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- Question 9 Exercise 7.2
- greatest term in the expansion $(x-y)="$ when $x=12$ and $y-4$. Solution: When we put $x=12$, then the give becounes $$ \begin{aligned} & \left(x \quad y=20(12-y)^{20}\right. \\ & =12^{2 n}\left(\begin{array}{ll} 1 & \frac{y}{12} \end{array}\right)^{31} \end{al
- Question 10 Exercise 7.3
- t{\frac{2}{3}}$. (ii) $1+\frac{5}{8}+\frac{5.8}{8.12}+\frac{5.8 .11}{8.12 .16}+\ldots$ Solution: The given series is binomial series. Let it be identical... ac{5}{8} \\ & \frac{n(n-1)}{2 !} x^2=\frac{5.8}{8.12} . \end{aligned} $$ Taking square of Eq.(1), we ... $$ \begin{aligned} & \frac{n-1}{2 n}=\frac{5.8}{8.12} \cdot \frac{64}{25}=\frac{16}{15} \\ & \Rightarr
- Question 2 Exercise 7.3
- 1.001 . \end{aligned} $$ (iii) Find cube root of 126 correct to five decimal places. Solution: The cube root of 126 can be written as: \begin{aligned} & =5\left[1+\frac{1}{3} \cdot \frac{1}{125}+\right. \\ & \frac{\frac{1}{3}\left(\frac{1}{3}-1\right)}{2 !}\left(\frac{1}{125}\right)^2+ \\ & \left.\frac{\frac{1}{3}\left(\fr
- Question 12 Exercise 7.1
- ====== Question 12 Exercise 7.1 ====== Solutions of Question 12 of Exercise 7.1 of Unit 07: Permutation, Combination and... KPTB or KPTBB) Peshawar, Pakistan. =====Question 12(i)===== Show by mathematical induction that $\dfr... s true for all $n \in \mathbf{N}$. =====Question 12(ii)===== Show by mathematical induction that $\df
- Question 1 Review Exercise 7
- the digits $3$ and $7$ must together? * (a) $120$ * (b) $180$ * %%(c)%% $144$ * %%(d... wer</btn><collapse id="a3" collapsed="true">(a): $120$</collapse> iv. Evaluate $\dfrac{(n+2) !(n-2) !... tee of 4 people will be selected from 8 girls and 12 boys in a class. How many different selections ar... * %%(a)%% $18$ * (b) $27$ * %%(c)%% $512$ * (d) $81$ \\ <btn type="link" collapse="a7
- Question 1 Exercise 7.3
- ht)}{3 !}(-x)^3+\ldots \end{aligned} $$ (iii) $(8+12 x)^{\frac{2}{3}}$ Solution: $$ \begin{aligned} & (8+12 x)^{\frac{2}{3}}=8^{\frac{2}{3}}\left(1+\frac{12 x}{8}\right)^{\frac{2}{3}} \\ & =\left(2^3\right)^{\
- Question 2 Review Exercise 7
- )^{8-4}(3 y)^4 \\ & T_5=70.2^4 \cdot 3^4 \cdot x^{12} \cdot y^4 \\ & =90720 x^{12} y^4 \end{aligned} $$ Hence the middle term of the expansion is $T_5=90720 x^{12} y^4$. ====Go To==== <text align="left"><btn
- Question 5 Exercise 7.2
- {(9-4) ! 4 !}(3 x)^{9-4}(-\dfrac{x^2}{2})^4 \\ & =126 \cdot 3^5 \cdot x^5 \cdot(-1)^4 \cdot \dfrac{(x^... {(9-5) ! 5 !}(3 x)^{9-5}(-\dfrac{x^2}{2})^5 \\ & =126 \cdot(-1)^5 \cdot 3^4 \cdot x^4 \cdot \dfrac{1}{
- Question 7 and 8 Exercise 7.3
- \left[1+\frac{x}{4}-\frac{3 x^2}{32}+\frac{5 x^3}{128}-\ldots\right]+ \\ & {\left[1-\frac{x}{4}-\frac{3 x^2}{32}-\frac{5 x^3}{128}+\ldots\right]} \\ & \end{aligned} $$ Adding an
- Question 12 Exercise 7.3
- ====== Question 12 Exercise 7.3 ====== Solutions of Question 12 of Exercise 7.3 of Unit 07: Permutation, Combination and
- Question 8 Exercise 7.1
- begin{align}1+2+2^2+2^3+\ldots+2^{k-1}-2^k & =2^k-12^k \\ & =2^k+2^k-1=2.2^k-1 \\ \Rightarrow & 1+2+2^
- Question 11 Exercise 7.1
- align="right"><btn type="success">[[math-11-kpk:sol:unit07:ex7-1-p12|Question 12 >]]</btn></text>
- Question 13 Exercise 7.1
- y">[[math-11-kpk:sol:unit07:ex7-1-p12 |< Question 12 ]]</btn></text> <text align="right"><btn type="s
- Question 11 Exercise 7.3
- ext align="right"><btn type="success">[[math-11-kpk:sol:unit07:ex7-3-p10|Question 12 >]]</btn></text>
- Question 13 Exercise 7.3
- y">[[math-11-kpk:sol:unit07:ex7-3-p10 |< Question 12 ]]</btn></text> <text align="right"><btn type="s