Mathematics 11 (Punjab Board)
Exercise 8.2 Solutions
Worked solutions for Punjab Board Class 11 Mathematics, Chapter 8: Mathematical Induction and Binomial Theorem. This exercise covers the binomial theorem for a positive integral index — expansions, approximations, specific terms, and middle terms.
Question 1: Expand the following using the binomial theorem: (i) \(\left(\frac{x}{2}-\frac{2}{x^2}\right)^6\)
\((a+b)^n\) mein \(a = x/2\), \(b = -2/x^2\) rakhenge. Saath terms likhenge aur har term ki powers simplify karenge. Negative \(b\) ki wajah se signs alternate honge.
\[
\left(\frac{x}{2}-\frac{2}{x^2}\right)^6 = \sum_{r=0}^{6}\binom{6}{r}\left(\frac{x}{2}\right)^{6-r}\left(-\frac{2}{x^2}\right)^r
\]
\[
= \frac{x^6}{64} - \frac{3x^3}{8} + \frac{15}{4} - \frac{20}{x^3} + \frac{60}{x^6} - \frac{96}{x^9} + \frac{64}{x^{12}}
\]
Answer: \(\frac{x^6}{64} - \frac{3x^3}{8} + \frac{15}{4} - \frac{20}{x^3} + \frac{60}{x^6} - \frac{96}{x^9} + \frac{64}{x^{12}}\).
Question 1(ii): Expand \(\left(2a-\frac{x}{a}\right)^7\) using the binomial theorem.
Yahan \(a = 2a\), \(b = -x/a\). Aath terms honge. Har term mein \(a\) aur \(x\) ki powers ko carefully simplify karenge.
\[
\left(2a-\frac{x}{a}\right)^7 = \sum_{r=0}^{7}\binom{7}{r}(2a)^{7-r}\left(-\frac{x}{a}\right)^r
\]
\[
= 128a^7 - 448a^5x + 672a^3x^2 - 560ax^3 + \frac{280x^4}{a} - \frac{84x^5}{a^3} + \frac{14x^6}{a^5} - \frac{x^7}{a^7}
\]
Answer: \(128a^7 - 448a^5x + 672a^3x^2 - 560ax^3 + \frac{280x^4}{a} - \frac{84x^5}{a^3} + \frac{14x^6}{a^5} - \frac{x^7}{a^7}\).
Question 1(iii): Expand \(\left(\sqrt{\frac{a}{x}}-\sqrt{\frac{x}{a}}\right)^6\) using the binomial theorem.
\(\sqrt{a/x} = a^{1/2}x^{-1/2}\) aur \(\sqrt{x/a} = x^{1/2}a^{-1/2}\) likhenge. Dono ka product 1 hota hai, is liye har term simplify ho kar \(a/x\) ki power ban jayega.
\[
\left(\sqrt{\frac{a}{x}}-\sqrt{\frac{x}{a}}\right)^6 = \sum_{r=0}^{6}\binom{6}{r}\left(\frac{a}{x}\right)^{\frac{6-r}{2}}\left(-\sqrt{\frac{x}{a}}\right)^r
\]
\[
= \frac{a^3}{x^3} - 6\frac{a^2}{x^2} + 15\frac{a}{x} - 20 + 15\frac{x}{a} - 6\frac{x^2}{a^2} + \frac{x^3}{a^3}
\]
Answer: \(\frac{a^3}{x^3} - 6\frac{a^2}{x^2} + 15\frac{a}{x} - 20 + 15\frac{x}{a} - 6\frac{x^2}{a^2} + \frac{x^3}{a^3}\).
Question 2: Find the approximate value of the following using the binomial theorem: (i) \((0.97)^3\)
\((0.97)^3 = (1-0.03)^3\) likhenge. Chhota number \(0.03\) hai, to expansion exact value dega (koi approximation error nahi, kyunke poori expansion likh rahe hain).
\[
(0.97)^3 = (1-0.03)^3 = 1 - 3(0.03) + 3(0.03)^2 - (0.03)^3
\]
\[
= 1 - 0.09 + 0.0027 - 0.000027 = 0.912673
\]
Answer: \(0.912673\).
Question 2(ii): Find the approximate value of \((2.02)^4\) using the binomial theorem.
\((2.02)^4 = (2+0.02)^4\). \(a = 2\), \(b = 0.02\) rakh kar paanch terms likhenge.
\[
(2.02)^4 = (2+0.02)^4 = 16 + 4(8)(0.02) + 6(4)(0.02)^2 + 4(2)(0.02)^3 + (0.02)^4
\]
\[
= 16 + 0.64 + 0.0096 + 0.000064 + 0.00000016 = 16.64966416
\]
Answer: \(16.64966416\).
Question 2(iii): Find the approximate value of \((9.98)^4\) using the binomial theorem.
\((9.98)^4 = (10-0.02)^4\). \(b = -0.02\) ki wajah se signs alternate honge.
\[
(9.98)^4 = (10-0.02)^4 = 10000 - 4(1000)(0.02) + 6(100)(0.02)^2 - 4(10)(0.02)^3 + (0.02)^4
\]
\[
= 10000 - 80 + 0.24 - 0.00032 + 0.00000016 = 9920.23968016
\]
Answer: \(9920.23968016\).
Question 2(iv): Find the approximate value of \((2.1)^5\) using the binomial theorem.
\((2.1)^5 = (2+0.1)^5\). Chhe terms likhenge aur add karenge.
\[
(2.1)^5 = (2+0.1)^5 = 32 + 5(16)(0.1) + 10(8)(0.1)^2 + 10(4)(0.1)^3 + 5(2)(0.1)^4 + (0.1)^5
\]
\[
= 32 + 8 + 0.8 + 0.04 + 0.001 + 0.00001 = 40.84101
\]
Answer: \(40.84101\).
Question 3: Simplify the following using the binomial theorem: (i) \((a+\sqrt{2}x)^4 + (a-\sqrt{2}x)^4\)
\((x+y)^n + (x-y)^n\) mein odd powers wale terms cancel ho jate hain — sirf even \(r\) wale terms bachte hain, aur har ek double ho jata hai. Is se calculation aadhi ho jati hai.
\[
(a+\sqrt{2}x)^4 + (a-\sqrt{2}x)^4 = 2\left[\binom{4}{0}a^4 + \binom{4}{2}a^2(\sqrt{2}x)^2 + \binom{4}{4}(\sqrt{2}x)^4\right]
\]
\[
= 2\left[a^4 + 6a^2(2x^2) + 4x^4\right] = 2a^4 + 24a^2x^2 + 8x^4
\]
Answer: \(2a^4 + 24a^2x^2 + 8x^4\).
Question 3(ii): Simplify \((2+\sqrt{3})^6 + (2-\sqrt{3})^6\).
Wohi trick: odd powers cancel, even powers double. \((\sqrt{3})^2 = 3\), \((\sqrt{3})^4 = 9\), \((\sqrt{3})^6 = 27\) yaad rakhen.
\[
(2+\sqrt{3})^6 + (2-\sqrt{3})^6 = 2\left[\binom{6}{0}2^6 + \binom{6}{2}2^4(\sqrt{3})^2 + \binom{6}{4}2^2(\sqrt{3})^4 + \binom{6}{6}(\sqrt{3})^6\right]
\]
\[
= 2\left[64 + 15(16)(3) + 15(4)(9) + 27\right] = 2(64 + 720 + 540 + 27) = 2(1351) = 2702
\]
Answer: \(2702\).
Question 4: Expand the following in ascending powers of \(x\): (i) \((2+x-x^2)^4\)
\((2 + (x-x^2))^4\) likhenge aur \((x-x^2)\) ko aik unit samajh kar binomial theorem lagayenge. Phir \((x-x^2)^2\), \((x-x^2)^3\) waghera expand karke \(x\) ki powers ke hisab se terms jama karenge.
\[
(2+x-x^2)^4 = 2^4 + 4(2^3)(x-x^2) + 6(2^2)(x-x^2)^2 + 4(2)(x-x^2)^3 + (x-x^2)^4
\]
\[
= 16 + 32(x-x^2) + 24(x^2-2x^3+x^4) + 8(x^3-3x^4+3x^5-x^6) + (x^4-4x^5+6x^6-4x^7+x^8)
\]
Collecting powers of \(x\): constant \(16\); \(x\): \(32\); \(x^2\): \(-32+24 = -8\); \(x^3\): \(-48+8 = -40\); \(x^4\): \(24-24+1 = 1\); \(x^5\): \(24-4 = 20\); \(x^6\): \(-8+6 = -2\); \(x^7\): \(-4\); \(x^8\): \(1\).
\[
= 16 + 32x - 8x^2 - 40x^3 + x^4 + 20x^5 - 2x^6 - 4x^7 + x^8
\]
Answer: \(16 + 32x - 8x^2 - 40x^3 + x^4 + 20x^5 - 2x^6 - 4x^7 + x^8\).
Question 4(ii): Expand \((1-x+x^2)^4\) in ascending powers of \(x\).
\((1-(x-x^2))^4\) likhenge. Wese hi expand karke powers jama karenge. Sign ka khaas khayal rakhen kyunke \(b = -(x-x^2)\) negative hai.
\[
(1-x+x^2)^4 = 1 - 4(x-x^2) + 6(x-x^2)^2 - 4(x-x^2)^3 + (x-x^2)^4
\]
\[
= 1 - 4x + 4x^2 + 6(x^2-2x^3+x^4) - 4(x^3-3x^4+3x^5-x^6) + (x^4-4x^5+6x^6-4x^7+x^8)
\]
Collecting: constant \(1\); \(x\): \(-4\); \(x^2\): \(4+6 = 10\); \(x^3\): \(-12-4 = -16\); \(x^4\): \(6+12+1 = 19\); \(x^5\): \(-12-4 = -16\); \(x^6\): \(4+6 = 10\); \(x^7\): \(-4\); \(x^8\): \(1\).
\[
= 1 - 4x + 10x^2 - 16x^3 + 19x^4 - 16x^5 + 10x^6 - 4x^7 + x^8
\]
Answer: \(1 - 4x + 10x^2 - 16x^3 + 19x^4 - 16x^5 + 10x^6 - 4x^7 + x^8\).
Question 5: Find the specified term in each expansion: (i) the term involving \(x^4\) in \((3-2x)^7\)
General term \(T_{r+1} = \binom{7}{r}3^{7-r}(-2x)^r\) likhenge. \(x^4\) ke liye \(r = 4\) hoga. Phir \(\binom{7}{4} = 35\), \(3^3 = 27\), \((-2)^4 = 16\) put karke multiply karenge.
\[
T_{r+1} = \binom{7}{r}3^{7-r}(-2x)^r; \quad \text{for } x^4: r = 4
\]
\[
T_5 = \binom{7}{4}\cdot 3^3\cdot (-2)^4 x^4 = 35\cdot 27\cdot 16\,x^4 = 15120x^4
\]
Answer: \(15120x^4\).
Question 5(ii): Find the term involving \(x^{-2}\) in \(\left(x-\frac{2}{x^2}\right)^{13}\).
General term mein \(x\) ki power \(13-3r\) hai. Ise \(-2\) ke barabar rakh kar \(r = 5\) nikalenge.
\[
T_{r+1} = \binom{13}{r}x^{13-r}\left(-\frac{2}{x^2}\right)^r = \binom{13}{r}(-2)^r x^{13-3r}
\]
For \(x^{-2}\): \(13 - 3r = -2 \Rightarrow r = 5\).
\[
T_6 = \binom{13}{5}(-2)^5 x^{-2} = 1287\cdot(-32)x^{-2} = -41184x^{-2}
\]
Answer: \(-41184x^{-2}\) (i.e. \(-\frac{41184}{x^2}\)).
Question 5(iii): Find the term involving \(a^4\) in \(\left(\frac{2}{x}-a\right)^9\).
General term mein \(a\) ki power \(r\) hai, to \(r = 4\). Saath hi \(x\) ki power \(-(9-4) = -5\) hogi.
\[
T_{r+1} = \binom{9}{r}\left(\frac{2}{x}\right)^{9-r}(-a)^r; \quad \text{for } a^4: r = 4
\]
\[
T_5 = \binom{9}{4}2^5(-1)^4\frac{a^4}{x^5} = 126\cdot 32\cdot\frac{a^4}{x^5} = \frac{4032a^4}{x^5}
\]
Answer: \(\frac{4032a^4}{x^5}\).
Question 5(iv): Find the term involving \(y^3\) in \((x-\sqrt{y})^{11}\).
\((-\sqrt{y})^r\) mein \(y\) ki power \(r/2\) hai. \(y^3\) ke liye \(r/2 = 3\), yani \(r = 6\).
\[
T_{r+1} = \binom{11}{r}x^{11-r}(-\sqrt{y})^r = \binom{11}{r}(-1)^r x^{11-r}y^{r/2}; \quad \text{for } y^3: r = 6
\]
\[
T_7 = \binom{11}{6}x^5y^3 = 462x^5y^3
\]
Answer: \(462x^5y^3\).
Question 6: Find the coefficient of the specified power: (i) coefficient of \(x^5\) in \(\left(x^2-\frac{3}{2x}\right)^{10}\)
General term mein \(x\) ki power \(20-3r\) hai. \(20-3r = 5\) se \(r = 5\). Phir \(\binom{10}{5} = 252\) aur \(\left(-\frac{3}{2}\right)^5 = -\frac{243}{32}\) multiply karenge.
\[
T_{r+1} = \binom{10}{r}(x^2)^{10-r}\left(-\frac{3}{2x}\right)^r = \binom{10}{r}\left(-\frac{3}{2}\right)^r x^{20-3r}
\]
For \(x^5\): \(20 - 3r = 5 \Rightarrow r = 5\).
\[
\text{Coefficient} = \binom{10}{5}\left(-\frac{3}{2}\right)^5 = 252\cdot\left(-\frac{243}{32}\right) = -\frac{15309}{8}
\]
Answer: \(-\frac{15309}{8}\).
Question 6(ii): Find the coefficient of \(x^n\) in \(\left(x^2-\frac{1}{x}\right)^{2n}\).
General term mein \(x\) ki power \(4n-3r\) hai. \(4n-3r = n\) se \(r = n\). Phir coefficient mein \(\binom{2n}{n}(-1)^n\) aayega.
\[
T_{r+1} = \binom{2n}{r}(x^2)^{2n-r}\left(-\frac{1}{x}\right)^r = \binom{2n}{r}(-1)^r x^{4n-3r}
\]
For \(x^n\): \(4n - 3r = n \Rightarrow r = n\).
\[
\text{Coefficient} = \binom{2n}{n}(-1)^n
\]
Answer: \((-1)^n\binom{2n}{n}\).
Question 7: Find the 6th term of \(\left(x^2-\frac{3}{2x}\right)^{10}\).
6th term ka matlab \(T_6\), yani \(r = 5\). Ye Question 6(i) wali hi calculation hai, bas poora term (coefficient ke saath \(x^5\)) likhna hai.
\[
T_6 = T_{5+1} = \binom{10}{5}(x^2)^5\left(-\frac{3}{2x}\right)^5 = 252\cdot x^{10}\cdot\left(-\frac{243}{32x^5}\right) = -\frac{15309}{8}x^5
\]
Answer: \(-\frac{15309}{8}x^5\).
Question 8: Find the term independent of \(x\): (i) in \(\left(x-\frac{2}{x}\right)^{10}\)
Note: the book prints \(\left(x-\frac{2}{x^2}\right)^{10}\), but with \(x^2\) in the denominator the equation \(10-3r = 0\) has no integer solution, so no term independent of \(x\) exists there. The intended form \(\left(x-\frac{2}{x}\right)^{10}\) gives the term below. Independent term ke liye \(x\) ki power 0 rakhenge: \(10-2r = 0\), yani \(r = 5\).
\[
T_{r+1} = \binom{10}{r}x^{10-r}\left(-\frac{2}{x}\right)^r = \binom{10}{r}(-2)^r x^{10-2r}
\]
Independent of \(x\): \(10 - 2r = 0 \Rightarrow r = 5\).
\[
T_6 = \binom{10}{5}(-2)^5 = 252\cdot(-32) = -8064
\]
Answer: \(-8064\).
Question 8(ii): Find the term independent of \(x\) in \(\left(\sqrt{x}+\frac{1}{2x^2}\right)^{10}\).
\(\sqrt{x} = x^{1/2}\) likhenge. General term mein \(x\) ki power \(\frac{10-r}{2} - 2r = \frac{10-5r}{2}\) hai. Ise 0 ke barabar rakh kar \(r = 2\).
\[
T_{r+1} = \binom{10}{r}(\sqrt{x})^{10-r}\left(\frac{1}{2x^2}\right)^r = \binom{10}{r}2^{-r}x^{\frac{10-5r}{2}}
\]
Independent of \(x\): \(\frac{10-5r}{2} = 0 \Rightarrow r = 2\).
\[
T_3 = \binom{10}{2}2^{-2} = 45\cdot\frac{1}{4} = \frac{45}{4}
\]
Answer: \(\frac{45}{4}\).
Question 8(iii): Find the term independent of \(x\) in \((1+x^3)^n\left(1+\frac{1}{x^2}\right)^n\).
Dono expansions ke general terms ka product lenge: \(\binom{n}{r}\binom{n}{s}x^{3r-2s}\). \(x\) ki power 0 hone ke liye \(3r = 2s\) chahiye, jiske integer solutions \(r = 2k\), \(s = 3k\) hain.
\[
\text{General term of product} = \binom{n}{r}\binom{n}{s}x^{3r-2s}, \quad 3r-2s = 0 \Rightarrow r = 2k,\ s = 3k
\]
Summing over all valid \(k = 0, 1, 2, \dots\):
\[
\text{Independent term} = \binom{n}{0}\binom{n}{0} + \binom{n}{2}\binom{n}{3} + \binom{n}{4}\binom{n}{6} + \dots = 1 + \binom{n}{2}\binom{n}{3} + \binom{n}{4}\binom{n}{6} + \dots
\]
Answer: \(1 + \binom{n}{2}\binom{n}{3} + \binom{n}{4}\binom{n}{6} + \dots\).
Question 9: Find the middle term(s): (i) in \(\left(\frac{1}{x}-\frac{x^2}{2}\right)^{12}\)
\(n = 12\) even hai, to sirf aik middle term hoga: \(T_{n/2+1} = T_7\) (yani \(r = 6\)).
\[
T_7 = \binom{12}{6}\left(\frac{1}{x}\right)^6\left(-\frac{x^2}{2}\right)^6 = 924\cdot\frac{1}{x^6}\cdot\frac{x^{12}}{64} = \frac{924}{64}x^6 = \frac{231}{16}x^6
\]
Answer: \(\frac{231}{16}x^6\).
Question 9(ii): Find the middle terms in \(\left(\frac{3x}{2}-\frac{1}{3x}\right)^{11}\).
\(n = 11\) odd hai, to do middle terms honge: \(T_6\) aur \(T_7\) (yani \(r = 5\) aur \(r = 6\)).
\[
T_6 = \binom{11}{5}\left(\frac{3x}{2}\right)^6\left(-\frac{1}{3x}\right)^5 = 462\cdot\frac{729}{64}x^6\cdot\left(-\frac{1}{243x^5}\right) = -\frac{693}{32}x
\]
\[
T_7 = \binom{11}{6}\left(\frac{3x}{2}\right)^5\left(-\frac{1}{3x}\right)^6 = 462\cdot\frac{243}{32}x^5\cdot\frac{1}{729x^6} = \frac{77}{16x}
\]
Answer: \(-\frac{693}{32}x\) and \(\frac{77}{16x}\).
Question 9(iii): Find the middle terms in \(\left(2x-\frac{1}{2x}\right)^{2n+1}\).
Power \(2n+1\) odd hai, to middle terms \(T_{n+1}\) aur \(T_{n+2}\) honge. General term likh kar \(r = n\) aur \(r = n+1\) rakhenge aur simplify karenge.
\[
T_{n+1} = \binom{2n+1}{n}(2x)^{n+1}\left(-\frac{1}{2x}\right)^n = \binom{2n+1}{n}2^{n+1}x^{n+1}(-1)^n2^{-n}x^{-n} = 2(-1)^n\binom{2n+1}{n}x
\]
\[
T_{n+2} = \binom{2n+1}{n+1}(2x)^n\left(-\frac{1}{2x}\right)^{n+1} = \frac{1}{2}(-1)^{n+1}\binom{2n+1}{n+1}\frac{1}{x}
\]
Answer: \(2(-1)^n\binom{2n+1}{n}x\) and \(\frac{(-1)^{n+1}}{2}\binom{2n+1}{n+1}\frac{1}{x}\).
Question 10: Prove that \(\binom{n}{1} + \binom{n}{2} + \binom{n}{3} + \dots + \binom{n}{n} = 2^n - 1\).
\((1+x)^n\) ke expansion mein \(x = 1\) rakhenge to saare binomial coefficients ka sum \(2^n\) mil jayega. Phir dono taraf se \(\binom{n}{0} = 1\) subtract kar denge.
\[
(1+x)^n = \binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \dots + \binom{n}{n}x^n
\]
Putting \(x = 1\): \(2^n = \binom{n}{0} + \binom{n}{1} + \dots + \binom{n}{n} = 1 + \binom{n}{1} + \dots + \binom{n}{n}\).
\[
\binom{n}{1} + \binom{n}{2} + \dots + \binom{n}{n} = 2^n - 1
\]
Answer: \(\binom{n}{1} + \binom{n}{2} + \dots + \binom{n}{n} = 2^n - 1\) (proved).
