[5 marks sum]

Question 15 Marks

If $a: b=c: d$, then prove that $\frac{a^2+c^2}{b^2+d^2}=\frac{a c}{b c}$

Answer

$\frac{ a }{ b }=\frac{ c }{ d } \Rightarrow a =\frac{ bc }{ d }$
To prove,
$\frac{ a ^2+ c ^2}{ b ^2+ d ^2}=\frac{ ac }{ bd }$
LHS
$\frac{a^2+c^2}{b^2+d^2} $
$=\frac{\left(\frac{b c}{d}\right)^2+c^2}{b^2+d^2} $
$=\frac{\frac{b^2 c^2}{d^2}+c^2}{b^2+d^2} $
$=\frac{c^2\left(b^2+d^2\right)}{d^2\left(b^2+d^2\right)} $
$=\frac{c^2}{d^2}$
$\text { RHS } $
$ \frac{ ac }{ bd }$
$=\frac{\frac{ bc }{ d c}}{ bd } $
$ =\frac{ bc ^2}{ bd ^2} $
$ =\frac{ c ^2}{ d ^2} $
$ \text { LHS }=\text { RHS }$

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Question 25 Marks

If $x =\frac{ pab }{ a + b }$, then prove that $\frac{ x + pa }{ x - pa }+\frac{ x + pb }{ x - pb }=\frac{2\left( a ^2- b ^2\right)}{ ab }$

Answer

$x=\frac{p a b}{a+b} \Rightarrow \frac{x}{p a}=\frac{b}{a+b}$
Applying componendo and dividendo
$\frac{ x + pa }{ x - pa }=\frac{ b + a + b }{ b - a - b }=\frac{2 b + a }{- a }\ldots(i)$
Again , $x=\frac{ pab }{ a + b } \Rightarrow \frac{ x }{ pb }=\frac{ a }{ a + b }$
Applying componendo and dividendo
$\frac{ x + pb }{ x - pb }=\frac{ a + a + b }{ a - a - b }=\frac{2 a + b }{- b }\ldots(ii)$
Adding $(i)$ and $(ii)$
$\frac{x+p a}{x-p a}+\frac{x+p b}{x-p b}=\frac{2 b+a}{-a}+\frac{2 a+b}{-b}=\frac{a-2 b}{a}+\frac{b-2 a}{b} $
$=\frac{-2 b^2+a b-2 a^2+a b}{a b} $
$=\frac{-2 b^2+2 a b-2 a^2}{a b} $
$=\frac{2\left(a^2-b^2\right)}{a b}=2=\text { RHS } $
$\text { LHS = RHS }$
Hence proved.

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Question 35 Marks

If $x=\frac{\sqrt[3]{m+1}+\sqrt[3]{m-1}}{\sqrt[3]{m+1}+\sqrt[3]{m-1}}$ then prove that $x^3-3 m x^2+3 x=m$

Answer

$\frac{x}{1}=\frac{\sqrt[3]{m+1}+\sqrt[3]{m-1}}{\sqrt[3]{m+1}+\sqrt[3]{m-1}}$
Applying componendo and dividendo
$\frac{x+1}{x-1}=\frac{\sqrt[3]{m+1}+\sqrt[3]{m-1}+\sqrt[3]{m+1}-\sqrt[3]{m-1}}{\sqrt[3]{m+1}+\sqrt[3]{m-1}-\sqrt[3]{m+1}+\sqrt[3]{m-1}} $
$\Rightarrow \frac{x+1}{x-1}=\frac{2 \sqrt[3]{m+1}}{2 \sqrt[3]{m-1}}$
Cubing both sides
$\Rightarrow \frac{( x +1)^3}{( x -1)^3}=\frac{8( m +1)}{8( m -1)} $
$\Rightarrow \frac{ x ^3+3 x ^2+3 x +1}{ x ^3-3 x ^2+3 x -1}=\frac{ m +1}{ m -1}$
$\Rightarrow(m-1)\left(x^3+3 x^2+3 x+1\right)=(m+1)\left(x^3-3 x^2+3 x-1\right) $
$\Rightarrow m x^3+3 m x^2+3 m x+m-x^3-3 x^2-3 x-1-m x^3-3 m x^2+3 m x-m+x^3-3 x^2+3 x-1 $
$\Rightarrow 6 m x^2+2 m-2 x^3-6 x=0 $
$\Rightarrow 3 m x^2+m-x^3-3 x=0 $
$\Rightarrow x^3-3 m x^2+3 x=m$
Hence Proved.

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Question 45 Marks

If $a, b, c$ and dare in continued proportion, then prove that
$(a+ d)(b+ c)-(a+ c)(b+ d)= (b-c)^2 $

Answer

$\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k $
$\Rightarrow c=k d $
$b=k c=k^2 d $
$a=k b=k^3 d $
$(a+d)(b+c)-(a+c)(b+d)=(b-c)^2$
$LHS$
$(a+d)(b+c)-(a+c)(b+d) $
$=a b+b d+a c+c d-a b-b c-a d-c d $
$=b d+c a-b c-a d $
$=k^2 d^2+k^4 d^4-k^3 d^2-k^3 d^2 $
$=k^2 d^2+k^4 d^4-2 k^3 d^2 $
$=k^2 d^2\left(1+k^2-2 k\right)$
RHS
$(b-c)^2=(b-c)(b-c)$
$=b^2-2 b c+c^2$
$=k^4 d^4-2 k^3 d^2+k^2 d^2$
$=k^2 d^2\left(k^2-2 k+1\right)$
LHS $=$ RHS. Hence, proved.

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Question 55 Marks

If $a, b, c$ and dare in continued proportion, then prove that
$\sqrt{(a+b+c)(b+c+d)}=\sqrt{a b}+\sqrt{b c}+\sqrt{c d}$

Answer

$\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k $
$ \Rightarrow c=k d $
$ b=k c=k^2 d$
$ a=k b=k^3 d $
$ \sqrt{(a+b+c)(b+c+d)}=\sqrt{a b}+\sqrt{b c}+\sqrt{c d} $
$\text { LHS }$
$\sqrt{(a+b+c)(b+c+d)} $
$ =\sqrt{\left(k^3 d+k^2 d+k d\right)\left(k^2 d+k d+d\right)} $
$=\sqrt{k d\left(k^2+k+1\right) \times d\left(k^2+k+1\right)} $
$ =\sqrt{k^2\left(k^2+k+1\right)^2} $
$ =d \sqrt{k}\left(k^2+k+1\right)$
$\text { RHS } $
$=\sqrt{a b}+\sqrt{b c}+\sqrt{c d} $
$ =\sqrt{k^3 d \times k^2 d}+\sqrt{k^2 d \times k d}+\sqrt{k d \times d} $
$=\sqrt{k^5 d^2}+\sqrt{k^3 d^2}+\sqrt{k} d^2$
$ =k^2 d \sqrt{k}+k d \sqrt{k}+d \sqrt{k} $
$=d \sqrt{k}\left(k^2+k+1\right)$

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Question 65 Marks

Find the fourth proportion to the following:
$(p^2q - qr^2 ), (pqr - pr^2 )$ and $(pq^2 - pr^2)$

Answer

$\text { Let } x \text { be the fourth proportion } $
$\left( p ^2 q - qr ^2\right):\left( pqr - pr ^2\right)::\left( pq ^2- pr ^2\right): x $
$\Rightarrow\left( p ^2 q - qr ^2\right) \times x =\left( pqr - pr ^2\right) \times\left( pq ^2- pr ^2\right) $
$\Rightarrow x =\frac{\left( pqr - pr ^2\right) \times\left( pq ^2- pr ^2\right)}{\left( p ^2 q - qr ^2\right)} $
$\Rightarrow x =\frac{\operatorname{pr}( q - r ) \times p \left( q ^2- r ^2\right)}{ q \left( p ^2- r ^2\right)} $
$\Rightarrow x =\frac{\operatorname{pr}( q - r ) \times p ( q - r )( q + r )}{ q \left( p ^2- r ^2\right)} $
$\Rightarrow x =\frac{ p ^2 r ( q - r )^2( q + r )}{ q \left( p ^2- r ^2\right)}$
The fourth proportion is $\frac{ p ^2 r ( q - r )^2( q + r )}{ q \left( p ^2- r ^2\right)}$

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Question 75 Marks

Given four quantities $p, q, r$ and s are in proportion, show that
$q^2(p - r) : rs (q - s) =(p^2- q^2- pq): ( r^2-s^2-rs)$.

Answer

$p, q, r$ and $s$ are $1 n$ proportion
then, $p: q:: r: s$
Let $\frac{ p }{ q }=\frac{ r }{ s }= k$
Then $p = kq$ and $r = ks$
Now, we have to prove that
$\frac{(p-r) q^2}{(q-s) r s}=\frac{p^2-q^2-p q}{r^2-s^2-r s}$
$\text { LHS }$
$ =\frac{(p-r) q^2}{(q-s) r s} $
$=\frac{(k q-k s) q^2}{(q-s) k s \times s} $
$ =\frac{k(q-s) q^2}{k s^2(q-s)} $
$ =\frac{q^2}{s^2}$
RHS
$=\frac{p^2-q^2-p q}{r^2-s^2-r s} $
$=\frac{k^2 q^2-q^2-k q \times q}{k^2 s^2-s^2-k s \times s} $
$=\frac{q^2\left(k^2-1-k\right)}{s^2\left(k^2-1-k\right)} $
$=\frac{q^2}{s^2} $
$\text { LHS }=\text { RHS }$

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Question 85 Marks

What quantity must be added to each term of the ratio $(p + q) : (p - q)$ to make it equal to $(p + q)^2 : (p - q)^2$ ?

Answer

Let $x$ be added from each term such that
$\frac{(p+q)+x}{(p-q)+x}=\frac{(p+q)^2}{(p-q)^2} $
$\Rightarrow(p+q+x)\left(p^2-2 p q+q^2\right)=\left(p^2+2 p q+q^2\right)(p-q+x) $
$\Rightarrow p^3+p^2 q+p^2 x+q^2 p+q^3+q^2 x-2 p^2 q-2 p q^2-2 p q x $
$\Rightarrow p^3-p^2 q+p^2 x+q^2 p-q^3+q^2 x+2 p^2 q-2 p q^2+2 p q x $
$\Rightarrow p^2 q+q^3-2 p^2 q-2 p q x=-p^2 q-q^3+2 p^2 q+2 p q x $
$\Rightarrow 2 q^3-p^2 q=4 p q x $
$\Rightarrow 2 q\left(q^2-p^2\right)=4 p q x $
$\Rightarrow x=\frac{q^2-p^2}{2 p} $
$\frac{q^2-p^2}{2 p} \text { should be added to each term. }$

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Mathematics [5 marks sum]

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