Question 14 Marks
If $x =\frac{8 a b}{ a + b }$ find the value of $\frac{x+4 a}{x-4 a}+\frac{x+4 b}{x-4 b}$
Answer
$
\begin{aligned}
& x =\frac{8 a b}{ a + b } \\
& \Rightarrow \frac{x}{4 a}=\frac{2 b}{ a + b }
\end{aligned}$
Applying componendo and dividendo,
$
\frac{x+4 a}{x-4 a}=\frac{2 b+a+b}{2 b-a-b}=\frac{3 b+a}{b-a}
$Again $\frac{x}{4 b}=\frac{2 a}{ a + b }$
Applying componendo and dividendo,
$
\frac{x+4 b}{x-4 b}=\frac{2 a+a+b}{2 a-a-b}=\frac{3 a+b}{a-b}$
Adding (i) and (ii)
$
\begin{aligned}
& \frac{x+4 a}{x-4 a}+\frac{x+4 b}{x-4 b} \\
& =\frac{3 b+a}{b-a}+\frac{3 a+b}{a-b}
\end{aligned}
$
$\begin{aligned} & =-\frac{a+3 b}{a-b}+\frac{3 a+b}{a-b} \\ & =\frac{-a-3 b+3 a+b}{a-b} \\ & =\frac{2 a-2 b}{a-b} \\ & =\frac{2(a-b)}{a-b} \\ & =2 .\end{aligned}$
View full question & answer→Question 24 Marks
If $x =\frac{2 a+b}{a+b}$ find the value of $\frac{x+a}{x-a}+\frac{x+b}{x-b}$
Answer
$
\begin{aligned}
& x=\frac{2 a+b}{ a + b } \\
& \Rightarrow \frac{x}{a}=\frac{2 b}{ a + b }
\end{aligned}$
Applying componendo and dividendo,
$
\frac{ x + a }{ x - a }=\frac{2 b+a+b}{2 b-a-b}=\frac{3 b+a}{ b - a }
$Again $\frac{x}{b}=\frac{2 a}{ a + b }$
Applying componendo and dividendo,
$
\frac{ x + b }{ x - b }=\frac{2 a+a+b}{2 a-a-b}=\frac{3 a+b}{ a - b }$
Adding (i) and (ii)
$
\begin{aligned}
& \frac{ x + a }{ x - a }+\frac{ x + b }{ x - b } \\
& =\frac{3 b+a}{ b - a }+\frac{3 a+b}{ a - b }
\end{aligned}
$
$\begin{aligned} & =-\frac{a+3 b}{ a - b }+\frac{3 a+b}{ a - b } \\ & =\frac{-a-3 b+3 a+b}{ a - b } \\ & =\frac{2 a-2 b}{ a - b } \\ & =\frac{2(a-b)}{ a - b } \\ & =2 .\end{aligned}$
View full question & answer→Question 34 Marks
If (a + 3b + 2c + 6d) (a – 3b – 2c + 6d) = (a + 3b – 2c – 6d) (a – 3b + 2c – 6d), prove that a : b :: c : d.
Answer
$
\begin{aligned}
& \frac{a+3 b+2 c+6 d}{a-3 b-2 c+6 d}=\frac{a+3 b-2 c-6 d}{a-3 b+2 c-6 d} \\
& \Rightarrow \frac{a+3 b+2 c+6 d}{a+3 b-2 c-6 d}=\frac{a-3 b+2 c-6 d}{a-3 b-2 c+6 d} \ldots \text { (by altenendo) }
\end{aligned}
$Applying componendo and dividendo
$
\begin{aligned}
& \frac{a+3 b+2 c+6 d+a+3 b-2 c-6 d}{a+3 b+2 c+6 d-a-3 b+2 c+6 d} \\
& =\frac{a-3 b+2 c-6 d+a-3 b-2 c+6 d}{a-3 b+2 c-6 d-a+3 b+2 c-6 d} \\
& \Rightarrow \frac{2(a+3 b)}{2(2 c+6 d)}=\frac{2(a-3 b)}{2(2 c-6 d)} \\
& \Rightarrow \frac{a+3 b}{2 c+6 d}=\frac{a-3 b}{2 c-6 d} \ldots \text { (Dividing by 2) } \\
& \Rightarrow \frac{a+3 b}{a-3 b}=\frac{2 c+6 d}{2 c-6 d} \ldots \text { (By alternendo) }
\end{aligned}
$Again applying componendo and dividendo
$
\frac{a+3 b+a-3 b}{a+3 b-a+3 b}=\frac{2 c+6 d+2 c 6 d}{2 c+6 d-2 c+6 d}
$
$\begin{aligned} & \Rightarrow \frac{2 a}{6 b}=\frac{4 c}{12 d}=\frac{2 c}{6 d} \\ & \Rightarrow \frac{a}{b}=\frac{c}{d} \cdot \quad \ldots\left[\text { Dividing by } \frac{2}{6}\right]\end{aligned}$
View full question & answer→Question 44 Marks
Give $\frac{x^3+12 x}{6 x^2+8}=\frac{y^3+27 y}{9 y^2+27}$ Using componendo and dividendo find $x: y$.
AnswerGive $\frac{x^3+12 x}{6 x^2+8}=\frac{y^3+27 y}{9 y^2+27}$
Using componendo-dividendo, we have
$
\begin{aligned}
& \frac{x^3+12 x+6 x^2+8}{x^3+12 x-6 x^2-8}=\frac{y^3+27 y+9 y^2+27}{y^3+27 y 9 y^2-27} \\
& \Rightarrow \frac{(x+2)^3}{(x 2)^3}=\frac{(y+3)^3}{(9 y-3)^3} \\
& \Rightarrow\left(\frac{x+2}{x-2}\right)^3=\left(\frac{y+3}{y-3}\right)^3 \\
& \Rightarrow \frac{x+2}{x-2}=\frac{y+3}{y-3}
\end{aligned}$
Again using componendo-dividendo, we get
$
\begin{aligned}
& \frac{x+2+x-2}{x+2-x+2}=\frac{y+3+y-3}{y+3-y+3} \\
& \Rightarrow \frac{2 x}{4}=\frac{2 y}{3} \\
& \Rightarrow \frac{x}{2}=\frac{y}{3} \\
& \Rightarrow \frac{x}{y}=\frac{2}{3}
\end{aligned}$
Thus the required ratio is $x: y=2: 3$.
View full question & answer→Question 54 Marks
Given that $\frac{a^3+3 a b^2}{b^3+3 a^2 b}=\frac{63}{62}$. Using componendo and dividendo find $a: b$.
AnswerGiven that $\frac{a^3+3 a b^2}{b^3+3 a^2 b}=\frac{63}{62}$
By componendo and dividendo
$
\begin{aligned}
& \frac{a^3+3 a b^2+b^3+3 a^2 b}{a^3+3 a b^2-b^3-3 a^2 b}=\frac{63+62}{63-62}=\frac{125}{1} \\
& \Rightarrow \frac{(a+b)^3}{(a b)^3}-\left(\frac{5}{1}\right)^3 \\
& \Rightarrow \frac{a+b}{a-b}=5 \\
& \Rightarrow a + b =5 a -5 b \\
& \Rightarrow 5 a - a -5 b - b =0 \\
& \Rightarrow 4 a -6 b =0 \\
& \Rightarrow 4 a =6 b \\
& \Rightarrow \frac{a}{b}=\frac{6}{4} \\
& \Rightarrow \frac{a}{b}=\frac{3}{2} \\
& a : b =3: 2 .
\end{aligned}
$
View full question & answer→Question 64 Marks
Given $x =\frac{\sqrt{a^2+b^2}+\sqrt{a^2-b^2}}{\sqrt{\wedge 2+b^2}-\sqrt{a^2-b^2}}$ Use componendo and dividendo to prove that $b ^2=\frac{2 a^2 x}{x^2+1}$
AnswerIf $\frac{x}{1}=\frac{\sqrt{a^2+b^2}+\sqrt{a^2-b^2}}{\sqrt{\wedge 2+b^2}-\sqrt{a^2-b^2}}$
Applying componendo and dividendo both sides
$
\begin{aligned}
& \frac{x+1}{x-1}=\frac{\sqrt{a^2+b^2}+\sqrt{a^2-b^2}+\sqrt{a^2+b^2}-\sqrt{a^2-b^2}}{\sqrt{a^2+b^2}+\sqrt{a^2-b^2}-\sqrt{a^2+b^2}+\sqrt{a^2-b^2}} \\
& \Rightarrow \frac{x+1}{x-1}=\frac{2 \sqrt{a^2+b^2}}{2 \sqrt{a^2+b^2}} \\
& \Rightarrow \frac{x+1}{x-1}=\frac{\sqrt{a^2+b^2}}{\sqrt{a^2+b^2}}
\end{aligned}
$Squaring both sides we have
$
\begin{aligned}
& \Rightarrow \frac{(x+1)^2}{(x-1)^2}=\frac{a^2+b^2}{a^2-b^2} \\
& \Rightarrow \frac{x^2+1+2 x}{x^2+1-2 x}=\frac{a^2+b^2}{a^2-b^2}
\end{aligned}$
Applying componendo and dividendo both sides
$\begin{aligned} & \Rightarrow \frac{x^2+1+2 x+x^2+1-2 x}{x^2+1+2 x-x^2-1+2 x}=\frac{a^2+b^2+a^2-b^2}{a^2+b^2-a^2+b^2} \\ & \frac{2 x^2+2}{4 x}=\frac{2 a^2}{2 b^2} \\ & \Rightarrow \frac{x^2+1}{2 x}=\frac{a^2}{b^2} \\ & \Rightarrow b ^2=\frac{2 a^2 x}{x^2+1} .\end{aligned}$
View full question & answer→Question 74 Marks
If $x =\frac{\sqrt{a+x}+\sqrt{a-1}}{\sqrt{a+1-\sqrt{a-1}}}$, using properties of proportion, show that $x^2-2 a x+1=0$
AnswerWe have $x =\frac{\sqrt{a+x}+\sqrt{a-1}}{\sqrt{a+1-\sqrt{a-1}}}$
$
\Rightarrow \frac{x+1}{x-1}=\frac{2 \sqrt{a+1}}{2 \sqrt{a-1}}
$
(Applying componendo and dividendo)
$
\begin{aligned}
& \Rightarrow \frac{(x+1)^2}{(x-1)^2}=\frac{a+1}{a-1} \\
& \Rightarrow \frac{(x+1)^2+(x-1)^2}{(x+1)^2-(x-1)^2}=\frac{2 a}{2}
\end{aligned}
$
(Again applying componendo and dividendo)
$
\begin{aligned}
& \Rightarrow \frac{x^2+1+2 x+x^2+1-2 x}{x^2+1+2 x-x^2-1+2 x}= a \\
& \Rightarrow \frac{2 x^2+2}{4 x}= a \\
& \Rightarrow \frac{2\left(x^2+1\right)}{4 x}= a \\
& \Rightarrow \frac{\left(x^2+1\right)}{2 x}= a \\
& \Rightarrow 2 ax = x ^2+1 \\
& \Rightarrow x ^2-2 ax +1=0
\end{aligned}$
Proved.
View full question & answer→Question 84 Marks
Solve for $x: 16\left(\frac{a-x}{a+x}\right)^3=\frac{a+x}{a-x}$
Answer
$
\begin{aligned}
& x: 16\left(\frac{a-x}{a+x}\right)^3=\frac{a+x}{a-x} \\
& \Rightarrow\left(\frac{a+x}{a-x}\right) \times\left(\frac{a+x}{a-x}\right)^3=16 \\
& \Rightarrow\left(\frac{a+x}{a-x}\right)^4=16=( \pm 2)^4 \\
& \Rightarrow \frac{a+x}{a-x}= \pm 2
\end{aligned}$
When $\frac{a+x}{a-x}=\frac{2}{1}$
Applying componendo and dividendo,
$
\begin{aligned}
& \frac{a+x+a-x}{a+x-a+x}=\frac{2+1}{2-1} \\
& \Rightarrow \frac{2 a}{2 x}=\frac{3}{1} \\
& \Rightarrow \frac{a}{x}=\frac{3}{1} \\
& \Rightarrow 3 x = a \\
& \therefore x =\frac{a}{3}
\end{aligned}$
When $\frac{a+x}{a-x}=\frac{-2}{1}$
Applying componendo and dividendo
$
\frac{a+x+a-x}{a+x-a+x}=\frac{-2+1}{-2-1}
$
$\Rightarrow \frac{2 a}{2 x}=\frac{-1}{3}$
$
\begin{aligned}
& \Rightarrow \frac{a}{x}=\frac{1}{3} \\
& \Rightarrow x =3 a
\end{aligned}$
Hence $x =\frac{a}{3}, 3 a$.
View full question & answer→Question 94 Marks
Solve $\frac{1+x+x^2}{1-x+x^2}=\frac{62(1+x)}{63(1+x)}$
Answer
$
\begin{aligned}
& \frac{1+x+x^2}{1-x+x^2}=\frac{62(1+x)}{63(1+x)} \\
& \Rightarrow \frac{(1-x)\left(1+x+x^2\right)}{(1+x)\left(1-x+x^2\right)}=\frac{62}{63} \\
& \Rightarrow \frac{(1+x)\left(1-x+x^2\right)}{(1-x)\left(1+x+x^2\right)}=\frac{62}{63} \\
& \Rightarrow \frac{1+x^3}{1-x^3}=\frac{63}{62}
\end{aligned}$
Applying componendo and dividendo,
$
\begin{aligned}
& \frac{1+x^3+1-x^3}{1+x^3-1+x^3}=\frac{63+62}{63-62} \\
& \Rightarrow \frac{2}{2 x^3}=\frac{125}{1} \\
& \Rightarrow \frac{1}{x^3}=\frac{125}{1} \\
& \Rightarrow x^3=\frac{1}{125} \\
& =\left(\frac{1}{5}\right)^3 \\
& \therefore x=\frac{1}{5} .
\end{aligned}
$
View full question & answer→Question 104 Marks
Find $x$ from the following equations : $\frac{\sqrt{a+x}+\sqrt{a-x}}{\sqrt{a+x}-\sqrt{a-x}}=\frac{c}{d}$
Answer$
\frac{\sqrt{a+x}+\sqrt{a-x}}{\sqrt{a+x}-\sqrt{a-x}}=\frac{c}{d}$
Applying componendo and dividendo,
$
\begin{aligned}
& \frac{\sqrt{a+x}+\sqrt{a-x}+\sqrt{a+x}-\sqrt{a-x}}{\sqrt{a+x}+\sqrt{a-x}-\sqrt{a+x}+\sqrt{a-x}}=\frac{c+d}{c-d} \\
\Rightarrow & \frac{2 \sqrt{a+x}}{2 \sqrt{a-x}}=\frac{c+d}{c-d} \\
\Rightarrow & \frac{\sqrt{a+x}}{\sqrt{a-x}}=\frac{c+d}{c-d}
\end{aligned}
$Squaring both sides
$
\frac{a+x}{a-x}-\frac{(c+d)^2}{(c-d)^2}$
Again applying componendo and dividendo
$
\begin{aligned}
& \frac{a+x+a-x}{a+x-a+x}=\frac{(c+d)^2+(c-d)^2}{(c+d)^2-(c-d)^2} \\
& \Rightarrow \frac{2 a}{2 x}=\frac{2\left(c^2+d^2\right)}{4 c d} \\
& \Rightarrow \frac{a}{x}=\frac{c^2+d^2}{2 c d} \\
& \Rightarrow x\left(c^2+d^2\right)=2 \text { acd } \\
& \Rightarrow x=\frac{2 a c d}{c^2+d^2} .
\end{aligned}
$
View full question & answer→Question 114 Marks
Find $x$ from the following equations $: \frac{3 x+\sqrt{9 x^2-5}}{3 x-\sqrt{9 x^2-5}}=\frac{5}{1}$
Answer$
\frac{3 x+\sqrt{9 x^2-5}}{3 x-\sqrt{9 x^2-5}}=\frac{5}{1}$
Applying componendo and dividendo
$
\frac{3 x+\sqrt{9 x^2-5}+3 x-\sqrt{9 x^2-5}}{3 x+\sqrt{9 x^2-5}-3 x+\sqrt{9 x^2-5}}=\frac{5+1}{5-1}
$
$
\begin{aligned}
& \frac{6 x}{2 \sqrt{9 x^2-5}}=\frac{6}{4} \\
& \Rightarrow \frac{3 x}{\sqrt{9 x^2-5}}=\frac{3}{2}
\end{aligned}$
Squaring both sides
$
\begin{aligned}
& \frac{9 x^2}{9 x^2-5}=\frac{9}{4} \\
& \Rightarrow 81 x ^2-45=36 x ^2 \\
& \Rightarrow 81 x ^2-36 x ^2=45 \\
& \Rightarrow 45 x ^2=45 \\
& \Rightarrow x ^2=1 \\
& \Rightarrow x = \pm 1 \\
& \therefore x =1,-1 \\
&
\end{aligned}
$Check:
(i) When $x=1$, then in the given equation
$
\frac{3 \times 1+\sqrt{9 \times 1-5}}{3 \times 1-\sqrt{9 \times 1-5}}
$
$
\begin{aligned}
& =\frac{3+\sqrt{4}}{3-\sqrt{4}} \\
& =\frac{3+2}{3-2} \\
& =\frac{5}{1}
\end{aligned}
$
which is given
$
\therefore x =1
$
(ii) When $x =-1$, then
$
\begin{aligned}
& \frac{3(-1)+\sqrt{9(-1)^2-5}}{3(-1)-\sqrt{9(-1)^2-5}} \\
& =\frac{-3+\sqrt{9-5}}{-3-\sqrt{9-5}} \\
& =\frac{-3+\sqrt{4}}{-3-\sqrt{4}} \\
& =\frac{-3+2}{-3-2} \\
& =\frac{-1}{-5}
\end{aligned}
$
$
=\frac{1}{5} \neq \frac{5}{1}
$
$\therefore x =-1$ is not its solution.
Hence $x=1$.
View full question & answer→Question 124 Marks
Find $x$ from the following equations:
$
\frac{\sqrt{12 x+1}+\sqrt{2 x-3}}{\sqrt{12 x+1}-\sqrt{2 x-3}}=\frac{3}{2}
$
Answer$
\frac{\sqrt{12 x+1}+\sqrt{2 x-3}}{\sqrt{12 x+1}-\sqrt{2 x-3}}=\frac{3}{2}$
Applying componendo and dividendo,
$
\begin{aligned}
& \frac{\sqrt{12 x+1}+\sqrt{2 x-3}+\sqrt{12 x+1}-\sqrt{2 x-3}}{\sqrt{2 x+1}+\sqrt{2 x-3}-\sqrt{12 x+1}+\sqrt{2 x-3}}=\frac{3+2}{3-2} \\
& \Rightarrow \frac{2 \sqrt{12 x+1}}{2 \sqrt{2 x-3}}=\frac{5}{1} \\
& \Rightarrow \frac{\sqrt{12 x+1}}{\sqrt{2 x-3}}=\frac{5}{1}
\end{aligned}
$Squaring both sides,
$
\begin{aligned}
& \frac{12 x+1}{2 x-3}=\frac{25}{1} \\
& \Rightarrow 50 x-75=12 x+1 \\
& \Rightarrow 50 x-12 x=1+75 \\
& \Rightarrow 38 x=76 \\
& \Rightarrow x=\frac{76}{38}=2 \\
& \therefore x=2
\end{aligned}
$
View full question & answer→Question 134 Marks
Find $x$ from the following equations : $\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}=\frac{a}{b}$
Answer$
\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}=\frac{a}{b}$
Applying componendo and dividendo,
$
\begin{aligned}
& \frac{\sqrt{1+x}+\sqrt{1-x}+\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}-\sqrt{1+x}+\sqrt{1-x}}=\frac{a+b}{a-b} \\
& \Rightarrow \frac{2 \sqrt{1+x}}{2 \sqrt{1-x}}=\frac{a+b}{a-b} \\
& \Rightarrow \frac{\sqrt{1+x}}{\sqrt{1-x}}=\frac{a+b}{a-b}
\end{aligned}$
Squaring both sides,
$
\frac{1+x}{1-x}=\frac{(a+b)^2}{(a-b)^2}$
Again applying componendo and dividendo,
$
\begin{aligned}
& \frac{1+x+1-x}{1+x-1+x} \\
& =\frac{(a+b)^2+(a-b)^2}{(a+b)^2-(a-b)^2} \\
& \Rightarrow \frac{2}{2 x}=\frac{2\left(a^2+b^2\right)}{4 a b}
\end{aligned}
$
$\begin{aligned} & \Rightarrow \frac{1}{x}=\frac{a^2+b^2}{2 a b} \\ & \therefore x=\frac{2 a b}{a^2+b^2} .\end{aligned}$
View full question & answer→Question 144 Marks
Find $x$ from the following equations $: \frac{\sqrt{x+4}+\sqrt{x-10}}{\sqrt{x+4}-\sqrt{x-10}}=\frac{5}{2}$
Answer$
\frac{\sqrt{x+4}+\sqrt{x-10}}{\sqrt{x+4}-\sqrt{x-10}}=\frac{5}{2}$
Applying componendo and dividendo,
$
\begin{aligned}
& \frac{\sqrt{x+4}+\sqrt{x-10}+\sqrt{x+4}-\sqrt{x-10}}{\sqrt{x+4}+\sqrt{x-10}-\sqrt{x+4}+\sqrt{x-10}}=\frac{5+2}{5-2} \\
& \Rightarrow \frac{2 \sqrt{x+4}}{2 \sqrt{x-10}}=\frac{7}{3} \\
& \Rightarrow \frac{\sqrt{x+4}}{\sqrt{x-10}}=\frac{7}{3}
\end{aligned}
$Squaring both sides,
$
\begin{aligned}
& \frac{x+4}{x-10}=\frac{49}{9} \\
& \Rightarrow 49 x-490=9 x+36 \\
& \Rightarrow 49 x-9 x=36+490 x \\
& \Rightarrow 40 x=526 \\
& \therefore x=\frac{526}{40}
\end{aligned}
$
$
=\frac{263}{20} .
$
View full question & answer→Question 154 Marks
Find $x$ from the following equations : $\frac{\sqrt{2-x}+\sqrt{2+x}}{\sqrt{2-x}-\sqrt{2+x}}=3$
Answer$
\frac{\sqrt{2-x}+\sqrt{2+x}}{\sqrt{2-x}-\sqrt{2+x}}=3
$Applying componendo and dividendo,
$
\begin{aligned}
& \frac{\sqrt{2-x}+\sqrt{2+x}+\sqrt{2-x}-\sqrt{2+x}}{\sqrt{2-x}+\sqrt{2+x}-\sqrt{2-x}+\sqrt{2+x}}=\frac{3+1}{3-1} \\
& \Rightarrow \frac{2 \sqrt{2-x}}{2 \sqrt{2+x}}=\frac{4}{2} \\
& \Rightarrow \frac{\sqrt{2-x}}{\sqrt{2+x}}=\frac{2}{1}
\end{aligned}$
Squaring both sides
$
\begin{aligned}
& \frac{2-x}{2+x}=\frac{4}{1} \\
& \Rightarrow 8+4 x=2-x \\
& 4 x+x=2 \\
& \Rightarrow 5 x=-6 \\
& \therefore x=\frac{-6}{5}
\end{aligned}
$
View full question & answer→Question 164 Marks
Solve $x : \frac{\sqrt{36 x+1}+6 \sqrt{x}}{\sqrt{36 x+1}-6 \sqrt{x}}=9$
Answer$
\frac{\sqrt{36 x+1}+6 \sqrt{x}}{\sqrt{36 x+1}-6 \sqrt{x}}=\frac{9}{1}$
Applying componendo and dividendo,
$
\begin{aligned}
& \frac{\sqrt{36 x+1}+6 \sqrt{x}+\sqrt{36 x+1}-6 \sqrt{x}}{\sqrt{36 x+1}+6 \sqrt{x}-\sqrt{36 x-1}+6 \sqrt{x}} \\
& =\frac{9+1}{9-1} \\
& \Rightarrow \frac{2 \sqrt{36 x+1}}{12 \sqrt{x}}=\frac{10}{8} \\
& \Rightarrow \frac{\sqrt{36 x+1}}{6 \sqrt{x}}=\frac{5}{4} \quad \ldots \text { (Squaring both sides) } \\
& \frac{36 x+1}{36 x}=\frac{25}{16} \\
& \Rightarrow 36 x \times 25=16(36 x+1) \\
& \Rightarrow 900 x=576 x+16 \\
& \Rightarrow 900 x-576 x=16 \\
& \Rightarrow 324=16
\end{aligned}
$
$
\therefore x=\frac{16}{324}
$
View full question & answer→Question 174 Marks
If $x =\frac{4 \sqrt{6}}{\sqrt{2}+\sqrt{3}}$ find the value of $\frac{x+2 \sqrt{2}}{x-2 \sqrt{2}}+\frac{x+2 \sqrt{3}}{x-2 \sqrt{3}}$
Answer
$
\begin{aligned}
& x=\frac{4 \sqrt{6}}{\sqrt{2}+\sqrt{3}} \\
& \Rightarrow \frac{4 \sqrt{2} \times \sqrt{3}}{\sqrt{2}+\sqrt{3}} \\
& \frac{x}{2 \sqrt{2}}=\frac{2 \sqrt{3}}{\sqrt{2}+\sqrt{3}}
\end{aligned}
$Applying componendo and dividendo,
$
\begin{aligned}
& \frac{x+2 \sqrt{2}}{x-2 \sqrt{2}} \\
& =\frac{2 \sqrt{3}+\sqrt{2}+\sqrt{3}}{2 \sqrt{3}-\sqrt{2}-\sqrt{3}} \\
& =\frac{3 \sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}
\end{aligned}
$Again $\frac{x}{2 \sqrt{3}}=\frac{2 \sqrt{2}}{\sqrt{2}+\sqrt{3}}$
Applying componendo and dividendo,$
\begin{aligned}
& \frac{x+2 \sqrt{3}}{x-2 \sqrt{3}} \\
& =\frac{2 \sqrt{2}+\sqrt{2}+\sqrt{3}}{2 \sqrt{2}-\sqrt{2}-\sqrt{3}} \\
& =\frac{3 \sqrt{2}+\sqrt{3}}{\sqrt{2}-\sqrt{3}}
\end{aligned}
$Adding (i) and (ii)
$
\begin{aligned}
& \frac{x+2 \sqrt{2}}{x-2 \sqrt{2}}+\frac{x+2 \sqrt{3}}{x-2 \sqrt{3}} \\
& =\frac{3 \sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\frac{3 \sqrt{2}+\sqrt{3}}{\sqrt{2}-\sqrt{3}} \\
& =\frac{3 \sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}-\frac{3 \sqrt{2}+\sqrt{3}}{\sqrt{3}-\sqrt{2}} \\
& =\frac{3 \sqrt{3}+\sqrt{2}-3 \sqrt{2}-\sqrt{3}}{\sqrt{3}-\sqrt{2}}
\end{aligned}
$
$\begin{aligned} & =\frac{2 \sqrt{3}-2 \sqrt{2}}{\sqrt{3}-\sqrt{2}} \\ & =\frac{2(\sqrt{3}-\sqrt{2})}{\sqrt{3}-\sqrt{2}} \\ & =2 .\end{aligned}$
View full question & answer→Question 184 Marks
If 2x – 1, 5x – 6, 6x + 2 and 15x – 9 are in proportion, find the value of x.
Answer
$
\begin{aligned}
& \because 2 x-1,5 x-6,6 x+2 \text { and } 15 x-9 \text { are in proportion. } \\
& \text { then }(2 x-1)(15 x-9)=(5 x-6)(6 x+2) \\
& \Rightarrow 30 x^2-18 x-15 x+9 \\
& \Rightarrow 30 x^2+10 x-36-12 \\
& \Rightarrow 30 x^2-33 x+9=30 x^2-26 x-12 \\
& \Rightarrow 30 x^2-33 x-30 x^2+26 x=-12-9 \\
& \Rightarrow-7=-21 \\
& \therefore x=\frac{-21}{-7}=3
\end{aligned}
$Hence $x=3$.
View full question & answer→Question 194 Marks
What number should be subtracted from each of the numbers 23, 30, 57 and 78 so that the remainders are in proportion ?
AnswerLet $x$ be subtracted from each term, then
$
\begin{aligned}
& 23-x_t 30-x_t 57-x \text { and } 78-x \text { are proportional } \\
& 23-x: 30-x:: 57-x: 78-x \\
& \Rightarrow \frac{23-x}{30-x}=\frac{57-x}{78-x} \\
& \Rightarrow(23-x)(78-x)=(30-x)(57-x) \\
& \Rightarrow 1794-23 x-78 x+x^2=1710-30 x-57 x+x^2 \\
& \Rightarrow x^2-101 x+1794=x^2-87 x+1710 \\
& \Rightarrow x^2-101 x+1794-x^2+87 x-1710=0 \\
& \Rightarrow-14 x+84=0 \\
& \Rightarrow 14 x=84 \\
& \therefore x=\frac{84}{14}=6
\end{aligned}
$Hence 6 is to be subtracted.
View full question & answer→Question 204 Marks
What number must be added to each of the numbers $5, 11, 19$ and $37$ so that they are in proportion?
AnswerLet x be added to $5, 11, 19$ and $37$ to make them in proportion.
$5 + x : 11 + x : : 19 + x : 37 + x$
$\Rightarrow (5 + x) (37 + x) = (11 + x) (19 + x)$
$\Rightarrow 185 + 5x + 37x + x^2 = 209 + 11x + 19x + x^2$
$\Rightarrow 185 + 42x + x^2 = 209 + 30x + x^2$
$\Rightarrow 42x - 30x + x^2 - x^2 = 209 - 185$
$\Rightarrow 12x = 24$
$\Rightarrow x = 2$
$\therefore$ Least number to be added $= 2.$
View full question & answer→Question 214 Marks
If a, 12, 16 and b are in continued proportion find a and b.
Answer$\because a , 12,16, b$ are in continued proportion, then
$
\begin{aligned}
& \frac{a}{12}=\frac{12}{16}=\frac{16}{b} \\
& \Rightarrow \frac{a}{12}=\frac{12}{16} \\
& \Rightarrow 16 a=144 \\
& \Rightarrow a=\frac{144}{16}=9 \\
& \text { and } \frac{12}{16}=\frac{16}{b} \\
& \Rightarrow 12 b=16 \times 16=256 \\
& b=\frac{256}{12}=\frac{64}{3}=21 \frac{1}{3}
\end{aligned}$
Hence $a=9, b=\frac{64}{3}$ or $21 \frac{1}{3}$.
View full question & answer→Question 224 Marks
If $a, b, c, d$ are in continued proportion, prove that:
$
\left(\frac{a-b}{c}+\frac{a-c}{b}\right)^2-\left(\frac{d-b}{c}+\frac{d-c}{b}\right)^2=(a-d)^2\left(\frac{1}{c^2}-\frac{1}{b^2}\right)
$
Answer$a, b, c, d$ are in continued proportion
$
\begin{aligned}
& \therefore \frac{a}{b}=\frac{b}{c}=\frac{c}{d}= k \text { (say) } \\
& \therefore c = dk , b ^{ ck }= dk \cdot k = dk ^2, \\
& a = bk = dk ^2 \cdot k = dk ^3 \\
& \text { L.H.S. }=\left(\frac{a-b}{c}+\frac{a-c}{b}\right)^2-\left(\frac{d-b}{c}+\frac{d-c}{b}\right)^2 \\
& =\left(\frac{d k^3-d k^2}{ dk }+\frac{d k^3-d k}{d k^2}\right)^2-\left(\frac{d-d k^2}{ dk }+\frac{d-d k}{d k^2}\right)^2 \\
& =\left(\frac{d k^2(k-1)}{ dk }+\frac{d k\left(k^2-1\right)}{d k^2}\right)^2-\left(\frac{d\left(1-k^2\right)}{ dk }+\frac{d\left(1-k^2\right)}{d k^2}\right. \\
& =\left(k(k-1)+\frac{k^2-1}{k}\right)^2-\left(\frac{1-k^2}{k}+\frac{1-k}{k^2}\right)^2 \\
& =\left(\frac{k^2(k-1)+\left(k^2-1\right)}{k}\right)^2-\left(\frac{k\left(1-k^2\right)+1-k}{k^2}\right)^2 \\
& =\frac{\left(k^3-1\right)^2}{k^2}-\frac{\left(-k^3+1\right)^2}{k^4}
\end{aligned}
$
$\begin{aligned} & =\frac{\left(k^3-1\right)^2}{k^2}-\frac{\left(1-k^3\right)^2}{k^4} \\ & =\left(\frac{k^3-1}{k^2}\right)^2\left(1-\frac{1}{k^2}\right) \\ & =\frac{\left(k^3-1\right)^2\left(k^2-1\right)}{k^4} \\ & =\frac{\left(k^3-1\right)^2\left(k^2-1\right)}{k^4} \\ & \text { R.H.S. }=(a-d)^2\left(\frac{1}{c^2}-\frac{1}{b^2}\right) \\ & =\left(d k^3-d\right)^2\left(\frac{1}{d^2 k^2}-\frac{1}{d^2 k^4}\right) \\ & =d^2\left(k^3-1\right)^2\left(\frac{k^2-1}{d^2 k^4}\right) \\ & =\frac{\left(k^3-1\right)^2\left(k^2-1\right)}{k^4} \\ & \therefore \text { L.H.S. = R.H.S. }\end{aligned}$
View full question & answer→Question 234 Marks
If a, b, c, d are in continued proportion, prove that: a : d = triplicate ratio of (a – b) : (b – c)
Answer$a, b, c, d$ are in continued proportion
$
\begin{aligned}
& \therefore \frac{a}{b}=\frac{b}{c}=\frac{c}{d}= k \text { (say) } \\
& \therefore c = dk _{ k } b = ck = dk \cdot k = dk ^2, \\
& a = bk = dk ^2 \cdot k = dk ^3 \\
& a : d =\text { triplicate ratio of }( a - b ):( b - C ) \\
& =( a - b )^3:( b - c )^3 \\
& \text { L.H.S. }= a : d \\
& =\frac{a}{d} \\
& =\frac{d k^3}{d} \\
& = k ^3 \\
& \text { R.H.S. }=\frac{(a-b)^3}{(b-c)^3} \\
& =\frac{\left(d k^3-d k^2\right)^3}{\left(d k^2-d k\right)^3} \\
& =\frac{d^3 k^6(k-1)^3}{d^3 k^3(k-1)^3} \\
& = k ^3 \\
& \therefore \text { L.H.S. = R.H.S. }
\end{aligned}
$
View full question & answer→Question 244 Marks
If $a, b, c, d$ are in continued proportion, prove that $: (a + d)(b + c) – (a + c)(b + d) = (b – c)^2$
Answer$a, b, c, d$ are in continued proportion
$\therefore \frac{a}{b}=\frac{b}{c}=\frac{c}{d}= k\ ($say$)$
$ \therefore c=d k^2 b=c k=d k \cdot k=dk^2$
$ a=b k=d k^2 \cdot k=d k^3$
$ \text { L.H.S. }=(a+d)(b+c)-(a+c)(b+d)$
$ =\left(d k^3+d\right)\left(d k^2+d k\right)-\left(d k^3+d k\right)\left(d k^2+d\right)$
$ =d\left(k^3+1\right) d k(k+1)-d k\left(k^2+1\right) d\left(k^2+1\right)$
$ =d^2 k(k+1)\left(k^3+1\right)-d^2 k\left(k^2+1\right)\left(k^2+1\right)$
$ =d^2 k\left[k^4+k^3+k+1-k^4-2 k^2-1\right]$
$ =d^2 k\left[k^3-2 k^2+k\right]$
$ =d^2 k^2\left[k^2-2 k+1\right]$
$ =d^2 k^2(k-1)^2$
$ \text { R.H.S. }=(b-c)^2$
$ =\left(d k^2-d k\right)^2$
$ =d^2 k^2(k-1)^2$
$ \therefore \text { L.H.S. - R.H.S. }$
View full question & answer→Question 254 Marks
If $a, b, c, d$ are in continued proportion$,$ prove that$: (a^2 – b^2) (c^2 – d^2) = (b^2 – c^2)^2$
Answer$a, b, c, d$ are in continued proportion
$\therefore \frac{a}{b}=\frac{b}{c}=\frac{c}{d}= k\ ($say$)$
$ \therefore c=d k^{\prime} b = ck = dk \cdot k = dk ^2,$
$ a = bk = dk ^2 \cdot k = dk ^3$
$ \text { L.H.S. }=\left( a ^2- b ^2\right)\left( c ^2- d ^2\right)$
$ =\left[\left( dk ^3\right)^2-\left( dk ^2\right)^2\right]\left[( dk )^2- d ^2\right]$
$ =\left( d ^2 k ^6- d ^2 k ^4\right)\left( d ^2 k ^2- d ^2\right)$
$ = d ^2 k ^4\left( k ^2-1\right) d ^2\left( k ^2-1\right)$
$ = d ^4 k ^4\left( k ^2-1\right)^2$
$ \text { R.H.S. }=\left(b^2- c ^2\right)^2$
$ =\left[\left( dk ^2\right)^2-( dk )^2\right]^2$
$ =\left[d^2 k ^2- d ^2 k ^2\right]^2$
$ =\left[ d ^2 k ^2\left( k ^2-1\right)\right]^2$
$ = d ^4 k ^4\left( k ^2-1\right)^2$
$ \therefore \text { L.H.S. }=\text { R.H.S. }$
View full question & answer→Question 264 Marks
If $a, b, c, d$ are in continued proportion, prove that:
$
\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\frac{a}{d}
$
Answer$a, b, c, d$ are in continued proportion
$
\begin{aligned}
& \therefore \frac{a}{b}=\frac{b}{c}=\frac{c}{d}= k \text { (say) } \\
& \therefore c = dk _{ t } b = ck = dk . k = dk ^2, \\
& a = bk = dk ^2 \cdot k = dk ^3 \\
& \text { L.H.S. }=\frac{a^3+b^3+c^3}{b^3+c^3+d^3} \\
& =\frac{(d k)^3+\left(d k^2\right)^3+(d k)^3}{\left(d k^2\right)^3+(d k)^3+d^3} \\
& =\frac{d^3 k^9+d^3 k^6+d^3 k^3}{d^3 k^6+d^3 k^3+d^3} \\
& =\frac{d^3 k^3\left(k^6+k^3+1\right)}{d^3\left(k^6+k^3+1\right)} \\
& = k ^3 \\
& \text { R.H.S. }=\frac{a}{d} \\
& =\frac{d k^3}{d} \\
& = k ^3 \\
& \therefore \text { L.H.S. }=\text { R.H.S. }
\end{aligned}
$
View full question & answer→Question 274 Marks
If $a, b, c$ are in continued proportion, prove that $: (a + b + c) (a – b + c) = a^2 + b^2 + c^2 $
AnswerAs $a, b, c$, are in continued proportion
Let $\frac{a}{b}=\frac{b}{c}= k$
$ \text { L.H.S. }=( a + b + c )( a - b + c )$
$ =\left(c k ^2+ ck + c \right)\left( ck ^2- ck + c \right)$
$ = c \left( k ^2+ k +1\right) c \left( k ^2- k +1\right)$
$ = c ^2\left( k ^2+ k +1\right)\left( k ^2- k +1\right)$
$ = c ^2\left( k ^4+ k ^2+1\right)$
$ \text { R.H.S. }= a ^2+ b ^2+ c ^2$
$ =\left(c k ^2\right)^2+(c k)^2+( c )^2$
$ = c ^2 k ^4+ c ^2 k ^2+ c ^2$
$ = c ^2\left( k ^4+ k ^2+1\right)$
$ \therefore \text { L.H.S. }=\text { R.H.S. }$
View full question & answer→Question 284 Marks
If $a, b, c$ are in continued proportion, prove that $:abc(a + b + c)^3 = (ab + bc + ca)^3$
AnswerAs $a, b, c,$ are in continued proportion
Let $\frac{a}{b}=\frac{b}{c}= k$
$ \text { L.H.S. }=a b c(a+b+c)^3$
$ =c k^2 \cdot c k \cdot c\left[c k^2+c k+c\right]^3$
$ =c^3 k^3\left[c\left(k^2+k+1\right)\right]^3$
$ =c^3 k^3 \cdot c^3 \cdot\left(k^2+k+1\right)^3$
$ =c 6 k^3\left(k^2+k+1\right)^3$
$ \text { R.H.S. }=(a b+b c+c a)^3$
$ =\left(c k^2 \cdot c k+c k \cdot c+c \cdot c k 2\right)^3$
$ =\left(c^2 k^3+c^2 k+c^2 k^2\right)^3$
$ =\left(c^2 k^3+c^2 k^2+c^2 k\right)^3$
$ =\left[c^2 k\left(k^2+k+1\right)\right]^3$
$ =c^6 k^3(k+k+1)^3$
$ \therefore \text { L.H.S. }=\text { R.H.S. }$
View full question & answer→Question 294 Marks
If $a, b, c$ are in continued proportion, prove that $: a^2 b^2 c^2 (a^{-4} + b^{-4} + c^{-4}) = b^{-2}(a^4 + b^4 + c^4)$
AnswerGiven : $a, b, c$ are in continued proportion.
$\frac{a}{b}=\frac{b}{c}= k$
$ \frac{a}{b}= k \therefore a = bk$
$ \frac{b}{c}= k \therefore b = ck$
$ \text { L.H.S. }= a ^2 b ^2 c ^2\left( a ^{-4}+ b ^{-4}+ c ^{-4}\right)$
$ \text { L.H.S. }=a^2 b^2 c^2\left[\frac{1}{a^4}+\frac{1}{b^4}+\frac{1}{c^4}\right]$
$ \text { L.H.S. }=\frac{a^2 b^2 c^2}{a^4}+\frac{a^2 b^2 c^2}{b^4}+\frac{a^2 b^2 c^2}{c^4}$
$ \text { L.H.S. }=\frac{b^2 c^2}{a^2}+\frac{c^2 a^2}{b^2}+\frac{a^2 b^2}{c^2}$
$ \text { L.H.S. }=\frac{(c k)^2 \cdot c^2}{\left(c k^2\right)^2}+\frac{c^2\left(c k^2\right)^2}{(c k)^2}+\frac{\left(c k^2\right)^2(c k)^2}{c^2}$
$ \text { L.H.S. }=\frac{c^2 k^2 \cdot c^2}{c^2 k^4}+\frac{c^2 \cdot c^2 k^4}{c^2 k^2}+\frac{c^2 k^4 \cdot c^2 k^2}{c^2}$
$ \text { L.H.S. }=\frac{c^2}{k^2}+\frac{c^2 k^2}{1}+\frac{c^2 k^6}{1}$
$ \text { L.H.S. }=c^2\left[\frac{1}{k^2}+k^2+k^6\right]$
$ \text { L.H.S. }=\frac{c^2}{k^2}\left[1+k^4+k^8\right]$
$ \text { R.H.S. }=b^{-2}\left[a^4+b^4+c^4\right]$
$ \text { R.H.S. }=\frac{1}{b^2}\left[a^4+b^4+c^4\right]$
$ \text { R.H.S. }=\frac{1}{c^2 k^2}\left[c^4 k^8+c^4 k^4+c^4\right]$
$\text { R.H.S. }=\frac{c^4}{c^2 k^2}\left[k^8+k^4+1\right]$
$ \text { R.H.S. }=\frac{c^2}{k^2}\left[1+k^4+k^8\right]$
$ \therefore \text { L.H.S. }=\text { R.H.S. }$
Hence proved.
View full question & answer→Question 304 Marks
If $a, b, c$ are in continued proportion, prove that: $a : c = (a^2 + b^2) : (b^2 + c^2)$
AnswerAs $a, b, c$, are in continued proportion
$
\begin{aligned}
& \text { Let } \frac{a}{b}=\frac{b}{c}= k \\
& \text { a }: c =\left( a ^2+ b ^2\right):\left( b ^2+ c ^2\right) \\
& \Rightarrow \frac{a}{c}=\frac{a^2+b^2}{b^2+c^2} \\
& \text { L.H.S. }=\frac{a}{c} \\
& =\frac{c k^2}{c} \\
& = k ^2 \\
& \text { R.H.S. }=\frac{\left(c k^2\right)^2+(c k)^2}{(c k)^2+c^2} \\
& =\frac{c^2 k^4+c^2 k^2}{c^2 k^2+c^2} \\
& =\frac{c^2 k^2\left(k^2+1\right)}{c^2\left(k^2+1\right)} \\
& = k ^2 \\
& \therefore \text { L.H.S. }=\text { R.H.S. }
\end{aligned}
$
View full question & answer→Question 314 Marks
If $a, b, c$ are in continued proportion, prove that:
$
\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{a}{b^2 c^2}+\frac{b}{c^2 a^2}+\frac{c}{a^2 b^2}
$
AnswerAs $a, b, c$, are in continued proportion
$
\begin{aligned}
& \text { Let } \frac{a}{b}=\frac{b}{c}= k \\
& \text { L.H.S. }=\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3} \\
& =\frac{1}{\left(c k^2\right)^3}+\frac{1}{(c k)^3}+\frac{1}{c^3} \\
& =\frac{1}{c^3 k^6}+\frac{1}{c^3 k^3}+\frac{1}{c^3} \\
& =\frac{1}{c^3}\left[\frac{1}{k^6}+\frac{1}{k^3}+\frac{1}{1}\right] \\
& \text { R.H.S. }=\frac{a}{b^2 c^2}+\frac{b}{c^2 a^2}+\frac{c}{a^2 b^2} \\
& =c \frac{k^2}{(c k)^2 c^2}+\frac{ ck }{c^2\left(c k^2\right)^2}+\frac{c}{\left(c k^2\right)^2(c k)^2} \\
& =\frac{c k^2}{c^4 k^2}+\frac{ ck }{c^4 k^4}+\frac{c}{c^4 k^6} \\
& =\frac{1}{c^3}+\frac{1}{c^3 k^3}+\frac{1}{c^3 k^6}
\end{aligned}
$
$\begin{aligned} & =\frac{1}{c^3}\left[1+\frac{1}{k^3}+\frac{1}{k^6}\right] \\ & =\frac{1}{c^3}\left[\frac{1}{k^6}+\frac{1}{k^3}+1\right] \\ & \therefore \text { L.H.S. = R.H.S. }\end{aligned}$
View full question & answer→Question 324 Marks
If $a, b, c$ are in continued proportion, prove that:
$
\frac{a+b}{b+c}=\frac{a^2(b-c)}{b^2(a-b)}
$
Answer$
\frac{a+b}{b+c}=\frac{a^2(b-c)}{b^2(a-b)}$
Since, $a, b, c$ are in continued proportion, $\frac{a}{b}=\frac{b}{c}$.
Let, $\frac{a}{b}=\frac{b}{c}=k$.Then, $a = bk$ and $b = ck$
Hence, $a = bk =( ck ) \cdot k = ck ^2$
$
\begin{aligned}
& \text { LHS }=\frac{a+b}{b+c} \\
& LHS =\frac{c k^2+c k}{c k+c} \\
& LHS =\frac{c k(k+1)}{c(k+1)} \\
& LHS =\frac{\not e k(\not k+\not 1)}{\not e(\not k+\not 1)}
\end{aligned}
$
$
\begin{aligned}
& \text { LHS }= k \ldots( I ) \\
& RHS =\frac{a^2(b-c)}{b^2(a-b)} \\
& RHS =\frac{\left(c k^2\right)^2(c k-c)}{(c k)^2\left(c k^2-c k\right)} \\
& \text { RHS }=\frac{c^2 k^4(c k-c)}{c^2 k^2\left(c k^2-c k\right)} \\
& \text { RHS }=\frac{c^3 k^4(k-1)}{c^3 k^3(k-1)} \\
& \text { RHS }=\frac{\not e^2 \not k^4(\not k-\not 1)}{\not e^3(\not k-\not 1)} \\
& \text { RHS }= k \ldots( II )
\end{aligned}
$
From (I) and (II),
$
\text { LHS }=\text { RHS }
$
View full question & answer→Question 334 Marks
If $a, b, c$ are in continued proportion, prove that:
$
\frac{p a^2+q a b+r b^2}{p b^2+q b c+r c^2}=\frac{a}{c}
$
AnswerGiven $a, b, c$ are in continued proportion
$
\begin{aligned}
& \frac{p a^2+q a b+r b^2}{p b^2+q b c+r c^2}=\frac{a}{c} \\
& \text { Let } \frac{a}{b}=\frac{b}{c}= k \\
& \Rightarrow a = bk \text { and } b = ck \\
& \Rightarrow a =( ck ) k = ck ^2
\end{aligned}
$
and $b = ck$
$
\text { L.H.S. }=\frac{a}{c}
$
$
\begin{aligned}
& =\frac{c k^2}{c} \\
& = k ^2
\end{aligned}
$
R.H.S. $=\frac{p\left(c k^2\right)^2+q\left(c k^2\right) c k+r(c k)^2}{p(c k)^2+q(c k) c+r c^2}$
$
\begin{aligned}
& =\frac{p c^2 k^4+q c^2 k^3+r c^2 k^2}{p c^2 k^2+q c^2 k+r c^2} \\
& =\frac{c^2 k^2}{c^2}\left[\frac{p k^2+q k+r}{p k^2+q k+r}\right] \\
& = k ^2
\end{aligned}$
From (ii) and (iii), L.H.S. = R.H.S.
$
\therefore b = ck , a = bk = ckk = ck ^2
$
$\begin{aligned} & \text { (i) L.H.S. } \\ & =\frac{ a + b }{ b + c } \\ & =\frac{c k^2+c k}{ ck + c } \\ & =\frac{c k(k+1)}{c(k+1)} \\ & = k \\ & \text { R.H.S. } \\ & =\frac{a^2(b-c)}{b^2(a-b)} \\ & =\frac{\left(c k^2\right)^2(c k-c)}{(c k)^2\left(c k^2-c k\right)} \\ & =\frac{c^2 k^4 c(k-1)}{c^2 k^2(k-1)} .\end{aligned}$
View full question & answer→Question 344 Marks
If $x , y , z$ are in continued proportion, prove that: $\frac{(x+y)^2}{(y+z)^2}=\frac{x}{z}$.
Answer$x, y, z$ are in continued proportion
Let $\frac{x}{y}=\frac{y}{z}= k$
Then $y = kz$
$
\begin{aligned}
& x = yk \\
& = kz x k \\
& = k ^2 z
\end{aligned}
$Now L.H.S.
$
\begin{aligned}
& =\frac{(x+y)^2}{(y+z)^2} \\
& =\frac{\left(k^2 z+k z\right)^2}{(k z+z)^2} \\
& =\frac{\{k z(k+1)\}^2}{\{z(k+1)\}^2} \\
& =\frac{k^2 z^2(k+1)^2}{z^2(k+1)^2} \\
& = k ^2 \\
& \text { R.H.S. }=\frac{x}{z} \\
& =\frac{k^2 z}{z} \\
& =\text { k }^2 \\
& \therefore \text { L.H.S. = R.H.S. }
\end{aligned}
$
View full question & answer→Question 354 Marks
If $a, b, c$ and $d$ are in proportion, prove that: $ a b c d\left[\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{d^2}\right]= a ^2+ b ^2+ c ^2+ d ^2\right. $
Answer
$\begin{aligned} & \because a _{,} b , c _r d \text { are in proportion } \\ & \frac{a}{b}=\frac{c}{d}= k (\text { say) } \\ & a = bk , c = dk . \\ & \text { L.H.S. }=a b c d\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{d^2}\right) \\ & =b k \cdot b \cdot d k \cdot d\left[\frac{1}{b^2 k^2}+\frac{1}{b^2}+\frac{1}{d^2 k^2}+\frac{1}{d^2}\right] \\ & =k^2 b^2 d^2\left[\frac{d^2+d^2 k^2+b^2+b^2 k^2}{b^2 d^2 k^2}\right] \\ & = d ^2\left(1+ k ^2\right)+ b ^2(1+ k 2) \\ & =\left(1+ k ^2\right)\left( b ^2+ d ^2\right) \\ & \text { R.H.S. }= a ^2+ b ^2+ c + d ^2 \\ & = b ^2 k ^2+ b ^2+ d ^2 k ^2+ d ^2 \\ & = b ^2\left( k ^2+1\right)+ d ^2\left( k ^2+1\right) \\ & =\left( k ^2+1\right)\left( b ^2+ d ^2\right) \\ & \therefore \text { L.H.S. }=\text { R.H.S. }\end{aligned}$
View full question & answer→Question 364 Marks
If $a, b, c$ and $d$ are in proportion, prove that:
$
\frac{a^2+b^2}{c^2+d^2}=\frac{ ab + ad - bc }{ bc + cd - ad }
$
Answer
$\begin{aligned} & \because a, b , c , d \text { are in proportion } \\ & \frac{c}{b}=\frac{ c }{d}= k ( say ) \\ & a = bk , c = dk . \\ & \text { L.H.S. }=\frac{a^2+b^2}{c^2+d^2} \\ & =\frac{b^2 k^2+b^2}{d^2 k^2+d^2} \\ & =\frac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)} \\ & =\frac{b^2}{d^2} \\ & \text { R.H.S. }=\frac{ ab + ad - bc }{ bc + cd - ad } \\ & =\frac{ bk \cdot b + bk \cdot d - b \cdot dk }{ b \cdot kd + dk \cdot d - bk \cdot d } \\ & =\frac{k\left(b^2+b d-b d\right)}{k\left(b d+d^2-b d\right)} \\ & =\frac{b^2}{d^2} \\ & \therefore \text { L.H.S. }=\text { R.H.S. }\end{aligned}$
View full question & answer→Question 374 Marks
If $a, b, c$ and $d$ are in proportion, prove that:
$
\frac{a^2+a b+b^2}{a^2-a b+b^2}=\frac{c^2+c d+d^2}{c^2-c d+d^2}
$
Answer
$\begin{aligned} & \because a , b , c , d \text { are in proportion } \\ & \frac{a}{b}=\frac{c}{d}= k \text { (say) } \\ & a = bk , c = dk . \\ & \text { L.H.S. }=\frac{a^2+a b+b^2}{a^2-a b+b^2} \\ & =\frac{b^2 k^2+b k \cdot b+b^2}{b^2 k^2-b k \cdot b+b^2} \\ & =\frac{b^2\left(k^2+k+1\right)}{b^2\left(k^2-k+1\right)} \\ & =\frac{k^2+k+1}{k^2-k+1} \\ & \text { R.H.S. }=\frac{c^2+c d+d^2}{c^2-c d+d^2} \\ & =\frac{d^2 k^2+d k d+d^2}{d^2 k^2-d k \cdot d+d^2} \\ & =\frac{d^2\left(k^2+k+1\right)}{d^2\left(k^2-k+1\right)} \\ & =\frac{k^2+k+1}{k^2-k+1} \\ & \therefore \text { L.H.S. = R.H.S. }\end{aligned}$
View full question & answer→Question 384 Marks
If $a, b, c$ and $d$ are in proportion, prove that:
$
\frac{(a+c)^3}{(b+d)^3}=\frac{a(a-c)^2}{b(b-d)^2}
$
Answer
$\begin{aligned} & \because a , b , c \text {, d are in proportion } \\ & \frac{a}{b}=\frac{c}{d}= k (\text { say) } \\ & a = bk , c = dk . \\ & \text { L.H.S. }=\frac{(a+c)^3}{(b+d)^3} \\ & =\frac{(b k+d k)^3}{(b+d)^3} \\ & =\frac{k^3(b+d)^3}{(b+d)^2} \\ & = k ^3 \\ & \text { R.H.S. }=\frac{a(a-c)^2}{b(b-d)^2} \\ & =\frac{b k(b k-d k)^2}{b(b-d)^2} \\ & =\frac{b k . k^2(b-d)^2}{b(b-d)^2} \\ & = k ^3 \\ & \therefore \text { L.H.S. }=\text { R.H.S. }\end{aligned}$
View full question & answer→Question 394 Marks
If $a, b, c$ and $d$ are in proportion, prove that:
$
\frac{a^2+a b}{c^2+c d}=\frac{b^2-2 a b}{d^2-2 c d}
$
Answer
$\begin{aligned} & \because a , b , c , d \text { are in proportion } \\ & \frac{a}{b}=\frac{c}{d}= k \text { (say) } \\ & a = bk , c = dk . \\ & \text { L.H.S. }=\frac{a^2+a b}{c^2+c d} \\ & =\frac{k^2 b^2+b k . b}{k^2 d^2+d k \cdot d} \\ & =\frac{k b^2(k+1)}{d^2 k(k+1)} \\ & =\frac{b^2}{d^2} \\ & \text { R.H.S. }=\frac{b^2-2 a b}{d^2-2 c d} \\ & =\frac{b^2-2 \cdot b k b}{d^2-2 d k d} \\ & =\frac{b^2(1-2 k)}{d^2(1-2 k)} \\ & =\frac{b^2}{d^2} \\ & \therefore \text { L.H.S. }=\text { R.H.S. }\end{aligned}$
View full question & answer→Question 404 Marks
If $a, b, c$ and d are in proportion, prove that: $(a^4 + c^4) : (b^4 + d^4) = a^2 c^2 : b^2 d^2.$
Answer$
\begin{aligned}
& \because a, b, c, d \text { are in proportion } \\
& \frac{a}{b}=\frac{c}{d}=k(\text { say }) \\
& a=b k, c=d k . \\
& \left(a^4+c^4\right):\left(b^4+d^4\right)=a^2 c^2: b^2 d^2
\end{aligned}
$
$
\text { L.H.S. }=\frac{a^4+c^4}{b^4+d^4}
$
$
=\frac{b^4 k^4+d^4 k^4}{b^4+d^4}
$
$
\begin{aligned}
& =\frac{k^4\left(b^4+d^4\right)}{\left(b^4+d^4\right)} \\
& = k ^4
\end{aligned}
$
$
\text { R.H.S. }=\frac{a^2 c^2}{b^2 d^2}
$
$
\begin{aligned}
& =\frac{k^2 b^2 \cdot k^2 d^2}{b^2 \cdot d^2} \\
& = k ^4
\end{aligned}
$Hence $ \text{L.H.S. = R.H.S.}$
View full question & answer→Question 414 Marks
If a, b, c and d are in proportion, prove that: (ma + nb) : b = (mc + nd) : d
Answer
$\begin{aligned} & \because a _{,} b , c _t d \text { are in proportion } \\ & \frac{a}{b}=\frac{c}{d}= k \text { (say) } \\ & a = bk , c = dk . \\ & ( ma + nb ): b =( mc + nd ): d \\ & \Rightarrow \frac{ ma + nb }{b}=\frac{ mc + nd }{d} \\ & \text { L.H.S. }=\frac{ mbk + nb }{b} \\ & =\frac{b( mk + n )}{b} \\ & = mk + n \\ & \text { R.H.S. }=\frac{ mc + nd }{d} \\ & =\frac{ mdk + nd }{d} \\ & =\frac{d( mk + n )}{d} \\ & = mk + n . \\ & \therefore \text { L.H.S. }=\text { R.H.S. }\end{aligned}$
View full question & answer→Question 424 Marks
If a, b, c and d are in proportion, prove that: (5a + 7b) (2c – 3d) = (5c + 7d) (2a – 3b).
AnswerIt is given that
$a, b, c, d$ are in proportion
Consider $\frac{a}{b}=\frac{c}{d}=k$
$a = bk _{ l } c = dk$
LHS $=(5 a+7 b)(2 c-3 d)$
LHS $=(5 b k+7 b)(2 d k-3 d) \ldots$ [Substituting the values]
LHS $=b(5 k+7) d(2 k-3) \quad \ldots[$ Taking out the common terms]
LHS $=b d(5 k+7)(2 k-3)$
LHS $=$ bd $[5 k(2 k-3)+7(2 k-3)]$
LHS $=b d\left(10 k^2-15 k+14 k-21\right)$LHS $=b d\left(10 k^2-k-21\right) \ldots(I)$
$
\text { RHS }=(5 c+7 d)(2 a-3 b)
$
RHS $=(5 dk +7 d )(2 bk -3 b ) \quad \ldots$ [Substituting the values]
RHS $=d(5 k+7) b(2 k-3) \ldots[$ Taking out the common terms]
$
\begin{aligned}
& \text { RHS }=b d(5 k+7)(2 k-3) \\
& \text { RHS }=\text { bd }[5 k(2 k-3)+7(2 k-3)] \\
& \text { RHS }=\text { bd }\left(10 k^2-15 k+14 k-21\right) \\
& \text { LHS }=b d\left(10 k^2-k-21\right) \ldots \text { (II) }
\end{aligned}
$From (I) and (II),Therefore, LHS $=$ RHS.
View full question & answer→Question 434 Marks
If $ax = by = cz$; prove that $\frac{x^2}{ yz }+\frac{y^2}{ zx }+\frac{z^2}{ xy }=\frac{ bc }{a^2}+\frac{ ca }{b^2}+\frac{ ab }{c^2}$
Answer
$\begin{aligned} & \text { Let } ax = by = cz = k \\ & \therefore x =\frac{k}{a}, y=\frac{k}{b}, z=\frac{k}{c} \\ & \text { L.H.S. }=\frac{x^2}{ yz }+\frac{y^2}{ zx }+\frac{z^2}{ xy } \\ & =\frac{\frac{k^2}{a^2}}{\frac{k}{b} \cdot \frac{k}{c}}+\frac{\frac{k^2}{b^2}}{\frac{k}{c} \cdot \frac{k}{a}}+\frac{\frac{k^2}{c^2}}{\frac{k}{a} \cdot \frac{k}{b}} \\ & =\frac{\frac{k^2}{a^2}}{\frac{k^2}{ bc }}+\frac{\frac{k^2}{b^2}}{\frac{k^2}{ ca }}+\frac{\frac{k^2}{c^2}}{\frac{k^2}{ ab }} \\ & =\frac{k^2}{a^2} \times \frac{ bc }{k^2}+\frac{k^2}{b^2} \times \frac{ ca }{k^2}+\frac{k^2}{c^2} \times \frac{ ab }{k^2} \\ & =\frac{ bc }{a^2}+\frac{ ca }{b^2}+\frac{ ab }{c^2} \\ & =\text { R.H.S. }\end{aligned}$
View full question & answer→Question 444 Marks
If $\frac{a}{c}=\frac{c}{d}=\frac{c}{f}$ prove that : $b d f\left[\frac{a+b}{b}+\frac{c+d}{d}+\frac{c+f}{f}\right]^3$ $=27(a+b)(c+d)(e+f)$
Answer
$\begin{aligned} & \frac{a}{c}=\frac{c}{d}=\frac{c}{f}= k \text { (say) } \\ & \therefore a = bk _{ l } c = dk _{ k } e = fk \\ & \text { L.H.S. }=b d f\left[\frac{a+b}{b}+\frac{c+d}{d}+\frac{c+f}{f}\right]^3 \\ & =b d f\left[\frac{b k+b}{b}+\frac{d k+d}{d}+\frac{f k+f}{f}\right]^3 \\ & =b d f\left[\frac{b(k+1)}{b}+\frac{d(k+1)}{d}+\frac{f(k+1)}{f}\right]^3 \\ & =\operatorname{bdf}( k +1+ k +1+ k +1)^3 \\ & =\operatorname{bdf}(3 k +3)^3=27 bdf ( k +1)^3 \\ & \text { R.H.S. }=27(a+b)(c+d)(e+f) \\ & =27(b k+b)(d k+d)(f k+f) \\ & =27 b ( k +1) d ( k +1) f ( k +1) \\ & =27 bdf ( k +1)^3 \\ & \therefore \text { L.H.S. }=\text { R.H.S. } \\ & \end{aligned}$
View full question & answer→Question 454 Marks
If $\frac{a}{c}=\frac{c}{d}=\frac{c}{f}$ prove that:
$
\frac{a^2}{b^2}+\frac{c^2}{d^2}+\frac{e^2}{f^2}=\frac{ ac }{ bd }+\frac{ ce }{ df }+\frac{ ae }{ df }
$
Answer
$\begin{aligned} & \frac{a}{c}=\frac{c}{d}=\frac{c}{f}= k \text { (say) } \\ & \therefore a = bk _1 c = dk _{ r } e = fk \\ & \text { L.H.S. }=\frac{a^2}{b^2}+\frac{c^2}{d^2}+\frac{e^2}{f^2} \\ & =\frac{b^2 k^2}{b^2}+\frac{d^2 k^2}{d^2}+\frac{f^2 k^2}{f^2} \\ & = k ^2+ k ^2+ k ^2 \\ & =3 k ^2 \\ & \text { R.H.S. }=\frac{ ac }{ bd }+\frac{ ce }{ df }+\frac{ ae }{ bf } \\ & =\frac{ bk \cdot dk }{ b \cdot d }+\frac{ dk \cdot fk }{ d \cdot f }+\frac{ bk \cdot fk }{ b \cdot f } \\ & =k^2+k^2+k^2 \\ & =3 k ^2 \\ & \therefore \text { L.H.S. }=\text { R.H.S. } \\ & \end{aligned}$
View full question & answer→Question 464 Marks
If $\frac{a}{c}=\frac{c}{d}=\frac{c}{f}$ prove that $: \frac{\left(a^3+c^3\right)^2}{\left(b^3+d^3\right)^2}=\frac{e^6}{f^6}$
Answer
$\begin{aligned} & \frac{a}{c}=\frac{c}{d}=\frac{c}{f}= k \text { (say) } \\ & \therefore a = bk , c = dk _{ r } e = fk \\ & \text { L.H.S. } \frac{\left(a^3+c^3\right)^2}{\left(b^3+d^3\right)^2} \\ & =\frac{\left(b^3 k^3+d^3 k^3\right)^2}{\left(b^3+d^3\right)^2} \\ & =\frac{\left[k^3\left(b^3+d^3\right)\right]^2}{\left(b^3+a^3\right)^2} \\ & =\frac{k^6\left(b^3+d^3\right)^2}{\left(b^3+d^3\right)^2}= k ^6 \\ & \text { R.H.S. }=\frac{e^6}{f^6} \\ & =f^6 \frac{k^6}{f^6}= k ^6 \\ & \therefore \text { L.H.S. }=\text { R.H.S. }\end{aligned}$
View full question & answer→Question 474 Marks
If $\frac{a}{c}=\frac{c}{d}=\frac{c}{f}$ prove that $:\left(b^2+d^2+f^2\right)\left(a^2+c^2+e^2\right)=(a b+$ $c d+e f)^2$
Answer
$\begin{aligned} & \frac{a}{c}=\frac{c}{d}=\frac{c}{f}= k \text { (say) } \\ & \therefore a=b k_t c=d k^2 e=f k \\ & \text { L.H.S. }=\left(b^2+d^2+f^2\right)\left(a^2+c^2+e^2\right) \\ & =\left(b^2+d^2+f^2\right)\left(b^2 k^2+d^2 k^2+f^2 k^2\right) \\ & =k^2\left(b^2+d^2+f^2\right) k^2\left(b^2+d^2+f^2\right) \\ & =k^2\left(b^2+d^2+f^2\right)^2 \\ & \text { R.H.S. }=(a b+c d+e f)^2 \\ & =(b . k b+d k \cdot d+f k . f)^2 \\ & =\left(k b^2+k d^2+k f^2\right) \\ & =k^2\left(b^2+d^2+f^2\right)^2 \\ & \therefore \text { L.H.S. }=\text { R.H.S. }\end{aligned}$
View full question & answer→Question 484 Marks
If $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}$, prove that $\frac{ ax - by }{(a+b)(x-y)}+\frac{ by - cz }{(b+c)(y-z)}+\frac{ cz - ax }{(c+a)(z-x)}=3$
Answer
$
\begin{aligned}
& \frac{x}{a}=\frac{y}{b}=\frac{z}{c} \\
& \therefore x = ak _t y = bk _{ r } z = ck
\end{aligned}
$
L.H.S.
$
\begin{aligned}
& \frac{ ax - by }{(a+b)(x-y)}+\frac{ by - cz }{(b+c)(y-z)}+\frac{ cz - ax }{(c+a)(z-x)} \\
& = \\
& \frac{ a \cdot ak - b \cdot bk }{(a+b)(a k-b k)}+\frac{ bk - c . ck }{(b+c)(b k-c k)}+\frac{ c . ck - a \cdot ak }{(c+a)(c k-a k)} \\
& =\frac{a^2 k-b^2 k}{(a+b) k(a-b)}+\frac{b^2 k^2}{(b+c) k(b-c)}+\frac{c^2 k-a^2 k}{(c+a)(c-a)} \\
& =\frac{k\left(a^2-b^2\right)}{k\left(a^2-b^2\right)}+\frac{k\left(b^2-c^2\right)}{k\left(b^2-c^2\right)}+\frac{k\left(c^2-a^2\right)}{k\left(c^2-a^2\right)} \\
& =1+1+1 \\
& =3 \\
& =\text { R.H.S. }
\end{aligned}
$
View full question & answer→Question 494 Marks
If $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}$, prove that $\left[\frac{a^2 x^2+b^2 y^2+c^2 z^2}{a^2 x+b^3 y+c^3 z}\right]^3=\frac{ xyz }{ abc }$
Answer
$\begin{aligned} & \frac{x}{a}=\frac{y}{b}=\frac{z}{c} \\ & \therefore x = ak _{ r } y = bk _{ r } z = c \end{aligned}$
$\begin{aligned} & \text { L.H.S. }=\left[\frac{a^2 x^2+b^2 y^2+c^2 z^2}{a^2 x+b^3 y+c^3 z}\right]^3 \\ & =\left[\frac{a^2 \cdot a^2 k^2+b^2 \cdot b^2 l^2+c^2 \cdot c^2 k^2}{a^3 \cdot a \cdot k+b^3 \cdot b k+c^3 \cdot c k}\right]^3 \\ & =\left[\frac{a^4 k^2+b^4 k^2+c^4 k^2}{a^4 k+b^4 k+c^4 k}\right]^3 \\ & =\left[\frac{k^2\left(a^4+b^4+c^4\right)}{k\left(a^4+b^4+c^4\right)}\right]^3= k ^3\end{aligned}$
$\begin{aligned} & \text { R.H.S. }=\frac{ xyz }{ abc } \\ & =\frac{ ak \cdot bk \cdot ck }{ abc }= k ^3 \\ & \therefore \text { L.H.S. }=\text { R.H.S. }\end{aligned}$
View full question & answer→Question 504 Marks
If $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}$, prove that $\frac{x^3}{a^2}+\frac{y^3}{b^2}+\frac{z^3}{c^2}=\frac{(x+y+z)^3}{(a+b+c)^2}$
Answer
$
\begin{aligned}
& \frac{x}{a}=\frac{y}{b}=\frac{z}{c} \\
& \therefore x = ak _1 y = bk _{ r } z = ck
\end{aligned}
$
$
\begin{aligned}
& \text { L.H.S. }=\frac{x^3}{a^2}+\frac{y^3}{b^2}+\frac{z^3}{c^2} \\
& =\frac{a^3 k^3}{a^2}+\frac{b^3 k^3}{b^2}+\frac{c^3 k^3}{c^2} \\
& =a k^3+b k^3+ ck ^3 \\
& = k ^3(a+b+c)
\end{aligned}
$
$
\begin{aligned}
& \text { R.H.S. }=\frac{(x+y+z)^3}{(a+b+c)^2} \\
& =\frac{(a k+b k+c k)^3}{(a+b+c)^2} \\
& =\frac{k^3(a+b+c)^3}{(a+b+c)} \\
& = k ^3(a+b+c)
\end{aligned}$
Hence L.H.S. = R.H.S.
View full question & answer→