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- Question 7 Exercise 7.2
- (i)===== Find $(2+\sqrt{3})^5+(2-\sqrt{3})^5$ ====Solution==== Using binomial formula \begin{align}(2+\sqrt{... 7(ii)===== $(1+\sqrt{2})^4-(1-\sqrt{2})^{-}$ ====Solution==== Using binomial formula \begin{align} (1+\sqrt... ==Question 7(iii)===== Find $(a+b)^5+(a-b)^5$ ====Solution==== (iii) $(a+b)^5+(a-b)^5$ Solution: Using binomial theorem $$ \begin{aligned} (a+b)^5+(a-b)^5&=\left[\le
- Question 1 Exercise 7.2
- sing Binomial theorem: $(x^2-\dfrac{1}{y})^4$ ====Solution==== Using binomial theorem \begin{align}(x^2-\dfr... Expand by using Binomial theorem: $(1+x y)^7$ ====Solution==== Using binonial theorem \begin{align} & (1+x y... theorem: $(\sqrt{y}+\dfrac{1}{\sqrt{y}})^5 $ ====Solution==== Using binomial theorem \begin{align}(\sqrt{y}
- Question 2 Exercise 7.2
- in the expansion $4^{th}$ term in $(2+a)^7$. ====Solution==== $\ln$ the above $n=7$, $a=2$ and $b=a$. Thus... }$ term in $(\dfrac{x}{2}-\dfrac{3}{y})^{10}$ ====Solution==== In the above $n=10$, $a=\dfrac{x}{2}$ and $b=... rd }}$ term in $(x^2+\dfrac{1}{\sqrt{x}})^4$. ====Solution==== In the above expansion $n=$ 21, $a=x$ and $b=
- Question 3 Exercise 7.2
- expansion $(\dfrac{4 x^2}{3}-\dfrac{3}{2 x})$ ====Solution==== In the above expansion $n=9, \quad a=\dfrac{4... x$ in the expansion $(x-\dfrac{3}{x^4})^{10}$ ====Solution==== In the above expansion $n=10, \quad a=x$ and ... $ in the expansion $(x-\dfrac{1}{x^2})^{2 !}$ ====Solution==== In the above expansion $n=$ 21, $a=x$ and $b=
- Question 4 Exercise 7.2
- the coefticient of $x^{23}$ in $(x^2-x)^{20}$ ====Solution==== In the above expansion $n=20, \quad a=x^2$ an... t of $\dfrac{1}{x^4}$ in $(2-\dfrac{1}{x})^x$ ====Solution==== In the above expansion $n=8, \quad a=2$ and $... icient of $a^6 b^3$ in $(2 a-\dfrac{b}{3})^9$ ====Solution==== In the above expansion $n=9 . \quad a=2a$ and
- Question 5 Exercise 7.2
- rm in the expansion of $(\dfrac{a}{x}+b x)^8$ ====Solution==== Since we see that $a=\dfrac{a}{x}$. $b=b x$ a... in the expansion of $(3 x-\dfrac{x^2}{2})^9$ ====Solution==== Since we see that $a=3 x$, $b=-\dfrac{x^2}{2}... the expansion of $(3 x^2-\dfrac{y}{3})^{10}$ ====Solution==== Since we see that $a=3 x^2$, $b=-\dfrac{y}{3}
- Question 1 Exercise 7.3
- rms in the expansion of (i) (1 - x) $\frac{1}{2}$ Solution: Using binomial theorem to tind the four terms $$... ^2+\frac{1}{16} x^3+\ldots \\ & \end{aligned} $$ Solution: Using binomial theorem $$ \begin{aligned} & (1-x... \end{aligned} $$ (iii) $(8+12 x)^{\frac{2}{3}}$ Solution: $$ \begin{aligned} & (8+12 x)^{\frac{2}{3}}=8^{\
- Question 2 Exercise 7.3
- 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 $$ \frac{1}{\sqrt{0.998}}=(0.9... cube root of 126 correct to five decimal places. Solution: The cube root of 126 can be written as: \begin{a
- Question 12 Exercise 7.1
- ction that $\dfrac{5^{2 n}-1}{24}$ is an integer. Solution: 1. For $n=1$, then $$\dfrac{5^{2 n}-1}{24}=\dfra... $\dfrac{10^{n+1}-9 n-10}{81}$ is an integer. ====Solution==== 1. For $n=1$ then \begin{align}\dfrac{10^{n+1
- Question 13 Exercise 7.1
- 13(i)===== $2^n>n \forall n \in \mathbf{N}$. ====Solution==== 1. For $n=1$ then $2^n=2^1=2$ and $n=1$. Cle... === $n$ ! $>n^2$ for every integer $n \geq 4$ ====Solution==== 1. For $n=4$ then $n !=4 !=24$ and $n^2=4^2=1
- Question 14 Exercise 7.1
- 2^{2 n-1}$ where $n$ is any positive integer. ====Solution==== 1. For $n=1$ then $$3^{2 n-1}+2^{2 n-1}=3^{2.... is a multiple of $3$ for all natural numbers. ====Solution==== 1. For $n=1$ then $$2^{2 n}-1=2^{2.1}-1=4-1=3
- Question 5 and 6 Exercise 7.3
- 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 $$ \begin{aligned} & \frac{x \sqrt{
- Question 7 and 8 Exercise 7.3
- x)^{\frac{1}{4}}=a-b x^2$ then find $a$, and $b$. Solution: We are taking L.H.S of the above given equation ... c{1}{2}}}{1-\frac{5 x}{6}}=\frac{15 x^2}{8} . $$ Solution: We are taking numerator in the L.H.S of the abov
- Question 10 Exercise 7.3
- {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 3 & 4 Review Exercise 7
- 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_