Question 13 MarksExpand using binomial theorem ${\left[ {1 + \frac{x}{2} - \frac{2}{x}} \right]^4},x \ne 0$View full question & answer→
Question 23 MarksFind n, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of ${\left( {\sqrt[4] 2 + \frac{1}{{\sqrt [4]{3} }}} \right)^n}$ is $\sqrt 6 :1$.View full question & answer→
Question 33 MarksFind the value of ${({a^2} + \sqrt {{a^2} - 1} )^4} + {({a^2} - \sqrt {{a^2} - 1} )^4}$View full question & answer→
Question 43 MarksEvaluate $(\sqrt{3}+\sqrt{2})^{6}-(\sqrt{3}-\sqrt{2})^{6}$.View full question & answer→
Question 53 MarksExpand the given expression ${\left( {x + \frac{1}{x}} \right)^6}$AnswerUsing binomial theorem for the expansion of ${\left( {x + \frac{1}{x}} \right)^6}$ we have ${\left( {x + \frac{1}{x}} \right)^6}$= ${ = ^6}{C_0}{(x)^6}{ + ^6}{C_1}{(x)^5}\left( {\frac{1}{x}} \right)$${ + ^6}{C_2}{(x)^4}{\left( {\frac{1}{x}} \right)^2}{ + ^6}{C_3}{(x)^3}{\left( {\frac{1}{x}} \right)^3}$ ${ + ^6}{C_4}{(x)^2}{\left( {\frac{1}{x}} \right)^4}{ + ^6}{C_5}(x){\left( {\frac{1}{x}} \right)^5}$${ + ^6}{C_6}{\left( {\frac{1}{6}} \right)^6}$ $ = {x^6} + 6 \cdot {x^5} \cdot \frac{1}{x} + 15 \cdot 4{x^4} \cdot \frac{1}{{{x^2}}} + $$20 \cdot {x^3} \cdot \frac{1}{{{x^3}}} + 15 \cdot {x^2} \cdot \frac{1}{{{x^4}}} + 6 \cdot x \cdot \frac{1}{{{x^5}}} + \frac{1}{{{x^6}}}$ $ = {x^6} + 6{x^4} + 15{x^2} + 20 + \frac{{15}}{{{x^2}}} + \frac{6}{{{x^4}}} + \frac{1}{{{x^6}}}$View full question & answer→
Question 63 MarksExpand the given expression ${\left( {\frac{x}{3} + \frac{1}{x}} \right)^5}$AnswerUsing binomial theorem for the expansion of ${\left( {\frac{x}{3} + \frac{1}{x}} \right)^5}$ we have ${\left( {\frac{x}{3} + \frac{1}{x}} \right)^5}$${ = ^5}{C_0}{\left( {\frac{x}{3}} \right)^5}{ + ^5}{C_1}{\left( {\frac{x}{3}} \right)^4}\left( {\frac{1}{x}} \right)$${ + ^5}{C_2}{\left( {\frac{x}{3}} \right)^3}{\left( {\frac{1}{x}} \right)^2}{ + ^5}{C_3}{\left( {\frac{x}{3}} \right)^2}{\left( {\frac{1}{x}} \right)^3}$ ${ + ^5}{C_4}\left( {\frac{x}{3}} \right){\left( {\frac{1}{x}} \right)^4}{ + ^5}{C_5}{\left( {\frac{1}{x}} \right)^5}$ $ = \frac{{{x^5}}}{{243}} + 5 \cdot \frac{{{x^4}}}{{81}} \cdot \frac{1}{x} + 10 \cdot \frac{{{x^3}}}{{27}} \cdot \frac{1}{{{x^2}}}$$ + 10 \cdot \frac{{{x^2}}}{9} \cdot \frac{1}{{{x^3}}} + 5 \cdot \frac{x}{3} \cdot \frac{1}{{{x^4}}} + \frac{1}{{{x^5}}}$ $ = \frac{{{x^5}}}{{243}} + \frac{5}{{81}}{x^3} + \frac{{10}}{{27}}x + \frac{{10}}{{9x}} + \frac{5}{{3{x^3}}} + \frac{1}{{{x^5}}}$View full question & answer→
Question 73 MarksExpand the given expression $(2x - 3)^6$AnswerUsing binomial theorem for the expansion of $(2x - 3)^6 $ we have ${(2x - 3)^6}{ = ^6}{C_0}{(2x)^6}{ + ^6}{C_1}{(2x)^5}( - 3)$${ + ^6}{C_2}{(2x)^4}{( - 3)^2}{ + ^6}{C_3}{(2x)^3}{( - 3)^3}$ ${ + ^6}{C_4}{(2x)^2}{( - 3)^4}{ + ^6}{C_5}2{x^5}{( - 3)^2}$${ + ^6}{C_6}{( - 3)^6}$ $= 64x^6 + 6.32x^5 (-3) + 15.16x^4.9 + 20.8x^3 (-27) + 15.4x^2.81 + 6.2x (-243) + 729$ $= 64x^6 - 576x^5 + 2160x^4 - 4320x^3 + 4860x^2 - 2916x + 729$View full question & answer→
Question 83 MarksExpand the given expression ${\left( {\frac{2}{x} - \frac{x}{2}} \right)^5}$AnswerUsing binomial theorem for the expansion of ${\left( {\frac{2}{x} - \frac{x}{2}} \right)^5}$ we have ${\left( {\frac{2}{x} - \frac{x}{2}} \right)^5}$${ = ^5}{C_0}{\left( {\frac{2}{x}} \right)^5}{ + ^5}{C_1}{\left( {\frac{2}{x}} \right)^4}\left( {\frac{{ - x}}{2}} \right)$${ + ^5}{C_2}{\left( {\frac{2}{x}} \right)^3}{\left( {\frac{{ - x}}{2}} \right)^2}{ + ^5}{C_3}{\left( {\frac{2}{x}} \right)^2}{\left( {\frac{{ - x}}{2}} \right)^3}$ ${ + ^5}{C_4}\left( {\frac{2}{x}} \right){\left( {\frac{{ - x}}{2}} \right)^4}{ + ^5}{C_5}{\left( {\frac{{ - x}}{2}} \right)^5}$ $ = \frac{{32}}{{{x^5}}} + 5 \cdot \frac{{16}}{{{x^4}}} \cdot \frac{{ - x}}{2} + 10 \cdot \frac{8}{{{x^3}}} \cdot \frac{{{x^2}}}{4}$$ + 10 \cdot \frac{4}{{{x^2}}} \cdot \frac{{ - {x^3}}}{8} + 5 \cdot \frac{2}{x} \cdot \frac{{{x^4}}}{{16}} + \frac{{ - {x^5}}}{{32}}$ $ = \frac{{32}}{{{x^5}}} - \frac{{40}}{{{x^3}}} + \frac{{20}}{x} - 5x + \frac{5}{8}{x^3} - \frac{{{x^5}}}{{32}}$View full question & answer→
Question 93 MarksShow that $9^{n+1} - 8n - 9$ is divisible by $64$ whenever $n$ is a positive integer.AnswerWe have $9^{n+1}=(1+8)^{n+1}$ $ ={ }^{n+1} C_0+{ }^{n+1} C_1(8)+{ }^{n+1} C_2(8)^2+{ }^{n+1} C_3(8)^3+\ldots+{ }^{n+1} C_{n+1}(8)^{n+1}$ $=1+(n+1) \times 8+{ }^{n+1} C_2(8)^2+{ }^{n+1} C_3(8)^3+\ldots+{ }^{n+1} C_{n+1}(8)^{n+1}$ $=1+8 n+8+{ }^{n+1} C_2(8)^2+{ }^{n+1} C_3(8)^3+\ldots+{ }^{n+1} C_{n+1}(8)^{n+1}$ $=9+8 n+{ }^{n+1} C_2(8)^2+{ }^{n+1} C_3(8)^3+\ldots \ldots \ldots+{ }^{n+1} C_{n+1}(8)^{n+1}$ $\Rightarrow 9^{n+1}-8 n-9={ }^{n+1} C_2(8)^2+{ }^{n+1} C_3(8)^3+\ldots+{ }^{n+1} C_{n+1}(8)^{n+1}$ $=64\left[{ }^{n+1} C_2+{ }^{n+1} C_3 \cdot 8+\ldots+{ }^{n+1} C_{n+1} \cdot 8^{n+1}\right]$ which show that $9^{n+1}-8^{n-9}$ is divisible by 64 wherever $n$ is a positive integerView full question & answer→
Question 103 MarksFind $(x + 1)^6 + (x - 1)^6.$ Hence or otherwise evaluate ${(\sqrt 2 + 1)^6} + {(\sqrt 2 - 1)^6}$Answer$(x + 1)^6 + (x - 1)^6 = $ $= {[^6}{C_0}{x^6}{ + ^6}{C_1}{x^5}{ + ^6}{C_2}{x^4}{ + ^6}{C_3}{x^3}{ + ^6}{C_4}{x^2}{ + ^6}{C_5}x{ + ^6}{C_6}]$ $ + {[^6}{C_0}{x^6}{ + ^6}{C_1}{x^5}( - 1){ + ^6}{C_2}{x^4}{( - 1)^2}{ + ^6}{C_3}{x^3}{( - 1)^3}$${ + ^6}{C_4}{x^2}{( - 1)^4}{ + ^6}{C_5}x{( - 1)^5}{ + ^6}{C_6}{( - 1)^6}]$ $= [x^6 + 6x^5 + 15x^4 + 20x^3 + 15x$^2$+ 6x + 1] + [x^6 - 6x^5 + 15x^4 - 20x^3 + 15x^2 - 6x + 1]$ $= 2x^6 + 30x^4 + 30x^2 + 2$ $= 2(x^6 + 15x^4 + 15x^2 + 1)$ Putting $x = \sqrt 2 $ ${(\sqrt 2 + 1)^6} + {(\sqrt 2 - 1)^6} = 2[{(\sqrt 2 )^6} + 15{(\sqrt 2 )^4} + 15{(\sqrt 2 )^2} + 1]$ $ = 2\left[ {8 + 15 \times 4 + 15 \times 2 + 1} \right]$ $= 2 [8 + 60 + 30 + 1]$ $ = 2 \times 99 = 198$View full question & answer→
Question 113 MarksFind $(a + b)^4 - (a - b)^4.$ Hence, evaluate ${(\sqrt 3 + \sqrt 2 )^4} - {(\sqrt 3 - \sqrt 2 )^4}$Answer$(a + b)^4 = {[^4}{C_0}{a^4}{ + ^4}{C_1}{a^3}b{ + ^4}{C_2}{a^2}{b^2}$${ + ^4}{C_3}a{b^3}{ + ^4}{C_4}{b^4}]$ and ${\left(a-b\right)^4{\;=\lbrack^4}C_0a^4-^4C_1a^3b+^4C_2a^2{b^2}}$${ - ^4}{C_3}a{b^3}{ + ^4}{C_4}{b^4}]$ ${\therefore\;(a+b)^4\;-\left(a-b\right)^4\;=}2\left[{}^4C_1a^3b\;+^4C_3ab^3\right]$ $=2\left[4a^3b\;+4ab^3\right]\;=\;8ab\left[a^2+b^2\right]$ $\therefore\left(\sqrt3\;+\sqrt2\right)^4\;-\;\left(\sqrt3\;-\sqrt2\right)^{4\;}=\;8.\sqrt3.\sqrt2\left[\left(\sqrt3\right)^2+\left(\sqrt2\right)^2\right]$ $=\;8.\sqrt3.\sqrt2\left[3+2\right]\;=\;40.\sqrt3.\sqrt2\;=\;40\sqrt6$View full question & answer→
Question 123 MarksExpand the given expression $(1 - 2x)^5$AnswerUsing binomial theorem for the expansion of $(1 - 2x)^5 $ we have ${(1 - 2x)^5}{ = ^5}{C_0}{ + ^5}{C_1}( - 2x)+\ ^5{C_2}{( - 2x)^2}{ + ^5}{C_3}{( - 2x)^3}{ + ^5}{C_4}{( - 2x)^4}$${ + ^5}{C_5}{( - 2x)^5}$ $= 1 + 5 (-2x) + 10(-2x)^2 + 10(-2x)^3 + 5(-2x)^4 + (-2x)^5$ $= 1 - 10x + 40x^2 - 80x^3 + 80x^4 - 32x^5$View full question & answer→