Search
You can find the results of your search below.
Matching pagenames:
Fulltext results:
- Question 1 Exercise 7.2 @math-11-kpk:sol:unit07
- =====Question 1(i)===== Expand by using Binomial theorem: $(x^2-\dfrac{1}{y})^4$ ====Solution==== Using binomial theorem \begin{align}(x^2-\dfrac{1}{y})^4&=(x^2)^4+{ }^4 ... =====Question 1(ii)===== Expand by using Binomial theorem: $(1+x y)^7$ ====Solution==== Using binonial theorem \begin{align} & (1+x y)^7=1+{ }^7 C_1(1)^6(x y)+{ }^7
- Unit 06: Permutation, Combination and Probability (Solutions) @math-11-kpk:sol
- onditional probability * Recognize the addition theorem ( or law) of probability. * Recognize multiplication theorem (or law) o probability. * Use theorem of addition and multiplication of probability to solve related pro
- Unit 07: Mathmatical Induction and Binomial Theorem (Solutions) @math-11-kpk:sol
- ===== Unit 07: Mathmatical Induction and Binomial Theorem (Solutions) ===== This is a seventh unit of the ... ll positive integer. * State and prove binomial theorem for positive integral index. * Expand $(x+y)^n$ using binomial theorem and find its general term. * Find the specified
- Question 2, Exercise 1.3 @math-11-kpk:sol:unit01
- $$p\left( z \right)={{z}^{3}}+6z+20$$ By factor theorem, $(z-a)$ is a factor of $P(z)$ iff $P(a)=0$. Put ... z \right)={{z}^{3}}-2{{z}^{2}}+z-2$$ \\ By factor theorem $(z-a)$ is a factor of $()$ iff $P(a)=0$.\\ Put $
- Question 9 & 10, Exercise 3.2 @math-11-kpk:sol:unit03
- ine joining the points internally. By using ratio theorem, we have \begin{align}\overrightarrow{OR}&=\dfrac... ine joining the points externally. By using ratio theorem, we have \begin{align}\overrightarrow{OR}&=\dfrac
- Question 11, Exercise 3.2 @math-11-kpk:sol:unit03
- CD}$ in the ratio $2:5$internally, then by ratio theorem, we have position vector $H$ is: \begin{align}\ov... EF}$ externally in the ratio $4:3$, then by ratio theorem, \begin{align}\overrightarrow{OK}&=\dfrac{3\overr
- Question 1 Exercise 7.3 @math-11-kpk:sol:unit07
- i) (1 - x) $\frac{1}{2}$ Solution: Using binomial theorem to tind the four terms $$ \begin{aligned} & (1-x)... s \\ & \end{aligned} $$ Solution: Using binomial theorem $$ \begin{aligned} & (1-x)^{\frac{3}{2}}=1-\frac{
- Multiple Choice Questions (MCQs)
- (B) formula * (C) rational fraction * (D) theorem </col> <col sm="6"> * An arrangement of the num
- Question 11, Exercise 3.3 @math-11-kpk:sol:unit03
- 14+21&=35\\ 35&=35\end{align} Thus by Pytagorous theorem, the vectors $\vec{a}, \vec{b}$ and $\vec{c}$ rep
- Question 11 Review Exercise 6 @math-11-kpk:sol:unit06
- Red })=\dfrac{1}{4}$$ Then by complementary event theorem: \begin{align} P(\text { not red })&=1-P(\text {
- Question 7 Exercise 7.2 @math-11-kpk:sol:unit07
- (iii) $(a+b)^5+(a-b)^5$ Solution: Using binomial theorem $$ \begin{aligned} (a+b)^5+(a-b)^5&=\left[\left(\
- Question 8 Exercise 7.2 @math-11-kpk:sol:unit07
- \end{aligned} Hence by perpe:ts of like binomial theorem, we hisu that: $p+1$ - 5.. 1 - 6 icrm is numerica
- Question 3 Exercise 7.3 @math-11-kpk:sol:unit07
- )^{-\frac{1}{2}} \text {. } $$ Applying binomial theorem, $$ \begin{aligned} & (1-x)^{\frac{1}{2}}(1+x)^{\
- Question 5 and 6 Exercise 7.3 @math-11-kpk:sol:unit07
- }\right)^{-2} \end{aligned} $$ Applying binomial theorem and neglecting $\frac{1}{x^3}$ etc $$ \begin{alig
- Question 7 and 8 Exercise 7.3 @math-11-kpk:sol:unit07
- f the above given equation and apply the binomial theorem $$ \begin{aligned} & (1+x)^{\frac{1}{4}}+(1-x)^{\