Mathematics 11 (Punjab Board)
Exercise 8.3 Solutions
Worked solutions for Punjab Board Class 11 Mathematics, Chapter 8: Mathematical Induction and Binomial Theorem. This exercise covers the binomial theorem when the index is a negative integer or a fraction — infinite series valid for \(|x| \lt 1\).
Question 1: Expand up to 4 terms, taking values of \(x\) such that the expansion in each case is valid: (i) \((1+x)^{-1/2}\)
\((1+x)^n\) ka formula \(n = -1/2\) ke saath lagayenge: \(1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \dots\). Expansion \(|x| \lt 1\) ke liye valid hai.
\[
(1+x)^{-\frac{1}{2}} = 1 - \frac{1}{2}x + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2!}x^2 + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)}{3!}x^3 + \dots
\]
\[
= 1 - \frac{x}{2} + \frac{3x^2}{8} - \frac{5x^3}{16} + \dots \qquad (|x| \lt 1)
\]
Answer: \(1 - \frac{x}{2} + \frac{3x^2}{8} - \frac{5x^3}{16} + \dots\), valid for \(|x| \lt 1\).
Question 1(ii): Expand \((4-3x)^{-1/2}\) up to 4 terms, stating where the expansion is valid.
Pehle constant part bahar nikalenge: \((4-3x)^{-1/2} = 4^{-1/2}\left(1-\frac{3x}{4}\right)^{-1/2} = \frac{1}{2}\left(1-\frac{3x}{4}\right)^{-1/2}\). Phir \(u = -3x/4\) ke saath expand karenge. Valid jab \(|3x/4| \lt 1\), yani \(|x| \lt 4/3\).
\[
(4-3x)^{-\frac{1}{2}} = \frac{1}{2}\left[1 + \frac{1}{2}\left(\frac{3x}{4}\right) + \frac{3}{8}\left(\frac{3x}{4}\right)^2 + \frac{5}{16}\left(\frac{3x}{4}\right)^3 + \dots\right]
\]
\[
= \frac{1}{2} + \frac{3x}{16} + \frac{27x^2}{256} + \frac{135x^3}{2048} + \dots \qquad \left(|x| \lt \tfrac{4}{3}\right)
\]
Answer: \(\frac{1}{2} + \frac{3x}{16} + \frac{27x^2}{256} + \frac{135x^3}{2048} + \dots\), valid for \(|x| \lt \frac{4}{3}\).
Question 1(iii): Expand \(\frac{(1-x)^{-1}}{(1+x)^2}\) up to 4 terms, stating where the expansion is valid.
\((1-x)^{-1}(1+x)^{-2}\) likhenge. Dono series ko 4 terms tak likh kar multiply karenge: \((1+x+x^2+x^3)(1-2x+3x^2-4x^3)\). Valid \(|x| \lt 1\) ke liye.
\[
(1-x)^{-1} = 1 + x + x^2 + x^3 + \dots, \qquad (1+x)^{-2} = 1 - 2x + 3x^2 - 4x^3 + \dots
\]
Multiplying and collecting up to \(x^3\): constant \(1\); \(x\): \(1-2 = -1\); \(x^2\): \(1-2+3 = 2\); \(x^3\): \(1-2+3-4 = -2\).
\[
= 1 - x + 2x^2 - 2x^3 + \dots \qquad (|x| \lt 1)
\]
Answer: \(1 - x + 2x^2 - 2x^3 + \dots\), valid for \(|x| \lt 1\).
Question 1(iv): Expand \(\frac{\sqrt{1+2x}}{1-x}\) up to 4 terms, stating where the expansion is valid.
\((1+2x)^{1/2}(1-x)^{-1}\) likhenge. \((1+2x)^{1/2} \approx 1 + x - \frac{x^2}{2} + \frac{x^3}{2}\) aur \((1-x)^{-1} = 1+x+x^2+x^3+\dots\). Multiply karke 4 terms rakhenge.
\[
(1+2x)^{\frac{1}{2}} = 1 + x - \frac{x^2}{2} + \frac{x^3}{2} + \dots, \qquad (1-x)^{-1} = 1 + x + x^2 + x^3 + \dots
\]
Multiplying: constant \(1\); \(x\): \(1+1 = 2\); \(x^2\): \(1+1-\frac{1}{2} = \frac{3}{2}\); \(x^3\): \(1+1-\frac{1}{2}+\frac{1}{2} = 2\).
\[
= 1 + 2x + \frac{3}{2}x^2 + 2x^3 + \dots \qquad (|x| \lt 1)
\]
Answer: \(1 + 2x + \frac{3}{2}x^2 + 2x^3 + \dots\), valid for \(|x| \lt 1\).
Question 2: Find the coefficient of \(x^n\) in the expansion of: (i) \(\frac{1+x^2}{(1+x)^2}\)
\((1+x^2)(1+x)^{-2}\) likhenge. \((1+x)^{-2} = \sum_{r\ge 0}(-1)^r(r+1)x^r\). \(x^n\) ka coefficient do hisson se: \(1\) se \((-1)^n(n+1)\) aur \(x^2\) se \((-1)^{n-2}(n-1)\) (jab \(n \ge 2\)).
\[
[x^n] = (-1)^n(n+1) + (-1)^{n-2}(n-1) = (-1)^n\big[(n+1)+(n-1)\big] = 2n(-1)^n \qquad (n \ge 2)
\]
Answer: \(2n(-1)^n\) for \(n \ge 2\) (and \(1\) for \(n = 0\), \(-2\) for \(n = 1\)).
Question 2(ii): Find the coefficient of \(x^n\) in the expansion of \(\frac{(1+x)^2}{(1-x)^2}\).
\((1+2x+x^2)(1-x)^{-2}\) likhenge. \((1-x)^{-2} = \sum_{r\ge 0}(r+1)x^r\). Teeno hisson ke coefficients jama karenge: \((n+1) + 2n + (n-1)\).
\[
[x^n] = (n+1) + 2(n) + (n-1) = 4n \qquad (n \ge 1)
\]
Answer: \(4n\) for \(n \ge 1\) (and \(1\) for \(n = 0\)).
Question 3: If \(x\) is so small that its square and higher powers can be neglected, show that: (i) \(\frac{1-x}{\sqrt{1+x}} \approx 1 - \frac{3}{2}x\)
\((1+x)^{-1/2} \approx 1 - x/2\) (sirf pehle do terms). Phir \((1-x)(1-x/2)\) multiply karke \(x^2\) wala term gira denge.
\[
\frac{1-x}{\sqrt{1+x}} = (1-x)(1+x)^{-\frac{1}{2}} \approx (1-x)\left(1-\frac{x}{2}\right) \approx 1 - \frac{x}{2} - x = 1 - \frac{3}{2}x
\]
(neglecting the \(x^2\) term).
Answer: \(\frac{1-x}{\sqrt{1+x}} \approx 1 - \frac{3}{2}x\) (proved).
Question 3(ii): If \(x\) is so small that its square and higher powers can be neglected, show that \(\frac{\sqrt{1+2x}}{\sqrt{1-x}} \approx 1 + \frac{3}{2}x\).
\((1+2x)^{1/2} \approx 1 + x\) aur \((1-x)^{-1/2} \approx 1 + x/2\). Dono ko multiply karke \(x^2\) gira denge.
\[
\frac{\sqrt{1+2x}}{\sqrt{1-x}} = (1+2x)^{\frac{1}{2}}(1-x)^{-\frac{1}{2}} \approx (1+x)\left(1+\frac{x}{2}\right) \approx 1 + \frac{x}{2} + x = 1 + \frac{3}{2}x
\]
Answer: \(\frac{\sqrt{1+2x}}{\sqrt{1-x}} \approx 1 + \frac{3}{2}x\) (proved).
Question 3(iii): If \(x\) is so small that its square and higher powers can be neglected, show that \(\frac{(9+7x)^{1/2}-(16+3x)^{1/4}}{4+5x} \approx \frac{1}{4} - \frac{17}{384}x\).
Numerator ke dono roots ko alag expand karenge: \((9+7x)^{1/2} = 3\left(1+\frac{7x}{9}\right)^{1/2} \approx 3 + \frac{7x}{6}\), aur \((16+3x)^{1/4} = 2\left(1+\frac{3x}{16}\right)^{1/4} \approx 2 + \frac{3x}{32}\). Phir denominator \((4+5x)^{-1} \approx \frac{1}{4}\left(1-\frac{5x}{4}\right)\) se multiply karenge.
\[
\text{Numerator} \approx \left(3+\frac{7x}{6}\right) - \left(2+\frac{3x}{32}\right) = 1 + \left(\frac{7}{6}-\frac{3}{32}\right)x = 1 + \frac{103}{96}x
\]
\[
\frac{1 + \frac{103}{96}x}{4+5x} \approx \frac{1}{4}\left(1+\frac{103}{96}x\right)\left(1-\frac{5x}{4}\right) \approx \frac{1}{4}\left[1 + \left(\frac{103}{96}-\frac{120}{96}\right)x\right] = \frac{1}{4} - \frac{17}{384}x
\]
Answer: \(\frac{(9+7x)^{1/2}-(16+3x)^{1/4}}{4+5x} \approx \frac{1}{4} - \frac{17}{384}x\) (proved).
Question 3(iv): If \(x\) is so small that its square and higher powers can be neglected, show that \(\frac{\sqrt{4+x}}{(1-x)^3} \approx 2 + \frac{25}{4}x\).
\(\sqrt{4+x} = 2\left(1+\frac{x}{4}\right)^{1/2} \approx 2 + \frac{x}{4}\), aur \((1-x)^{-3} \approx 1 + 3x\). Multiply karke \(x^2\) gira denge.
\[
\frac{\sqrt{4+x}}{(1-x)^3} = (4+x)^{\frac{1}{2}}(1-x)^{-3} \approx \left(2+\frac{x}{4}\right)(1+3x) \approx 2 + 6x + \frac{x}{4} = 2 + \frac{25}{4}x
\]
Answer: \(\frac{\sqrt{4+x}}{(1-x)^3} \approx 2 + \frac{25}{4}x\) (proved).
Question 4: If \(x\) is so small that its cube and higher powers can be neglected, show that: (i) \(\sqrt{1-x-2x^2} \approx 1 - \frac{1}{2}x - \frac{9}{8}x^2\)
\((1-x-2x^2)^{1/2} \approx 1 + ax + bx^2\) maan kar dono taraf square karenge aur coefficients compare karenge: \(2a = -1\) se \(a = -1/2\), phir \(2b + a^2 = -2\) se \(b = -9/8\).
Let \((1-x-2x^2)^{1/2} \approx 1 + ax + bx^2\). Squaring both sides (up to \(x^2\)):
\[
1 + 2ax + (2b+a^2)x^2 \approx 1 - x - 2x^2
\]
Comparing coefficients: \(2a = -1 \Rightarrow a = -\frac{1}{2}\); \(2b + a^2 = -2 \Rightarrow 2b = -2-\frac{1}{4} = -\frac{9}{4} \Rightarrow b = -\frac{9}{8}\).
Answer: \(\sqrt{1-x-2x^2} \approx 1 - \frac{1}{2}x - \frac{9}{8}x^2\) (proved).
Question 4(ii): If \(x\) is so small that its cube and higher powers can be neglected, show that \(\frac{\sqrt{1+x}}{\sqrt{1-x}} \approx 1 + x + \frac{1}{2}x^2\).
\((1+x)^{1/2} \approx 1 + \frac{x}{2} - \frac{x^2}{8}\) aur \((1-x)^{-1/2} \approx 1 + \frac{x}{2} + \frac{3x^2}{8}\). Dono ko \(x^2\) tak multiply karenge.
\[
\left(1+\frac{x}{2}-\frac{x^2}{8}\right)\left(1+\frac{x}{2}+\frac{3x^2}{8}\right) \approx 1 + x + \left(\frac{3}{8}-\frac{1}{8}+\frac{1}{4}\right)x^2 = 1 + x + \frac{1}{2}x^2
\]
Answer: \(\frac{\sqrt{1+x}}{\sqrt{1-x}} \approx 1 + x + \frac{1}{2}x^2\) (proved).
Question 5: If \(x\) is very nearly equal to 1, prove that \(px^p - qx^q \approx (p-q)x^{p+q}\).
\(x = 1+h\) rakhenge jahan \(h\) chhota hai. \((1+h)^p \approx 1+ph\) use karke LHS expand karenge, aur RHS mein \((1+h)^{p+q} \approx 1+(p+q)h\) use karke dono ko match karenge.
Put \(x = 1+h\) with \(h\) small. Then \(x^p = (1+h)^p \approx 1 + ph\) and \(x^q \approx 1 + qh\).
\[
px^p - qx^q \approx p(1+ph) - q(1+qh) = (p-q) + (p^2-q^2)h = (p-q)\big[1+(p+q)h\big]
\]
\[
(p-q)x^{p+q} = (p-q)(1+h)^{p+q} \approx (p-q)\big[1+(p+q)h\big]
\]
Both sides agree up to the first power of \(h\).
Answer: \(px^p - qx^q \approx (p-q)x^{p+q}\) (proved).
Question 6: Identify the following series as a binomial expansion and find its sum: \(1 - \frac{1}{2}\left(\frac{1}{4}\right) + \frac{1\cdot 3}{2\cdot 4}\left(\frac{1}{4}\right)^2 - \frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\left(\frac{1}{4}\right)^3 + \dots\)
Ye \((1+x)^{-1/2}\) ka expansion hai: pehla term 1, doosra \(-\frac{1}{2}x\), teesra \(\frac{1\cdot 3}{2\cdot 4}x^2\). \(x = \frac{1}{4}\) rakhne par saare terms match karte hain. Sum \(= \left(1+\frac{1}{4}\right)^{-1/2}\).
\[
(1+x)^{-\frac{1}{2}} = 1 - \frac{1}{2}x + \frac{1\cdot 3}{2\cdot 4}x^2 - \frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}x^3 + \dots
\]
Comparing with the given series, \(x = \frac{1}{4}\). Hence:
\[
\text{Sum} = \left(1+\frac{1}{4}\right)^{-\frac{1}{2}} = \left(\frac{5}{4}\right)^{-\frac{1}{2}} = \left(\frac{4}{5}\right)^{\frac{1}{2}} = \frac{2}{\sqrt{5}}
\]
Answer: The series is the expansion of \(\left(1+\frac{1}{4}\right)^{-1/2}\); its sum is \(\frac{2}{\sqrt{5}}\).
Question 7: Use the binomial theorem to show that \(1 + \frac{1}{4} + \frac{1\cdot 3}{4\cdot 8} + \frac{1\cdot 3\cdot 5}{4\cdot 8\cdot 12} + \dots = \sqrt{2}\).
Ye \((1+x)^{-1/2}\) ka expansion hai jismein \(x = -\frac{1}{2}\): doosra term \(-\frac{1}{2}\left(-\frac{1}{2}\right) = \frac{1}{4}\) ✓, teesra \(\frac{1\cdot 3}{2\cdot 4}\left(\frac{1}{4}\right) = \frac{3}{32} = \frac{1\cdot 3}{4\cdot 8}\) ✓.
\[
1 + \frac{1}{4} + \frac{1\cdot 3}{4\cdot 8} + \dots = \left(1-\frac{1}{2}\right)^{-\frac{1}{2}} = \left(\frac{1}{2}\right)^{-\frac{1}{2}} = \sqrt{2}
\]
Answer: The series sums to \(\sqrt{2}\) (proved).
Question 8: If \(y = \frac{1}{2}\left(\frac{2}{3}\right) + \frac{1\cdot 3}{2\cdot 4}\left(\frac{2}{3}\right)^2 + \frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\left(\frac{2}{3}\right)^3 + \dots\), prove that \(y^2 + 2y - 2 = 0\).
Ye \((1-x)^{-1/2} - 1\) ka expansion hai jismein \(x = \frac{2}{3}\). To \(1+y = \left(1-\frac{2}{3}\right)^{-1/2} = \sqrt{3}\). Dono taraf square karke required equation mil jayegi.
\[
1 + y = \left(1-\frac{2}{3}\right)^{-\frac{1}{2}} = \left(\frac{1}{3}\right)^{-\frac{1}{2}} = \sqrt{3}
\]
\[
(1+y)^2 = 3 \Rightarrow 1 + 2y + y^2 = 3 \Rightarrow y^2 + 2y - 2 = 0
\]
Answer: \(y^2 + 2y - 2 = 0\) (proved).
Question 9: If \(2y = \frac{1}{2^2} + \frac{1\cdot 3}{2!}\cdot\frac{1}{2^4} + \frac{1\cdot 3\cdot 5}{3!}\cdot\frac{1}{2^6} + \dots\), prove that \(4y^2 + 4y - 1 = 0\).
Note karen: \(\frac{1\cdot 3}{2!}\cdot\frac{1}{2^4} = \frac{1\cdot 3}{2\cdot 4}\left(\frac{1}{2}\right)^2\) aur \(\frac{1\cdot 3\cdot 5}{3!}\cdot\frac{1}{2^6} = \frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\left(\frac{1}{2}\right)^3\). To ye \((1-x)^{-1/2} - 1\) hai jismein \(x = \frac{1}{2}\). Phir \(2y+1 = \sqrt{2}\).
\[
1 + 2y = \left(1-\frac{1}{2}\right)^{-\frac{1}{2}} = \left(\frac{1}{2}\right)^{-\frac{1}{2}} = \sqrt{2}
\]
\[
(1+2y)^2 = 2 \Rightarrow 1 + 4y + 4y^2 = 2 \Rightarrow 4y^2 + 4y - 1 = 0
\]
Answer: \(4y^2 + 4y - 1 = 0\) (proved).
Question 10: Show that the coefficient of \(x^r\) in \(\frac{x}{(1-px)(1-qx)}\) is \(\frac{p^r-q^r}{p-q}\).
Partial fractions use karenge: \(\frac{1}{(1-px)(1-qx)} = \frac{1}{p-q}\left[\frac{p}{1-px} - \frac{q}{1-qx}\right]\). Phir \(\frac{1}{1-px} = \sum_{n\ge 0}p^nx^n\) expand karke \(x^{r-1}\) ka coefficient nikalenge (bahar wala \(x\) power aik barha dega).
\[
\frac{x}{(1-px)(1-qx)} = \frac{x}{p-q}\left[\frac{p}{1-px}-\frac{q}{1-qx}\right] = \frac{x}{p-q}\sum_{n\ge 0}\left(p^{n+1}-q^{n+1}\right)x^n
\]
\[
= \sum_{n\ge 0}\frac{p^{n+1}-q^{n+1}}{p-q}\,x^{n+1}
\]
The coefficient of \(x^r\) (putting \(n+1 = r\)) is \(\frac{p^r-q^r}{p-q}\).
Answer: The coefficient of \(x^r\) is \(\frac{p^r-q^r}{p-q}\) (proved).
