(Each question 3 marks) - MATHS STD 11 Science Questions - Vidyadip

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21 questions · self-marked practice — reveal the answer and mark yourself.

Question 13 Marks

Expand using binomial theorem ${\left[ {1 + \frac{x}{2} - \frac{2}{x}} \right]^4},x \ne 0$

Answer

We have ${\left[ {1 + \frac{x}{2} - \frac{2}{x}} \right]^4} = {\left[ {1 + \left( {\frac{x}{2} - \frac{2}{x}} \right)} \right]^4}$
${ = ^4}{C_0}{ + ^4}{C_1}\left( {\frac{x}{2} - \frac{2}{x}} \right){ + ^4}{C_2}{\left( {\frac{x}{2} - \frac{2}{x}} \right)^2}$${ + ^4}{C_3}{\left( {\frac{x}{2} - \frac{2}{x}} \right)^3}{ + ^4}{C_4}{\left( {\frac{x}{2} - \frac{2}{x}} \right)^4}$
$ = 1 + 4\left( {\frac{x}{2} - \frac{2}{x}} \right) + 6\left( {\frac{{{x^2}}}{4} + \frac{4}{{{x^2}}} - 2} \right)$$ + 4\left( {\frac{{{x^3}}}{8} - \frac{8}{{{x^3}}} - \frac{{3x}}{2} + \frac{6}{x}} \right)$
$ + \left[ {^4{C_0}{{\left( {\frac{x}{2}} \right)}^4}{ - ^4}{C_1}{{\left( {\frac{x}{2}} \right)}^3}\left( {\frac{2}{x}} \right){ + ^4}{C_2}} \right.$${\left( {\frac{x}{2}} \right)^2}$$\left. {{{\left( {\frac{2}{x}} \right)}^2}{ - ^4}{C_3}\left( {\frac{x}{2}} \right){{\left( {\frac{2}{x}} \right)}^3}{ + ^4}{C_4}{{\left( {\frac{2}{x}} \right)}^4}} \right]$
$ = 1 + \left( {2x - \frac{8}{x}} \right) + \left( {\frac{3}{2}{x^2} + \frac{{24}}{{{x^2}}} - 12} \right)$$ + \left( {\frac{{{x^3}}}{2} - \frac{{32}}{{{x^3}}} - 6x + \frac{{24}}{x}} \right)$
$ + \left( {\frac{{{x^4}}}{{16}} - {x^2} + 6 - \frac{{16}}{{{x^2}}} + \frac{{16}}{{{x^4}}}} \right)$
$ = - 5 - 4x + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{2} + \frac{{{x^4}}}{{16}} + \frac{{16}}{x} + \frac{8}{{{x^2}}}$$ - \frac{{32}}{{{x^3}}} + \frac{{16}}{{{x^4}}}$

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Question 23 Marks

Find 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 $\sqrt 6 :1$.

Answer

We have ${\left( {\sqrt[4] 2 + \frac{1}{{\sqrt [4]{3} }}} \right)^n}$
$5^{th}$ term from the beginning ${ = ^n}{C_4}{(\sqrt[4]{2})^{n - 4}}{\left( {\frac{1}{{\sqrt [4]{3} }}} \right)^4}$
$5^{th}$ term from the end $= (n + 1 - 5 + 1)^{th}$ term from beginning
$= (n - 3)^{th}$ term from beginning
${ = ^n}{C_{n - 4}}{(\sqrt[4]{2})^4}{\left( {\frac{1}{{\sqrt[4]{3}}}} \right)^{n - 4}}$
Now $\frac{{^n{C_4}{{(\sqrt[4]{2})}^{n - 4}}{{\left( {\frac{1}{{\sqrt[4]{3}}}} \right)}^4}}}{{^n{C_{n - 4}}{{(\sqrt[4]{2})}^4}{{\left( {\frac{1}{{\sqrt [4]3 }}} \right)}^{n - 4}}}} = \frac{{\sqrt 6 }}{1}$
$ \Rightarrow (2)^\frac{{n - 8}}{4} \cdot (3)^\frac{{n - 8}}{4} = {2^{\frac{1}{2}}} \times {3^{\frac{1}{2}}}$
$\frac{{n - 8}}{4} = \frac{1}{2} \Rightarrow n - 8 = 2$
$ \Rightarrow n = 10$

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Question 33 Marks

Find the value of ${({a^2} + \sqrt {{a^2} - 1} )^4} + {({a^2} - \sqrt {{a^2} - 1} )^4}$

Answer

Putting $a^2 = x$ and $\sqrt {{a^2} - 1} = y$ we have
${({a^2} + \sqrt {{a^2} - 1} )^4} + {({a^2} - \sqrt {{a^2} - 1} )^4}$$ = {(x + y)^4} + {(x - y)^4}$
$ = {[^4}{C_0}{x^4}{ + ^4}{C_1}{x^3}y{ + ^4}{C_2}{x^2}{y^2}$${ + ^4}{C_3}x{y^3}{ + ^4}{C_4}{y^4}] + {[^4}{C_4}{x^4}{ - ^4}{C_1}{x^3}y$${ + ^4}{C_2}{x^2}{y^2}{ - ^4}{C_3}x{y^3}{ + ^4}{C_4}{y^4}]$
$ = 2{[^4}{C_0}{x^4}{ + ^4}{C_2}{x^2}{y^2}{ + ^4}{C_4}{y^4}]$$ = 2[{x^4} + 6{x^2}{y^2} + {y^4}]$
$ = 2[{({a^2})^4} + 6{({a^2})^2}{(\sqrt {{a^2} - 1} )^2}$$ + {(\sqrt {{a^2} - 1} )^4}]$
$ = 2[{a^8} + 6{a^4}({a^2} - 1) + {({a^2} - 1)^2}]$
$ = 2[{a^8} + 6{a^6} - 6{a^4} + {a^4} - 2{a^2} + 1]$
$ = 2[{a^8} + 6{a^6} - 5{a^4} - 2{a^2} + 1]$

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Question 43 Marks

Evaluate $(\sqrt{3}+\sqrt{2})^{6}-(\sqrt{3}-\sqrt{2})^{6}$.

Answer

Using Binomial Theorem.
This can be done as
$(a+b)^6={ }^6 C_0 a^6+{ }^6 C_1 a^5 b+{ }^6 C_2 a^4 b^2+{ }^6 C_3 a^3 b^3+{ }^6 C_4 a^2 b^4+{ }^6 C_5 a^1 b^5+{ }^6 C_6 b^6$
$=a^6+6 a^5 b+15 a^4 b^2+20 a^3 b^3+15 a^2 b^4+6 a b^5+b^6$
$(a-b)^6={ }^6 C_0 a^6-{ }^6 C_1 a^5 b+{ }^6 C_2 a^4 b^2-{ }^6 C_3 a^3 b^3+{ }^6 C_4 a^2 b^4-{ }^6 C_5 a^1 b^5+{ }^6 C_6 b^6$
$=a^6-6 a^5 b+15 a^4 b^2-20 a^3 b^3+15 a^2 b^4-6 a b^5+b^6$
Therefore $(a+b)^6-(a-b)^6=2\left[6 a^5 b+20 a^3 b^3+6 a b^5\right]$
Puting a = $\sqrt3$ and b = $\sqrt2$, we obtain
$(\sqrt{3}+\sqrt{2})^{6}-(\sqrt{3}-\sqrt{2})^{6}=2\left[6(\sqrt{3})^{5}(\sqrt{2})+20(\sqrt{3})^{3}(\sqrt{2})^{3}+6(\sqrt{3})(\sqrt{2})^{5}\right]$
$=2[54 \sqrt{6}+120 \sqrt{6}+24 \sqrt{6}]$
$=2 \times 198 \sqrt{6}$
$=396 \sqrt{6}$

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Question 53 Marks

If a and b are distinct integers, prove that$a - b$ is a factor of $a^n - b^n$, whenever $n$ is a positive integer.
[Hint write $a^n = (a – b + b)^n$ and expand]

Answer

We have $a^n = [(a - b) +b]^n$
${ = ^n}{C_0}{(a - b)^n}{ + ^n}{C_1}{(a - b)^{n - 1}} \cdot b$${ + ^n}{C_2}{(a - b)^{n - 2}} \cdot {b^2} + .... + $$^n{C_{n - 1}}(a - b){b^{n - 1}}{ + ^n}{C_n}{b^n}$
$ = {(a - b)^n}{ + ^n}{C_1}{(a - b)^{n - 1}} \cdot b$${ + ^n}{C_2}{(a - b)^{n - 2}} \cdot {b^2} + .... + $$^n{C_{n - 1}}(a - b){b^{n - 1}} +{b^n}$
${\Rightarrow\;a^n-b^n={(a-b)^n}+^nC_1{(a-b)^{n-1}}\cdot b}$${ + ^n}{C_2}{(a - b)^{n - 2}} \cdot {b^2} + ....{ + ^n}{C_{n - 1}}(a - b)$${b^{n - 1}}$
$ = (a - b)[{(a - b)^{n - 1}}{ + ^n}{C_1}{(a - b)^{n - 2}} \cdot b$${ + ^n}{C_2}{(a - b)^{n - 3}} \cdot {b^2} + .....{ + ^n}{C_{n - 1}}{b^{n - 1}}]$
Which shows that $(a - b)$ is a factor of $a^n - b^n$.

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Question 63 Marks

Find a if the coefficient of $x^2$ and $x^3$ in the expansion of $(3 + ax)^9$ are equal.

Answer

Here $(3 + ax)^9 { = ^9}{C_0}{(3)^9}{ + ^9}{C_1}{(3)^8}(ax)$${ + ^9}{C_2}{(3)^7}{(ax)^2}{ + ^9}{C_3}{(3)^6}{(ax)^3} + ...$
${ = ^9}{C_0}{(3)^9}{ + ^9}{C_1}{(3)^8} \cdot a \cdot x$${ + ^9}{C_2}{(3)^7}{(a)^2} \cdot {x^2}{ + ^9}{C_3}{(3)^6} \cdot {a^3}{x^3} + ...$
$\therefore $ Coefficient of ${x^2}{ = ^9}{C_2}{(3)^7}{a^2}$
Coefficient of ${x^3}{ = ^9}{C_3}{(3)^6}{a^3}$
It is given that
$^9{C_2}{(3)^7}{a^2}{ = ^9}{C_3}{(3)^6}{a^3}$$ \Rightarrow 36 \cdot {3^7}{a^2} = 84 \cdot {3^6} \cdot {a^3}$
$ \Rightarrow a = \frac{{36 \cdot {3^7}}}{{84 \cdot {3^6}}} = \frac{{108}}{{84}} = \frac{9}{7}$.

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Question 73 Marks

Find the expansion of $(3x^2 - 2ax + 3a^2)^3$ using binomial theorem.

Answer

We have
$(3x^2 - 2ax + 3a^2)^3 = [(3x^2 - 2ax) + 3a^2)]^3$
${ = ^3}{C_0}{(3{x^2} - 2ax)^3}{ + ^3}{C_1}{(3{x^2} - 2ax)^2}$$(3{a^2}){ + ^3}{C_2}(3{x^2} - 2ax){(3{a^2})^2}{ + ^3}{C_3}{(3{a^2})^3}$
$ = {(3{x^2} - 2ax)^3} + 3 \times 3{a^2}{(3{x^2} - 2ax)^2}$$ + 3 \times 9{a^4}(3{x^2} - 2ax) + 27{a^6}$
$ = (27{x^6} - 8{a^3}{x^3} - 54a{x^5} + 36{a^2}{x^4})$$ + 9{a^2}(9{x^4} + 4{a^2}{x^2} - 12a{x^3})$$ + 27{a^4}(3{x^2} - 2ax) + 27{a^6}$
$ = 27{x^6} - 8{a^3}{x^3} - 54a{x^5} + 36{a^2}{x^4}$$ + 81{a^2}{x^4} + 36{a^4}{x^2} - 108{a^3}{x^3} + 81{a^4}{x^2}$$ - 54{a^5}x + 27{a^6}$
$ = 27{x^6} - 54a{x^5} + 117{a^2}{x^4} - 116{a^3}{x^3}$$ + 117{a^4}{x^2} - 54{a^5}x27{a^6}$

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Question 83 Marks

In the expansion of $(1+a)^{m+n}$, prove that the coefficients of $a^m$ and $a^n$ are equal.

Answer

In the expansion of $(1+a)^{m+n}$, the general term is given by
$T_{r+1}={ }^{m+n} C_r(a)^r$
For coefficient of $a^m$, put $r=m$.
Then, $\mathrm{T}_{\mathrm{m}+1}={ }^{\mathrm{m}+\mathrm{n}} \mathrm{C}_{\mathrm{m}} \mathrm{a}^{\mathrm{m}}$
$\therefore$ coefficient of $a^m={ }^{m+n} C_m \ldots$ (i)
For coefficient of $a^n$, put $r=n$, then
$T_{n+1}={ }^{m+n} C_n a^n$
$\therefore \text { coefficient of an }={ }^{m+n} C_n$
$={ }^{m+n} C_{m+n-n}\left[\because{ }^n C_r={ }^n C_{n-r}\right]$
$={ }^{m+n} C_m \ldots \text { (ii) }$
From Eqs. (i) and (ii), the coefficient of $a^m$ is equal to coefficient of $a^n$. Hence proved.

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Question 93 Marks

Find the middle terms in the expansion of : ${\left( {\frac{x}{3} + 9y} \right)^{10}}$

Answer

Here n = 10, which is even.
So the middle term is $\left( {\frac{{10}}{2} + 1} \right)$ i.e. $6^{th}$ term
The general term in the expansion of ${\left( {\frac{x}{3} + 9y} \right)^{10}}$ is
${T_{r + 1}}{ = ^{10}}{C_r}{\left( {\frac{x}{3}} \right)^{10 - r}} \cdot {(9y)^r}$ . . . (i)
Putting r = 5 in (i)
${T_6}{ = ^{10}}{C_5}{\left( {\frac{x}{3}} \right)^{10 - 5}} \cdot {(9y)^5}$
${ = ^{10}}{C_5}\frac{{{x^5}}}{{{3^5}}} \cdot {9^5} \cdot {y^5}$$ = 252 \times \frac{{{x^5}}}{{243}} \times 59049{y^5}$
= 61236 $x^5y^5$

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Question 103 Marks

Find the middle term in the expansion of $(3 - \frac { x ^ { 3 } } { 6 })^7$.

Answer

Here, $n = 7$ (odd)
$\therefore$ Miiddle terms are obtained $\left( \frac { n + 1 } { 2 } \right)$th and $\left( \frac { n + 3 } { 2 } \right)$th terms i.e., $\left( \frac { 7 + 1 } { 2 } \right)$th term and $\left( \frac { 7 + 3 } { 2 } \right)$th term i.e., 4th and 5th terms.
Now, $T_4 = T_{3+1} = ^7C_3 (3)^{7-3} \left( - \frac { x ^ { 3 } } { 6 } \right) ^ { 3 }$
[$\because T_{r+1} = ^nC_rx^{n-r} y^r$]
= $\frac { 7 \times 6 \times 5 } { 6 } \times \frac { 3 ^ { 4 } ( - 1 ) ^ { 3 } \times x ^ { 9 } } { 6 ^ { 3 } }$
= $\frac { 35 \times 81 \times ( - 1 ) \times x ^ { 9 } } { 6 \times 6 \times 6 }$
= - $\frac { 35 \times 27 \times x ^ { 9 } } { 2 \times 6 \times 6 }$ = - $\frac { 105 } { 8 } x ^ { 9 }$
and $T_5 = T_{4+1} = ^7C_4(3)^{7-4} \left( - \frac { x ^ { 3 } } { 6 } \right) ^ { 4 }$
= $^7C_3 \frac { 3 ^ { 3 } ( - 1 ) ^ { 4 } \left( x ^ { 3 } \right) ^ { 4 } } { 6 ^ { 4 } }$ [$\because ^nC_r = ^nC_{r-1}$]
= $\frac { 7 \times 6 \times 5 } { 6 } \times \frac { 3 \times 3 \times 3 \times x ^ { 12 } } { 6 \times 6 \times 6 \times 6 } = \frac { 35 } { 48 } x ^ { 12 }$

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Question 113 Marks

Find the $13^{th}$ term in the expansion of ${\left[ {9x - \frac{1}{{3\sqrt x }}} \right]^{18}},x \ne 0$.

Answer

Here general term in the expansion of ${\left[ {9x - \frac{1}{{3\sqrt x }}} \right]^{18}}$
${T_{r + 1}}{ = ^{18}}{C_r}{(9x)^{18 - r}}{\left( { - \frac{1}{{3\sqrt x }}} \right)^r}$ . . . (i)
Putting r = 12 in (i)
${T_{13}}{ = ^{18}}{C_{12}}{(9x)^{18 - 12}}{\left( { - \frac{1}{{3\sqrt x }}} \right)^{12}}$
${ = ^{18}}{C_{12}}{9^6}{x^6} \cdot {( - 1)^{12}} \cdot \frac{1}{{{3^{12}} \cdot {x^6}}}$
${ = ^{18}}{C_{12}}\frac{{{9^6}}}{{{3^{12}}}}{ = ^{18}}{C_{12}} = 18564$

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Question 123 Marks

Find the $4^{th}$ term in the expansion of $(x - 2y)^{12}$.

Answer

Here general term in the expansion of $(x - 2y)^{12}$ is
${T_{r + 1}}{ = ^{12}}{C_r}{(x)^{12 - r}} \cdot {( - 2y)^r}$
$ = {( - 1)^r}^{12}{C_r}{2^r} \cdot {x^{12 - r}} \cdot {y^r}$
Putting r = 3
$\therefore {T_3} = {( - 1)^3}^{12}{C_3}{2^3}{x^{12 - 3}} \cdot {y^3}$$ = { - ^{12}}{C_3} \cdot 8{x^9}{y^3}$
$ = - 220 \times 8{x^9}{y^3} = - 1760{x^9}{y^3}$

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Question 133 Marks

Find a positive value of $m$ for which the coefficient of $x^2$ in the expansion $(1+x)^m$ is 6 .

Answer

Here coefficient of $x^2$ in $(1+x)^m={ }^m C_2=6$
$\therefore \frac{m(m-1)}{2}=6 \Rightarrow m^2-m=12 \Rightarrow m^2-m-12=0$
$\Rightarrow m^2-4 m+3 m-12=0 \Rightarrow m(m-4)+3(m-4)=0$
$\Rightarrow(m-4)(m+3)=0$
Either $m-4=0$ or $m+3=0$
$\Rightarrow m=4 \text { or } m=-3$
So positive value of $m$ is 4 .

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Question 143 Marks

Expand the given expression ${\left( {x + \frac{1}{x}} \right)^6}$

Answer

Using 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}}}$

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Question 153 Marks

Expand the given expression ${\left( {\frac{x}{3} + \frac{1}{x}} \right)^5}$

Answer

Using 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}}}$

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Question 163 Marks

Expand the given expression $(2x - 3)^6$

Answer

Using 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}$
$= 64x^6 + 6.32x^5 (-3) + 15.16x^4.9 + 20.8x^3 (-27) + 15.4x^2.81 + 6.2x (-243) + 729$
$= 64x^6 - 576x^5 + 2160x^4 - 4320x^3 + 4860x^2 - 2916x + 729$

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Question 173 Marks

Expand the given expression ${\left( {\frac{2}{x} - \frac{x}{2}} \right)^5}$

Answer

Using 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}}$

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Question 183 Marks

Show that $9^{n+1} - 8n - 9$ is divisible by $64$ whenever $n$ is a positive integer.

Answer

We 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 integer

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Question 193 Marks

Find $(x + 1)^6 + (x - 1)^6$. 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$

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Question 203 Marks

Find $(a + b)^4 - (a - b)^4.$ Hence, evaluate ${(\sqrt 3 + \sqrt 2 )^4} - {(\sqrt 3 - \sqrt 2 )^4}$

Answer

$(a+b) 4=\left[{ }^4 C_0 a^4+{ }^4 C_1 a^3 b+{ }^4 C_2 a^2 b^2+{ }^4 C_3 a b^3+{ }^4 C_4 b^4\right]$
$\text { and }(a-b)^4=\left[{ }^4 C_0 a^4-{ }^4 C_1 a^3 b+{ }^4 C_2 a^2 b^2-{ }^4 C_3 a b^3+{ }^4 C_4 b^4\right]$
$\therefore(a+b)^4-(a-b)^4=2\left[{ }^4 C_1 a^3 b+{ }^4 C_3 a b^3\right]$
$=2\left[4 a^3 b+4 a b^3\right]=8 a b\left[a^2+b^2\right]$
$\therefore(\sqrt{3}+\sqrt{2})^4-(\sqrt{3}-\sqrt{2})^4=8 \cdot \sqrt{3} \cdot \sqrt{2}\left[(\sqrt{3})^2+(\sqrt{2})^2\right]$
$=8 \cdot \sqrt{3} \cdot \sqrt{2}[3+2]=40 \cdot \sqrt{3} \cdot \sqrt{2}=40 \sqrt{6}$

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Question 213 Marks

Expand the given expression $(1 - 2x)^5$

Answer

Using 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}$
$= 1 + 5 (-2x) + 10(-2x)^2 + 10(-2x)^3 + 5(-2x)^4 + (-2x)^5$
$= 1 - 10x + 40x^2 - 80x^3 + 80x^4 - 32x^5$

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