Anand Classes provides NCERT Solutions for Class 11 Maths Chapter 8 – Binomial Theorem (Miscellaneous Exercise) with step-by-step explanations based on the latest NCERT and CBSE syllabus. These detailed solutions help students master important concepts such as binomial expansion, general term, middle term, coefficients, and properties of binomial expressions. Each problem is solved clearly to strengthen mathematical understanding and prepare students for CBSE board exams, JEE Main, JEE Advanced, NDA, and other competitive exams. Click the print button to download study material and notes in PDF format.
Question.1 : If $a$ and $b$ are distinct integers, prove that $a – b$ is a factor of $a^n – b^n$, whenever $n$ is a positive integer.
Solution :
To prove that $(a – b)$ divides $(a^n – b^n)$.
We know that
$$a = (a – b) + b$$
Then,
$$a^n = \big((a – b) + b\big)^n$$
Using the Binomial Theorem,
$$a^n = \sum_{k=0}^{n} {}^{n}C_{k}\; (a – b)^{n-k} \;b^{k}$$
Expanding,
$a^n = {}^{n}C_{0}(a – b)^{n} + {}^{n}C_{1}(a – b)^{n-1}b + {}^{n}C_{2}(a – b)^{n-2}b^{2} + \dots + {}^{n}C_{n-1}(a – b)b^{n-1} + {}^{n}C_{n}b^{n}$
Since ${}^{n}C_{0} = 1$ and ${}^{n}C_{n} = 1$,
$$a^n = (a – b)^{n} + n(a – b)^{n-1}b + {}^{n}C_{2}(a – b)^{n-2}b^{2} + \dots + {}^{n}C_{n-1}(a – b)b^{n-1} + b^{n}$$
Subtracting $b^{n}$ from both sides,
$$a^n – b^n = (a – b)^{n} + n(a – b)^{n-1}b + {}^{n}C_{2}(a – b)^{n-2}b^{2} + \dots + {}^{n}C_{n-1}(a – b)b^{n-1}$$
Now take $(a – b)$ common,
$$a^n – b^n = (a – b)\Big[(a – b)^{n-1} + n(a – b)^{n-2}b + {}^{n}C_{2}(a – b)^{n-3}b^{2} + \dots + {}^{n}C_{n-1}b^{n-1}\Big]$$
Let
$$k = (a – b)^{n-1} + n(a – b)^{n-2}b + {}^{n}C_{2}(a – b)^{n-3}b^{2} + \dots + {}^{n}C_{n-1}b^{n-1}$$
Hence,
$$a^n – b^n = (a – b)k$$
Therefore, $(a – b)$ is a factor of $(a^n – b^n)$ whenever $n$ is a positive integer.
✅ Hence Proved.
Question.2 : Evaluate $(\sqrt{3} + \sqrt{2})^6 – (\sqrt{3} – \sqrt{2})^6$.
Solution :
Using the Binomial Theorem, the expressions $(a + b)^6$ and $(a – b)^6$ can be expanded as follows:
$(a + b)^6 = {}^{6}C_{0}a^{6} + {}^{6}C_{1}a^{5}b + {}^{6}C_{2}a^{4}b^{2} + {}^{6}C_{3}a^{3}b^{3} + {}^{6}C_{4}a^{2}b^{4} + {}^{6}C_{5}ab^{5} + {}^{6}C_{6}b^{6}$
$$(a – b)^6 = {}^{6}C_{0}a^{6} – {}^{6}C_{1}a^{5}b + {}^{6}C_{2}a^{4}b^{2} – {}^{6}C_{3}a^{3}b^{3} + {}^{6}C_{4}a^{2}b^{4} – {}^{6}C_{5}ab^{5} + {}^{6}C_{6}b^{6}$$
Now, subtracting the second expansion from the first:
$$(a + b)^6 – (a – b)^6 = 2\left[{}^{6}C_{1}a^{5}b + {}^{6}C_{3}a^{3}b^{3} + {}^{6}C_{5}ab^{5}\right]$$
Substituting $a = \sqrt{3}$ and $b = \sqrt{2}$, we get
$$(\sqrt{3} + \sqrt{2})^6 – (\sqrt{3} – \sqrt{2})^6 = 2\left[6(\sqrt{3})^5(\sqrt{2}) + 20(\sqrt{3})^3(\sqrt{2})^3 + 6(\sqrt{3})(\sqrt{2})^5\right]$$
Simplify each term:
$(\sqrt{3})^5 = 9\sqrt{3}, \quad (\sqrt{2})^5 = 4\sqrt{2}, \quad (\sqrt{3})^3(\sqrt{2})^3 = 6\sqrt{6}$
So,
$$(\sqrt{3} + \sqrt{2})^6 – (\sqrt{3} – \sqrt{2})^6 = 2\left[6(9\sqrt{3}\sqrt{2}) + 20(6\sqrt{6}) + 6(4\sqrt{2}\sqrt{3})\right]$$
$$(\sqrt{3} + \sqrt{2})^6 – (\sqrt{3} – \sqrt{2})^6 = 2\left[54\sqrt{6} + 120\sqrt{6} + 24\sqrt{6}\right]$$
$$(\sqrt{3} + \sqrt{2})^6 – (\sqrt{3} – \sqrt{2})^6 = 2(\sqrt{6})(198)$$
$$(\sqrt{3} + \sqrt{2})^6 – (\sqrt{3} – \sqrt{2})^6 = 396\sqrt{6}$$
✅ Final Answer:
$$\boxed{396\sqrt{6}}$$
Question.3 : Find the value of
$$(a^2 + \sqrt{a^2 – 1})^4 + (a^2 – \sqrt{a^2 – 1})^4$$
Using the Binomial Theorem:
$$(x + y)^4 = {}^{4}C_{0}x^4 + {}^{4}C_{1}x^3y + {}^{4}C_{2}x^2y^2 + {}^{4}C_{3}xy^3 + {}^{4}C_{4}y^4$$
$$(x – y)^4 = {}^{4}C_{0}x^4 – {}^{4}C_{1}x^3y + {}^{4}C_{2}x^2y^2 – {}^{4}C_{3}xy^3 + {}^{4}C_{4}y^4$$
Adding both equations:
$$(x + y)^4 + (x – y)^4 = 2\big({}^{4}C_{0}x^4 + {}^{4}C_{2}x^2y^2 + {}^{4}C_{4}y^4\big)$$
Substitute the values:
$${}^{4}C_{0} = 1,\quad {}^{4}C_{2} = 6,\quad {}^{4}C_{4} = 1$$
Hence,
$$(x + y)^4 + (x – y)^4 = 2(x^4 + 6x^2y^2 + y^4)$$
Now, let
$$x = a^2 \quad \text{and} \quad y = \sqrt{a^2 – 1}$$
Substitute these values:
$$(a^2 + \sqrt{a^2 – 1})^4 + (a^2 – \sqrt{a^2 – 1})^4 = 2\big((a^2)^4 + 6(a^2)^2(\sqrt{a^2 – 1})^2 + (\sqrt{a^2 – 1})^4\big)$$
Simplify step by step:
$$(a^2 + \sqrt{a^2 – 1})^4 + (a^2 – \sqrt{a^2 – 1})^4 = 2\big(a^8 + 6a^4(a^2 – 1) + (a^2 – 1)^2\big)$$
$$(a^2 + \sqrt{a^2 – 1})^4 + (a^2 – \sqrt{a^2 – 1})^4 = 2\big(a^8 + 6a^6 – 6a^4 + a^4 – 2a^2 + 1\big)$$
$$(a^2 + \sqrt{a^2 – 1})^4 + (a^2 – \sqrt{a^2 – 1})^4 = 2\big(a^8 + 6a^6 – 5a^4 – 2a^2 + 1\big)$$
Final Answer:
$$\boxed{2a^8 + 12a^6 – 10a^4 – 4a^2 + 2}$$
Question.4 : Find an approximation of $(0.99)^5$ using the first three terms of its expansion.
Solution:
We can write
$$0.99 = 1 – 0.01$$
Now, applying the Binomial Theorem,
$$(1 – 0.01)^5 = {}^{5}C_{0}(1)^5 – {}^{5}C_{1}(1)^4(0.01) + {}^{5}C_{2}(1)^3(0.01)^2 – {}^{5}C_{3}(1)^2(0.01)^3 + \cdots$$
Taking the first three terms of this expansion, we get:
$$(0.99)^5 = {}^{5}C_{0}(1)^5 – {}^{5}C_{1}(1)^4(0.01) + {}^{5}C_{2}(1)^3(0.01)^2$$
Substitute the binomial coefficients:
$$(0.99)^5 = 1 – 5(0.01) + 10(0.01)^2$$
Simplify step by step:
$$(0.99)^5 = 1 – 0.05 + 0.001$$
$$(0.99)^5 = 0.951$$
Approximation:
$$\boxed{(0.99)^5 \approx 0.951}$$
Question.5 : Expand using Binomial Theorem $$(1+\frac{x}{2}-\frac{2}{x})^4, \quad x\neq 0$$
Solution:
Step 1: Grouping terms for binomial expansion
$$(1+\frac{x}{2}-\frac{2}{x})^4 = \Big[(1+\frac{x}{2}) – \frac{2}{x}\Big]^4$$
Comparing with the standard binomial form $(a+b)^n$, we have:
$$\mathbf{a = (1+\frac{x}{2})}, \quad \mathbf{b = -\frac{2}{x}}, \quad \mathbf{n = 4}$$
Step 2: Apply Binomial Theorem using $^nC_r$ notation
$$(a+b)^n = \sum_{r=0}^{n} {^nC_r} a^{n-r} b^r$$
$$(1+\frac{x}{2}-\frac{2}{x})^4 = {^4C_0} (1+\frac{x}{2})^4 – {^4C_1} (1+\frac{x}{2})^3 \frac{2}{x} $$$$+ {^4C_2} (1+\frac{x}{2})^2 \left(\frac{2}{x}\right)^2 – {^4C_3} (1+\frac{x}{2}) \left(\frac{2}{x}\right)^3 + {^4C_4} \left(\frac{2}{x}\right)^4$$
Simplifying the coefficients:
$$(1+\frac{x}{2}-\frac{2}{x})^4 = (1+\frac{x}{2})^4 – \frac{8}{x} (1+\frac{x}{2})^3 $$ $$ + \frac{24}{x^2} (1+\frac{x}{2})^2 – 4 \left(\frac{8}{x^3} + \frac{4}{x^2}\right) + \frac{16}{x^4}$$
Step 3: Expand powers of $(1+\frac{x}{2})$
$$(1+\frac{x}{2})^2 = 1 + x + \frac{x^2}{4}$$
$$(1+\frac{x}{2})^3 = 1 + \frac{3x}{2} + \frac{3x^2}{4} + \frac{x^3}{8}$$
$$(1+\frac{x}{2})^4 = 1 + 2x + \frac{3x^2}{2} + \frac{x^3}{2} + \frac{x^4}{16}$$
Step 4: Substitute back into main expression
$$(1+\frac{x}{2}-\frac{2}{x})^4 = 1 + 2x + \frac{3x^2}{2} + \frac{x^3}{2} + \frac{x^4}{16}$$
$$- \frac{8}{x} \Big(1 + \frac{3x}{2} + \frac{3x^2}{4} + \frac{x^3}{8}\Big)$$
$$+ \frac{24}{x^2} \Big(1 + x + \frac{x^2}{4}\Big)$$
$$- 4 \Big(\frac{8}{x^3} + \frac{4}{x^2}\Big) + \frac{16}{x^4}$$
Step 5: Simplify terms
$$(1+\frac{x}{2}-\frac{2}{x})^4 = \frac{16}{x} + \frac{8}{x^2} – \frac{32}{x^3} + \frac{16}{x^4} – 4x + \frac{x^2}{2} + \frac{x^3}{2} + \frac{x^4}{16} – 5$$
✅ Final Answer
$$\boxed{\frac{16}{x} + \frac{8}{x^2} – \frac{32}{x^3} + \frac{16}{x^4} – 4x + \frac{x^2}{2} + \frac{x^3}{2} + \frac{x^4}{16} – 5}$$
Question.6 : Find the expansion of $$(3x^2 – 2ax + 3a^2)^3$$ using the Binomial Theorem.
Solution:
Step 1: Grouping terms for binomial expansion
We can write:
$$(3x^2 – 2ax + 3a^2)^3 = \Big[3x^2 – (2ax – 3a^2)\Big]^3$$
Comparing with the standard binomial form $(a+b)^n$, we have:
$$\mathbf{a = 3x^2}, \quad \mathbf{b = – (2ax – 3a^2)}, \quad \mathbf{n = 3}$$
Step 2: Apply Binomial Theorem using $^nC_r$ notation
$$(a+b)^n = \sum_{r=0}^{n} {^nC_r} a^{n-r} b^r$$
So:
$$(3x^2 – (2ax – 3a^2))^3 = {^3C_0} (3x^2)^3 + {^3C_1} (3x^2)^2 \big(- (2ax – 3a^2)\big)$$ $$ + {^3C_2} (3x^2) \big(- (2ax – 3a^2)\big)^2 + {^3C_3} \big(- (2ax – 3a^2)\big)^3$$
Step 3: Simplify coefficients
$$(3x^2)^3 = 27x^6$$
$${^3C_1} (3x^2)^2 (- (2ax – 3a^2)) = 3 \cdot 9x^4 \cdot (-1)(2ax – 3a^2) = -54ax^5 + 81a^2x^4$$
$${^3C_2} (3x^2) \big(- (2ax – 3a^2)\big)^2 = 3 \cdot 3x^2 \cdot (2ax – 3a^2)^2 = 9x^2 (2ax – 3a^2)^2$$
$${^3C_3} \big(- (2ax – 3a^2)\big)^3 = – (2ax – 3a^2)^3$$
So the expression becomes:
$$27x^6 – 54ax^5 + 81a^2x^4 + 9x^2 (2ax – 3a^2)^2 – (2ax – 3a^2)^3$$
Step 4: Expand $(2ax – 3a^2)^2$ and $(2ax – 3a^2)^3$
$$(2ax – 3a^2)^2 = (2ax)^2 – 2(2ax)(3a^2) + (3a^2)^2 = 4a^2x^2 – 12a^3x + 9a^4$$
$$(2ax – 3a^2)^3 = (2ax)^3 – 3(2ax)^2(3a^2) + 3(2ax)(3a^2)^2 – (3a^2)^3$$
$$(2ax – 3a^2)^3 = 8a^3x^3 – 36a^4x^2 + 54a^5x – 27a^6$$
Step 5: Substitute back into main expression
$$(3x^2 – (2ax – 3a^2))^3 =27x^6 – 54ax^5 + 81a^2x^4 + 9x^2(4a^2x^2 – 12a^3x + 9a^4) $$ $$- (8a^3x^3 – 36a^4x^2 + 54a^5x – 27a^6)$$
Simplify each term:
$$(3x^2 – (2ax – 3a^2))^3 =27x^6 – 54ax^5 + 81a^2x^4 + (36a^2x^4 – 108a^3x^3 + 81a^4x^2) $$ $$ – 8a^3x^3 + 36a^4x^2 – 54a^5x + 27a^6$$
Combine like terms:
- $x^6$: $27x^6$
- $x^5$: $-54ax^5$
- $x^4$: $81a^2x^4 + 36a^2x^4 = 117a^2x^4$
- $x^3$: $-108a^3x^3 – 8a^3x^3 = -116a^3x^3$
- $x^2$: $81a^4x^2 + 36a^4x^2 = 117a^4x^2$
- $x^1$: $-54a^4x$
- constants: $27a^6$
So the final expansion is:
$(3x^2 – (2ax – 3a^2))^3 = 27x^6 – 54ax^5 + 117a^2x^4 – 116a^3x^3 + 117a^4x^2 – 54a^5x + 27a^6$
Hence final result :
$$\boxed{27x^6 – 54ax^5 + 117a^2x^4 – 116a^3x^3 + 117a^4x^2 – 54a^5x + 27a^6}$$
