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- Question 1 Exercise 7.3
- irst four terms in the expansion of (i) (1 - x) $\frac{1}{2}$ Solution: Using binomial theorem to tind the four terms $$ \begin{aligned} & (1-x)^{\frac{1}{2}}=1+\frac{1}{2} x+ \\ & \frac{\frac{1}{2}\left(-\frac{1}{2}-1\right)}{2 !}(-x)^2 \end{aligned} $$ $$ \begin{align
- Question 5 and 6 Exercise 7.3
- igher of $x$ may be negleeled. then show that $$ \frac{(8+3 x)^{\frac{2}{3}}}{(2+3 x) \sqrt{4-5 x}}=1-\frac{5 x}{8} $$ Solution: Given that: $$ \frac{\sqrt[4]{3}-3 x j^{\frac{2}{3}}}{2 \cdot 3 x+4-5 x} $$ $$ \b
- Question 2 Exercise 7.3
- & \sqrt{26}=\sqrt{25+1} \\ & =\sqrt{25} \sqrt{1+\frac{1}{25}}=5\left[1+\frac{1}{25}\right]^{\frac{1}{2}} \end{aligned} $$ Using binomial expansion $$ \begin{aligned} & \sqrt{26}=5\left[1+\frac{1}{25}\right]^{\frac{1}{2}} \\ & =5\left[1+\frac{
- Question 10 Exercise 7.3
- Q10 Find the sum of the following series: (i) $1-\frac{1}{2^2}+\frac{1.3}{2 !} \cdot \frac{1}{2^4}+\ldots$ Solution: The given series is binomial series. Let it be identica... of $(1+x)^n$ that is $$ \begin{aligned} & 1+n x+\frac{n(n-1)}{2 !} x^2 \\ & +\frac{n(n-1(n-2))}{3 !} x^
- Question 12 Exercise 7.3
- (KPTB or KPTBB) Peshawar, Pakistan. Q12 If $2 y=\frac{1}{2^2}+\frac{1.3}{2 !} \cdot \frac{1}{2^4}+\frac{1.3 \cdot 5}{3 !} \cdot \frac{1}{2^6}+\ldots$ then show that $4 y^2+4 y-1=0$. Solution: W
- Question 13 Exercise 7.3
- hat $n^{\text {th }}$ root of $1+x$ is equal to $\frac{2 n+(n+1) x}{2 n+(n-1) x}$ Solution: We have show that $$ \begin{aligned} & (1+x)^{\frac{1}{n}}=\frac{2 n+(n+1) x}{2 n+(n-1) x} \\ & \frac{2 n+(n+1) x}{2 n+(n-1) x} \\ & =1+\frac{1}{n} x+\frac{\frac{1}{n}\
- Question 3 Exercise 7.3
- or KPTBB) Peshawar, Pakistan. Q3 Expand $\sqrt{\frac{1-x}{1+x}}$ up-to $x^3$. Solution: Given $\sqrt{\frac{1-x}{1+x}}$ $$ =(1-x)^{\frac{1}{2}}(1+x)^{-\frac{1}{2}} \text {. } $$ Applying binomial theorem, $$ \begin{aligned} & (1-x)^{\frac{
- Question 7 and 8 Exercise 7.3
- x^4$ and higher powers are neglected and $(1-x)^{\frac{1}{4}}+(1-x)^{\frac{1}{4}}=a-b x^2$ then find $a$, and $b$. Solution: We are taking L.H.S of the above ... the binomial theorem $$ \begin{aligned} & (1+x)^{\frac{1}{4}}+(1-x)^{\frac{1}{4}} \\ & =\left[1+\frac{x}{4}+\frac{\frac{1}{4}\left(\frac{1}{4}-1\right)}{2 !}
- Question 11 Exercise 7.3
- d (KPTB or KPTBB) Peshawar, Pakistan. Q11 If $y=\frac{1}{2^2}+\frac{1.3}{2 !} \cdot \frac{1}{2^4}+\frac{1 \cdot 3 \cdot 5}{3 !} \cdot \frac{1}{2^6}+\ldots$ then show that $y^2+2 y-1=0$. Solutio
- Question 9 Exercise 7.2
- right. \\ & =12^{2 n}\left(\begin{array}{ll} 1 & \frac{y}{12} \end{array}\right)^{31} \end{aligned} $$ Now we theck $\frac{(n+1) \cdot x}{1+|x|}$ for $\left(\frac{1}{12}\right)^2 \cdot$ Here $n=20$ and $\left.\left|x^{\prime}=\right|-\frac{y}{12} \right\rvert\,=\frac{1}{3}$ for $y=4$ $\Ri
- Question 8 Exercise 7.2
- rically greatest term in $(3-2 x)^{10}$. when $x=\frac{3}{4}$. Solution: We first write in form $\left(3-2,1^{10}=3^{10}\left(1-\frac{3 x}{2}\right)^{10}\right.$. The numerically greatest term in $\left(1-\frac{3 x}{2}\right)^{10}$ is: \begin{aligned} & \frac{(1+n) i}{1+i} \quad \frac{1+10}{1}{ }^4 \text { here }
- Question 5 & 6 Review Exercise 7
- is the constant term in the expansion of $\left(\frac{2}{x^2}+\frac{x^2}{2}\right)^{10}$ ? Solution: Here $n=10, a^{\prime}=\frac{2}{x^2}$ and $b=\frac{x^2}{2}$. Let $T_{r+1}$ be the term independent of $x$ that is: $$ \begin{aligned
- Question 4 Exercise 7.3
- of $x$ may be neglected, then show that $$ \sqrt{\frac{1-3 x}{1+4 x}}=1-\frac{7 x}{2} $$ Solution: Given that $$ \sqrt{\frac{1-3 x}{1-4 x}}=(1-3 x)^{\frac{1}{2}}(1+4 x)^{-\frac{1}{2}} $$ Applying binomial expansion and neglecti
- Question 3 & 4 Review Exercise 7
- above expansion is: $$ \begin{aligned} & T_{3+1}=\frac{7 !}{(7-3) ! 3 !}(2 x)^{7 \cdot 3}(-4 y)^3 \\ & =\frac{7 !}{(7-3) ! 3 !} \cdot\left(2^4\right) \cdot(-4)... expansion, that is: $$ \begin{aligned} & T_{r+1}=\frac{4 !}{(4-r)+\infty !}(a x)^{4-r}(2 y)^r \\ & =\frac{4 !}{(4-r) ! r !} 2^r a^4 x^{4-r} y^r \end{aligned}
- Question 9 Exercise 7.3
- the coefficient of $x^{\prime \prime}$ in $\left(\frac{1+x}{1-x}\right)^2$. Solution: Given that: $$ \begin{aligned} & \left(\frac{1+x}{1-x}\right)^2=(1+x)^2(1-x)^{-2} \\ & =\left(... n{aligned} & =\left(x^2+2 x+1\right)[1+2 x+ \\ & \frac{-2(-2-1)}{2 !}(-x)^2 \\ & \left.+\frac{\cdots 2(-2-1)(-2-2)}{3 !}(-x)^3+\cdots\right] \\ & =\left(i^2+2