + a)100 + (x β a)100 = 2[100C0 x100. 100C2 x98 . a2 + . . . . . . + 100C100 a100]
β΄ Total Terms = 51
Example.5 : Find the coefficient of t4Β in the expansion ofΒ [(1-t6)/(1 β t)].
Solution:
β [(1-t6)/(1 β t)] = (1 β t18 β 3t6 + 3t12) (1 β t)-3
Coefficient of t in (1 β t)-3 = 3 + 4 β 1
C4 = 6C2 =15
The coefficient of xr in (1 β x)-n = (r + n β 1) Cr
Example.6: Find the coefficient of x4 in the expansion of (1 + x + x2Β +x3)11.
Solution:
By expanding the given equation using the expansion formula, we can get the coefficient x4
i.e. 1 + x + x2 + x3 = (1 + x) + x2 (1 + x) = (1 + x) (1 + x2)
β (1 + x + x2 + x3) x11 = (1+x)11 (1+x2)11
= 1+ 11C1 x2 + 11C2 x2 + 11C3 x3 + 11C4 x4 . . . . . . .
= 1 + 11C1 x2 + 11C2 x4 + . . . . . .
To find the term in from the product of two brackets on the right-hand-side, consider the following products terms as
= 1 Γ 11C2 x4 + 11C2 x2 Γ 11C1 x2 + 11C4 x4
= 11C2 + 11C2 Γ 11C1 + 11C4 ] x4
β [55 + 605 + 330] x4 = 990x4
β΄ The coefficient of x4 is 990.
Example.7 : Find the number of terms free from the radical sign in the expansion ofΒ (β5 +Β 4βn)100.
Solution:
Tr+1 = 100Cr . 5(100 β r)/2 nr/4
Where r = 0, 1, 2, . . . . . . , 100
r must be 0, 4, 8, β¦ 100
The number of rational terms = 26
Example.8 : Find the degree of the polynomial [x + {β(3(3-1))}1/2]5Β + [x + {β(3(3-1))}1/2]5.
Solution:
[x + { β(3(3-1)) }1/2 ]5:
= 2 [5C0 x5 + 5C2 x5 (x3 β 1) + 5C4 . x . (x3 β 1)2]
Therefore, the highest power = 7.
Example.9 : Find the last three digits of 2726.
Solution:
By reducing 2726 into the form (730 β 1)n and using simple binomial expansion, we will get the required digits.
We have 272 = 729
Now 2726 = (729)13 = (730 β 1)13
= 13C0 (730)13 β 13C1 (730)12 + 13C2 (730)11 β . . . . . β 13C10 (730)3 + 13C11(730)2 β 13C12 (730) + 1
= 1000m + [(13 Γ 12)]/2] Γ (14)2 β (13) Γ (730) + 1
Where βmβ is a positive integer
= 1000m + 15288 β 9490 = 1000m + 5799
Thus, the last three digits of 17256 are 799.
π Why Choose Neeraj Anandβs Book?
βοΈ Comprehensive Explanation β Step-by-step solutions
βοΈ JEE-Focused Approach β Covers JEE Mains & Advanced syllabus
βοΈ Practice Questions β Includes previous year JEE questions
βοΈ Tricks & Shortcuts β Solve problems faster and efficiently
π₯ Download the PDF & Start Practicing Today!
π Master Binomial Theorem & Score High in JEE!