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- Unit 10: Trigonometric Identities of Sum and Difference of Angles (Solutions)
- Question 9 & 10, Exercise 1.1
- Question 10, Exercise 2.1
- Question 11, Exercise 2.1
- Question 8,9 & 10, Exercise 2.2
- Question 12, Exercise 2.2
- Question 9 & 10, Exercise 3.2
- Question 9 & 10, Exercise 3.3
- Question 10 Review Exercise 3
- Question 10 Exercise 4.2
- Question 14 Exercise 4.2
- Question 9 & 10 Exercise 4.3
- Question 10 Exercise 4.4
- Question 9 & 10 Exercise 4.5
- Question 15 & 16 Exercise 4.5
- Question 10 Review Exercise
- Question 10 Exercise 6.2
- Question 14 and 15 Exercise 6.2
- Question 10 Exercise 6.5
- Question 9 & 10 Review Exercise 6
- Question 10 Exercise 7.1
- Question 10 Exercise 7.2
- Question 10 Exercise 7.3
- Question 12 Exercise 7.3
- Question 9 and 10 Review Exercise 7
- Question 1, Exercise 10.1
- Question 2, Exercise 10.1
- Question 3, Exercise 10.1
- Question, Exercise 10.1
- Question 5, Exercise 10.1
- Question 6, Exercise 10.1
- Question 7, Exercise 10.1
- Question 8, Exercise 10.1
- Question 9 and 10, Exercise 10.1
- Question11 and 12, Exercise 10.1
- Question 13, Exercise 10.1
- Question 1, Exercise 10.2
- Question 2, Exercise 10.2
- Question 3, Exercise 10.2
- Question 4 and 5, Exercise 10.2
- Question 6, Exercise 10.2
- Question 7, Exercise 10.2
- Question 8 and 9, Exercise 10.2
- Question 1, Exercise 10.3
- Question 2, Exercise 10.3
- Question 3, Exercise 10.3
- Question 5, Exercise 10.3
- Question 5, Exercise 10.3
- Question 1, Review Exercise 10
- Question 2 and 3, Review Exercise 10
- Question 4 & 5, Review Exercise 10
- Question 6 & 7, Review Exercise 10
- Question 8 & 9, Review Exercise 10
Fulltext results:
- Question 12 Exercise 7.1 @math-11-kpk:sol:unit07
- ===== Show by mathematical induction that $\dfrac{10^{n+1}-9 n-10}{81}$ is an integer. ====Solution==== 1. For $n=1$ then \begin{align}\dfrac{10^{n+1}-9 n-10}{81}&=\dfrac{10^{i+1}-9.1-10}{81} \\ & =\dfrac{100-9-10}{81}=1 \in \mathbb{Z}\end{align}
- Question 13 Exercise 6.2 @math-11-kpk:sol:unit06
- e total number of letters in 'Excellence' are: $n=10$, out of which $m_1=4$ are $E, m_2=2$ are $L$ and... m_3 \end{array}\right)\\&=\left(\begin{array}{c} 10 \\ 4,2,2 \end{array}\right) \\ & =\dfrac{10 !}{4 ! \cdot 2 ! \cdot 2 !}\\ &=37,800 \end{align} Begin ... e total number of letters in 'Excellence' are: $n=10$, out of which $m_1=4$ are $E$, $m_2=2$ are $L
- Question 13 & 14 Exercise 4.5 @math-11-kpk:sol:unit04
- from which it is dropped. If it is dropped from $10 \mathrm{ft}$, how far does it travel from the mom... s eighth bounce? ====Solution==== \begin{align}S&=10+[10(\dfrac{1}{2})+10(\dfrac{1}{2})]+ \\ & {[10(\dfrac{1}{2})^2+10(\dfrac{1}{2})^2]+\ldots} \\ & +[10(\dfr
- Question 3 Exercise 7.2 @math-11-kpk:sol:unit07
- dent of $x$ in the expansion $(x-\dfrac{3}{x^4})^{10}$ ====Solution==== In the above expansion $n=10, \quad a=x$ and $b=-\dfrac{3}{x^4}$. Let $T_{r+1}$ b... given expansion is: \begin{align} T_{r+1}&=\dfrac{10 !}{(10-r) ! r !}(x)^{10 \cdot r}(-\dfrac{3}{x^4})^r \\ & =\dfrac{10 !}{(10-r) ! r !} \cdot(-3)^r x^{1
- Question 5 & 6 Review Exercise 7 @math-11-kpk:sol:unit07
- ion 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 ... f $x$ that is: $$ \begin{aligned} & T_{r+1}=\frac{10 !}{(10-r) ! r !}\left(\frac{2}{x^2}\right)^{10 r}\left(\frac{x^2}{2}\right)^r \\ & =\frac{10 !}{(10-r
- Question 5 Exercise 7.2 @math-11-kpk:sol:unit07
- e total number of terms in the expansion are $9+1=10$. So in this we have two middle terms that are $... ^4 \cdot \dfrac{1}{2^5} \cdot(x^2)^5\\ &=-\dfrac{5103}{16} x^{14} \end{align} Thus the two middle term... dfrac{15309}{8} x^{13} \text { and } T_6=-\dfrac{5103}{16} x^{14}$$ =====Question 5(iii)===== Find m... e term in the expansion of $(3 x^2-\dfrac{y}{3})^{10}$ ====Solution==== Since we see that $a=3 x^2$, $
- Question 8 Exercise 7.2 @math-11-kpk:sol:unit07
- Q8 Find numerically 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}\ri
- Question 4 Review Exercise @math-11-kpk:sol:unit05
- Sum the series: $\dfrac{1}{1.4 .7}+\dfrac{1}{4.7 .10}+\dfrac{1}{7.10 .13}+\ldots$ ====Solution==== In the denominator Each term is the product of the succ... dfrac{1}{7})+(\dfrac{1}{4}-\dfrac{2}{7}+\dfrac{1}{10})+ \\ &\cdots+(\dfrac{1}{3 n-2}-\dfrac{2}{3 n+1}+... rac{1}{18}[(1+\dfrac{1}{4}+\dfrac{1}{7}+\dfrac{1}{10} \cdots+\dfrac{1}{3 n-2}) \\ & -(\dfrac{2}{4}+\df
- Question 5 Exercise 6.4 @math-11-kpk:sol:unit06
- n. ====Solution==== Total number of persons $=6+4=10$. Total number of ways to select $5$ out of these $10$ are: \begin{align}{ }^{10)} C_5 &=\dfrac{10 !}{(10-5) ! 5 !}\\ &=252\\ n(S)&=252\end{align} When $3$ men and $2$ women. By mu
- Question 5 & 6 Exercise 4.5 @math-11-kpk:sol:unit04
- tan. =====Question 5===== Find $r$ such that $S_{10}=244 S_5$. ====Solution==== We know that $$S_n=\dfrac{a_1(r^n-1)}{r-1}$$\\ then\\ $$S_{10}=\dfrac{a_1(r^{10}-1)}{r-1} \quad \text{and}\quad S_5=\dfrac{a_1(r^5-1)}{r-1}$$\\ Putting both the $S_{10}$ and $S_S$ in the given equation, we get\\ \begi
- Question 7 Exercise 6.4 @math-11-kpk:sol:unit06
- 6$\}$, then from sample space, we see that $n(B)=10$. Thus the probability of getting a number less than 6 is: $$P(B)=\dfrac{n(B)}{n(S)}=\dfrac{10}{36}=\dfrac{5}{18}$$ =====Question 7(iii)===== T... nd the probability of getting a sum greater than $10.$ ====Solution==== The sample space rolling a pa... ts are $$n(S)=6 \times 6=36$$ A sum greater than $10$ Let $D=\{$ a sum greater than 10$\}$, then fro
- Question 5, Exercise 10.3 @math-11-kpk:sol:unit10
- ====== Question 5, Exercise 10.3 ====== Solutions of Question 5 of Exercise 10.3 of Unit 10: Trigonometric Identities of Sum and Difference of Angles. This is unit of A Textbook of... ==Question 5(iii)===== Prove the identity $\sin {{10}^{\circ }}\sin {{30}^{\circ }}\sin {{50}^{\circ }
- Question 5, Exercise 10.3 @math-11-kpk:sol:unit10
- ====== Question 5, Exercise 10.3 ====== Solutions of Question 5 of Exercise 10.3 of Unit 10: Trigonometric Identities of Sum and Difference of Angles. This is unit of A Textbook of... ==Question 5(iii)===== Prove the identity $\sin {{10}^{\circ }}\sin {{30}^{\circ }}\sin {{50}^{\circ }
- Question 16 & 17, Exercise 2.2 @math-11-kpk:sol:unit02
- d{matrix} \right]$$ $$|A|=6+4$$ $$\Rightarrow |A|=10\ldots (1)$$ $$A^{-1}=\dfrac{1}{|A|}AdjA$$ $$AdjA=... 3 \\ \end{matrix} \right]$$ $$A^{-1}=\dfrac{1}{10}\left[ \begin{matrix} 2 & 1 \\ -4 & 3 \\ ... } \right]$$ $$=\left[ \begin{matrix} \dfrac{2}{10} & \dfrac{1}{10} \\ -\dfrac{4}{10} & \dfrac{3}{10} \\ \end{matrix} \right]$$ $$A^{-1}=\left[ \be
- Question 1 Exercise 4.5 @math-11-kpk:sol:unit04
- )^{n-1}&=2^9 \\ \Rightarrow n-1&=9 \text { or } n=10 \\ \text {. Now }\quad S_n&=\dfrac{a_1(r^n-1)}{r... lign} becomes in the given case\\ \begin{align}S_{10}&=\dfrac{3[2^{10}-1]}{2-1} \\ \Rightarrow \quad S_{10}&=3(2^{10}-1)\end{align}\\ is the required sum. =====Question 1