📚 Binomial Theorem Solved Problems for JEE Exam – Class 11 Math
📝 Author: Neeraj Anand
🏛 Published by: ANAND TECHNICAL PUBLISHERS
📖 Available at: ANAND CLASSES
🔍 For CBSE Board & JEE Mains/Advanced Aspirants
🔢 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)ⁿ and is widely used in problem-solving.
In this article, we will cover:
✔️ Basic & Advanced Problems on Binomial Expansion
✔️ Application-Based Problems for JEE
✔️ Shortcuts & Tricks for Fast Calculation
✔️ Important Questions with Solutions
📥 Download the PDF by Neeraj Anand (ANAND CLASSES) for detailed explanations and extra practice questions!
Solved Examples
Example.1 :: Find the positive value of λ for which the coefficient of x2 in the expression x2[√x + (λ/x2)]10 is 720.
Solution:
⇒ x2 [10Cr . (√x)10-r . (λ/x2)r] = x2 [10Cr . λr . x(10-r)/2 . x-2r]
= x2 [10Cr . λr . x(10-5r)/2]
Therefore, r = 2
Hence, 10C2 . λ2 = 720
⇒ λ2 = 16
⇒ λ = ±4.
Example.2: Let (x + 10)50 + (x – 10)50 = a0 +a1x + a2 x2 + . . . . . + a50 x50 for all x ∈R, then a2/a0 is equal to?
Solution:
⇒ (x + 10)50 + (x – 10)50:
a2 = 2 × 50C2 × 1048
a0 = 2 × 1050
⇒ a2/a0 = 50C2/102 = 12.25.
Example.3 : Find the coefficient of x9 in the expansion of (1 + x) (1 + x2 ) (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
∴ The coefficient of x9 = 1 + 1 + 1 + . . . . + 8 times = 8.
Example.4 : Find the total number of terms in the expansion of (x + a)100 + (x – a)100.
Solution:
⇒ (x + a)100 + (x – a)100 = 2[100C0 x100. 100C2 x98 . a2 + . . . . . . + 100C100 a100]
∴ Total Terms = 51
Example.5 : Find the coefficient of t4 in the expansion of [(1-t6)/(1 – t)].
Solution:
⇒ [(1-t6)/(1 – t)] = (1 – t18 – 3t6 + 3t12) (1 – t)-3
Coefficient of t in (1 – t)-3 = 3 + 4 – 1
C4 = 6C2 =15
The coefficient of xr in (1 – x)-n = (r + n – 1) Cr
Example.6: Find the coefficient of x4 in the expansion of (1 + x + x2 +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)
⇒ (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 × 11C2 x4 + 11C2 x2 × 11C1 x2 + 11C4 x4
= 11C2 + 11C2 × 11C1 + 11C4 ] x4
⇒ [55 + 605 + 330] x4 = 990x4
∴ The coefficient of x4 is 990.
Example.7 : Find the number of terms free from the radical sign in the expansion of (√5 + 4√n)100.
Solution:
Tr+1 = 100Cr . 5(100 – r)/2 nr/4
Where r = 0, 1, 2, . . . . . . , 100
r must be 0, 4, 8, … 100
The number of rational terms = 26
Example.8 : Find the degree of the polynomial [x + {√(3(3-1))}1/2]5 + [x + {√(3(3-1))}1/2]5.
Solution:
[x + { √(3(3-1)) }1/2 ]5:
= 2 [5C0 x5 + 5C2 x5 (x3 – 1) + 5C4 . x . (x3 – 1)2]
Therefore, the highest power = 7.
Example.9 : Find the last three digits of 2726.
Solution:
By reducing 2726 into the form (730 – 1)n and using simple binomial expansion, we will get the required digits.
We have 272 = 729
Now 2726 = (729)13 = (730 – 1)13
= 13C0 (730)13 – 13C1 (730)12 + 13C2 (730)11 – . . . . . – 13C10 (730)3 + 13C11(730)2 – 13C12 (730) + 1
= 1000m + [(13 × 12)]/2] × (14)2 – (13) × (730) + 1
Where ‘m’ is a positive integer
= 1000m + 15288 – 9490 = 1000m + 5799
Thus, the last three digits of 17256 are 799.
📚 Why Choose Neeraj Anand’s Book?
✔️ Comprehensive Explanation – Step-by-step solutions
✔️ JEE-Focused Approach – Covers JEE Mains & Advanced syllabus
✔️ Practice Questions – Includes previous year JEE questions
✔️ Tricks & Shortcuts – Solve problems faster and efficiently
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