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- Question 11 Review Exercise 7
- ====== Question 11 Review Exercise 7 ====== Solutions of Question 11 of Review Exercise 7 of Unit 07: Permut
- Question 9 and 10 Review Exercise 7
- ====== Question 9 and 10 Review Exercise 7 ====== Solutions of Question 9 and 10 of Review Exercise 7 of Uni
- Question 5 & 6 Review Exercise 7
- ====== Question 5 & 6 Review Exercise 7 ====== Solutions of Question 5 & 6 of Review Exercise 7 of Unit 07: ... $\left(\frac{2}{x^2}+\frac{x^2}{2}\right)^{10}$ ? Solution: Here $n=10, a^{\prime}=\frac{2}{x^2}$ and $b=\fr... ^5$ using the first three terms of its expansion. Solution: We can write $$ (0.99)^5=(1-0.01)^5 $$ Using bi
- Question 7 & 8 Review Exercise 7
- ====== Question 7 & 8 Review Exercise 7 ====== Solutions of Question 7 & 8 of Review Exercise 7 of Unit 07: ... tege: n. prove that $7^n-3^n$ is divisible by 4 . Solution: We using mathematical induction to prove the giv... n x)$, for all natural number $n$ where $x>-1$. - Solution: We try to prove this using mathernatical inducti
- Question 2 Review Exercise 7
- ====== Question 2 Review Exercise 7 ====== Solutions of Question 2 of Review Exercise 7 of Unit 07: Permutat... in the expansion of $\left(2 x^3+3 y\right)^8$ ? Solution: In the above $a=2 x^3$, $b=3 y$ and $n=8$. Since
- Question 3 & 4 Review Exercise 7
- ====== Question 3 & 4 Review Exercise 7 ====== Solutions of Question 3 & 4 of Review Exercise 7 of Unit 07: ... e fourth term in the expansion of $(2 x-4 y)^7$ ? Solution: Here $n=7, a=2 x$ and $b=-4 y$. The fourth tenn ... term in the expansion of $(a x+2 y)^4$. Find $a$. Solution: Here $n=7, a^{\prime}=2 x$ and $b=-4 y$. Let $T_
- Question 11 Exercise 7.3
- ====== Question 11 Exercise 7.3 ====== Solutions of Question 11 of Exercise 7.3 of Unit 07: Permutation, Com... rac{1}{2^6}+\ldots$ then show that $y^2+2 y-1=0$. Solution: We are given $y=\frac{1}{2^2}+\frac{1.3}{2 !} \c
- Question 1 Review Exercise 7
- ====== Question 1 Review Exercise 7 ====== Solutions of Question 1 of Review Exercise 7 of Unit 07: Permutat
- Question 10 Exercise 7.3
- ====== Question 10 Exercise 7.3 ====== Solutions of Question 10 of Exercise 7.3 of Unit 07: Permutation, Com... {2^2}+\frac{1.3}{2 !} \cdot \frac{1}{2^4}+\ldots$ Solution: The given series is binomial series. Let it be i... frac{5.8}{8.12}+\frac{5.8 .11}{8.12 .16}+\ldots$ Solution: The given series is binomial series. Let it be i
- Question 9 Exercise 7.3
- ====== Question 9 Exercise 7.3 ====== Solutions of Question 9 of Exercise 7.3 of Unit 07: Permutation, Combi... ime \prime}$ in $\left(\frac{1+x}{1-x}\right)^2$. Solution: Given that: $$ \begin{aligned} & \left(\frac{1+x
- Question 7 and 8 Exercise 7.3
- ====== Question 7 and 8 Exercise 7.3 ====== Solutions of Question 7 and 8 of Exercise 7.3 of Unit 07: Permut... x)^{\frac{1}{4}}=a-b x^2$ then find $a$, and $b$. Solution: We are taking L.H.S of the above given equation ... c{1}{2}}}{1-\frac{5 x}{6}}=\frac{15 x^2}{8} . $$ Solution: We are taking numerator in the L.H.S of the abov
- Question 4 Exercise 7.3
- ====== Question 4 Exercise 7.3 ====== Solutions of Question 4 of Exercise 7.3 of Unit 07: Permutation, Combi... $$ \sqrt{\frac{1-3 x}{1+4 x}}=1-\frac{7 x}{2} $$ Solution: Given that $$ \sqrt{\frac{1-3 x}{1-4 x}}=(1-3 x)
- Question 5 and 6 Exercise 7.3
- ====== Question 5 and 6 Exercise 7.3 ====== Solutions of Question 5 and 6 of Exercise 7.3 of Unit 07: Permut... 2}{3}}}{(2+3 x) \sqrt{4-5 x}}=1-\frac{5 x}{8} $$ Solution: Given that: $$ \frac{\sqrt[4]{3}-3 x j^{\frac{2}... value of: $$ \frac{x \sqrt{x^2-2 x}}{(x+1)^2} $$ Solution: We are given $$ \begin{aligned} & \frac{x \sqrt{
- Question 3 Exercise 7.3
- ====== Question 3 Exercise 7.3 ====== Solutions of Question 3 of Exercise 7.3 of Unit 07: Permutation, Combi... Q3 Expand $\sqrt{\frac{1-x}{1+x}}$ up-to $x^3$. Solution: Given $\sqrt{\frac{1-x}{1+x}}$ $$ =(1-x)^{\frac{
- Question 2 Exercise 7.3
- ====== Question 2 Exercise 7.3 ====== Solutions of Question 2 of Exercise 7.3 of Unit 07: Permutation, Combi... correct to three decimal places. (i) $\sqrt{26}$ Solution: We are given $$ \begin{aligned} & \sqrt{26}=\sqr... & \end{aligned} $$ (ii) $\frac{1}{\sqrt{0.998}}$ Solution: We are given that $$ \frac{1}{\sqrt{0.998}}=(0.9... cube root of 126 correct to five decimal places. Solution: The cube root of 126 can be written as: \begin{a