NCERT Class 11 Mathematics — Binomial Theorem, Miscellaneous Exercise on Chapter 7. All 6 questions solved.

Six questions, four distinct techniques:

• Rewriting to expose a factor (question 1): choose the split of the binomial so that the factor you want appears in every term but one.
• Pairing conjugate surds (questions 2 and 3): $(A + B)^n \pm (A – B)^n$ cancels half the terms, and the surviving half is rational.
• Truncating for an approximation (question 4): keep as many terms as the required accuracy demands and discard the rest.
• Grouping a trinomial into a binomial (questions 5 and 6): a three-term expression becomes $(P + Q)^n$ by bracketing two of the terms together.

Key insight. The binomial theorem takes two terms, so a trinomial must first be grouped into two. Which two you bracket together is a choice, and it decides how much work follows — in question 6, treating $3x^2 + 3a^2$ as one part and $-2ax$ as the other keeps the symmetry of the expression intact.

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.

(Hint: write $a^n = (a – b + b)^n$ and expand.)

Solution. Follow the hint. Set $A = a – b$, so that $a = A + b$, and expand $a^n = (A + b)^n$ by the binomial theorem:

$$a^n = (A + b)^n = {}^n\mathrm{C}_0 A^n + {}^n\mathrm{C}_1 A^{n-1}b + \cdots + {}^n\mathrm{C}_{n-1} A\,b^{n-1} + {}^n\mathrm{C}_n b^n$$

The last term is $b^n$, and every other term contains at least one factor of $A$. Subtracting $b^n$ from both sides therefore removes the only term without an $A$:

$$a^n – b^n = {}^n\mathrm{C}_0 A^n + {}^n\mathrm{C}_1 A^{n-1}b + \cdots + {}^n\mathrm{C}_{n-1}A\,b^{n-1}$$

Every remaining term has $A$ as a factor, so factor it out:

$$a^n – b^n = A\left[{}^n\mathrm{C}_0 A^{n-1} + {}^n\mathrm{C}_1 A^{n-2}b + \cdots + {}^n\mathrm{C}_{n-1}b^{n-1}\right]$$

The bracket is an integer, since $a$ and $b$ are integers, so

$$a^n – b^n = (a – b) \times \text{an integer}$$

that is, $a – b$ is a factor of $a^n – b^n$. $\blacksquare$

The hypothesis that $a$ and $b$ are distinct matters only so that $a – b \ne 0$ and “factor” is meaningful.

Question 2

Evaluate $\left(\sqrt3 + \sqrt2\right)^6 – \left(\sqrt3 – \sqrt2\right)^6$.

Solution. Expand the general pair first. With $a = \sqrt3$, $b = \sqrt2$ and $n = 6$, the two expansions differ only in the signs of the odd-power terms, so subtracting cancels the even ones and doubles the odd ones:

$$(a+b)^6 – (a-b)^6 = 2\left[{}^6\mathrm{C}_1 a^5 b + {}^6\mathrm{C}_3 a^3b^3 + {}^6\mathrm{C}_5 ab^5\right]$$

$$= 2\left[6a^5b + 20a^3b^3 + 6ab^5\right]$$

Now substitute. With $a = \sqrt3$: $a^3 = 3\sqrt3$, $a^5 = 9\sqrt3$. With $b = \sqrt2$: $b^3 = 2\sqrt2$, $b^5 = 4\sqrt2$.

$$6a^5b = 6(9\sqrt3)(\sqrt2) = 54\sqrt6$$ $$20a^3b^3 = 20(3\sqrt3)(2\sqrt2) = 120\sqrt6$$ $$6ab^5 = 6(\sqrt3)(4\sqrt2) = 24\sqrt6$$

Adding and doubling:

$$2(54 + 120 + 24)\sqrt6 = 2 \times 198\,\sqrt6 = 396\sqrt6$$

Question 3

Find the value of $\left(a^2 + \sqrt{a^2 – 1}\right)^4 + \left(a^2 – \sqrt{a^2 – 1}\right)^4$.

Solution. Write $A = a^2$ and $B = \sqrt{a^2 – 1}$, so that $B^2 = a^2 – 1$ — this is what makes the radical disappear. Adding the two fourth powers cancels the odd-power terms:

$$(A+B)^4 + (A-B)^4 = 2\left[A^4 + 6A^2B^2 + B^4\right]$$

Now substitute $A^2 = a^4$, $A^4 = a^8$, $B^2 = a^2 – 1$ and $B^4 = (a^2-1)^2 = a^4 – 2a^2 + 1$:

$$= 2\left[a^8 + 6a^4(a^2 – 1) + (a^4 – 2a^2 + 1)\right]$$

$$= 2\left[a^8 + 6a^6 – 6a^4 + a^4 – 2a^2 + 1\right]$$

$$= 2\left[a^8 + 6a^6 – 5a^4 – 2a^2 + 1\right]$$

$$= 2a^8 + 12a^6 – 10a^4 – 4a^2 + 2$$

Every surd has vanished, which is the point of the pairing — only even powers of $B$ survive, and $B^2$ is a polynomial.

Question 4

Find an approximation of $(0.99)^5$ using the first three terms of its expansion.

Solution. Write $0.99 = 1 – 0.01$ and expand with $n = 5$:

$$(1 – 0.01)^5 = {}^5\mathrm{C}_0 – {}^5\mathrm{C}_1(0.01) + {}^5\mathrm{C}_2(0.01)^2 – \cdots$$

The first three terms are

$$1 – 5(0.01) + 10(0.0001) = 1 – 0.05 + 0.001 = 0.951$$

The next term is $-10(0.01)^3 = -0.00001$, which is why three terms already give four-figure accuracy: the exact value is $0.95099005$, so the approximation $0.9510$ is correct to four decimal places.

Question 5

Expand using the binomial theorem $\left(1 + \dfrac{x}{2} – \dfrac{2}{x}\right)^4$, $x \ne 0$.

Solution. Three terms, so group two of them. Take

$$P = 1 + \frac x2, \qquad Q = -\frac2x$$

and expand $(P + Q)^4$:

$$(P+Q)^4 = P^4 + 4P^3Q + 6P^2Q^2 + 4PQ^3 + Q^4$$

Now compute the powers of $P = 1 + \tfrac x2$:

$$P^2 = 1 + x + \frac{x^2}{4}, \qquad P^3 = 1 + \frac{3x}{2} + \frac{3x^2}{4} + \frac{x^3}{8}$$ $$P^4 = 1 + 2x + \frac{3x^2}{2} + \frac{x^3}{2} + \frac{x^4}{16}$$

and the powers of $Q$:

$$Q^2 = \frac{4}{x^2}, \qquad Q^3 = -\frac{8}{x^3}, \qquad Q^4 = \frac{16}{x^4}$$

Substituting term by term:

$$4P^3Q = -\frac{8}{x}\left(1 + \frac{3x}{2} + \frac{3x^2}{4} + \frac{x^3}{8}\right) = -\frac8x – 12 – 6x – x^2$$

$$6P^2Q^2 = \frac{24}{x^2}\left(1 + x + \frac{x^2}{4}\right) = \frac{24}{x^2} + \frac{24}{x} + 6$$

$$4PQ^3 = -\frac{32}{x^3}\left(1 + \frac x2\right) = -\frac{32}{x^3} – \frac{16}{x^2}$$

Now add everything, collecting by power of $x$:

• $x^4$: $\dfrac{1}{16}$
• $x^3$: $\dfrac12$
• $x^2$: $\dfrac32 – 1 = \dfrac12$
• $x^1$: $2 – 6 = -4$
• constant: $1 – 12 + 6 = -5$
• $x^{-1}$: $-8 + 24 = 16$
• $x^{-2}$: $24 – 16 = 8$
• $x^{-3}$: $-32$
• $x^{-4}$: $16$

$$\frac{x^4}{16} + \frac{x^3}{2} + \frac{x^2}{2} – 4x – 5 + \frac{16}{x} + \frac{8}{x^2} – \frac{32}{x^3} + \frac{16}{x^4}$$

Question 6

Find the expansion of $\left(3x^2 – 2ax + 3a^2\right)^3$ using the binomial theorem.

Solution. Group the two terms that share the same structure:

$$P = 3x^2 + 3a^2, \qquad Q = -2ax$$

Then, with $n = 3$:

$$(P+Q)^3 = P^3 + 3P^2Q + 3PQ^2 + Q^3$$

$P^3$: factor out the $3$ first — $P = 3(x^2 + a^2)$, so $P^3 = 27(x^2+a^2)^3$, and

$$(x^2+a^2)^3 = x^6 + 3x^4a^2 + 3x^2a^4 + a^6$$

$$P^3 = 27x^6 + 81a^2x^4 + 81a^4x^2 + 27a^6$$

$3P^2Q$: $P^2 = 9(x^2+a^2)^2 = 9(x^4 + 2a^2x^2 + a^4)$, so

$$3P^2Q = 3 \times 9(x^4 + 2a^2x^2 + a^4) \times (-2ax) = -54a\left(x^5 + 2a^2x^3 + a^4x\right)$$

$$= -54ax^5 – 108a^3x^3 – 54a^5x$$

$3PQ^2$: $Q^2 = 4a^2x^2$, so

$$3PQ^2 = 3 \times 3(x^2+a^2) \times 4a^2x^2 = 36a^2x^2(x^2 + a^2) = 36a^2x^4 + 36a^4x^2$$

$Q^3$: $(-2ax)^3 = -8a^3x^3$.

Now add, collecting by power of $x$:

• $x^6$: $27$
• $x^5$: $-54a$
• $x^4$: $81a^2 + 36a^2 = 117a^2$
• $x^3$: $-108a^3 – 8a^3 = -116a^3$
• $x^2$: $81a^4 + 36a^4 = 117a^4$
• $x^1$: $-54a^5$
• constant: $27a^6$

$$27x^6 – 54ax^5 + 117a^2x^4 – 116a^3x^3 + 117a^4x^2 – 54a^5x + 27a^6$$

The result is palindromic in the coefficients ($27, -54, 117, -116, 117, -54, 27$), which it had to be: the original trinomial is unchanged when $x$ and $a$ are swapped. That symmetry is a free check on all seven coefficients.

Common mistakes

• Question 1, expanding $(a-b)^n$ instead. The hint is to expand $a^n$ as $\big((a-b) + b\big)^n$, which puts $a – b$ inside the binomial where it can be factored out.
• Questions 2 and 3, expanding both sixth or fourth powers in full. Half the terms cancel. Writing down only the surviving ones — odd powers for a difference, even powers for a sum — halves the work and removes most of the sign risk.
• Question 3, forgetting that $B^2 = a^2 – 1$, not $a^2$. The whole point of the substitution is that squaring the radical removes it; carrying $\sqrt{a^2-1}$ through the algebra defeats the method.
• Question 4, using more terms than asked or fewer than needed. Three terms give $0.951$; two would give $0.95$, which is not accurate to four places.
• Question 5, grouping the wrong pair. Any grouping works in principle, but $\left(1 + \tfrac x2\right)$ as one part keeps the powers of $P$ manageable. Taking $P = \tfrac x2 – \tfrac2x$ leaves messier intermediate expansions.
• Question 5, dropping a term when collecting. Nine distinct powers of $x$ appear, from $x^4$ down to $x^{-4}$. Listing them as a column, as above, is the reliable way to gather them.
• Question 6, expanding the trinomial directly as a product of three brackets. It works but produces $27$ terms to combine. Grouping into a binomial reduces it to four.

Practise next

• Exercise 8.1 — the next chapter, Sequences and Series, where the expansions here give way to arithmetic and geometric progressions.
• Exercise 7.1 — worth working again alongside questions 2 and 3, since questions 11 and 12 there are the same conjugate-pairing technique at lower powers.