[5 marks sum] - MATHEMATICS STD 9 Questions - Vidyadip

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[5 marks sum]

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21 questions · timed · auto-graded

Question 15 Marks

Solve for $x:2^{2x- 1} -9 \times 2^{x - 2} + 1= 0$

Answer

$ 2^{2 x-1}-9 \times 2^{x-2}+1=0$
$2^{2 x} \cdot 2^{-1}-9 x^{2 x} \cdot 2^{-2}+1=0 $
Let $2^x=t$, so $2^{2 x}=t^2$
So, $2^{2 x} \cdot 2^{-1}-9 \times 2^x \cdot 2^{-2}+1=0$
becomes $\frac{t^2}{2}-9 \times \frac{t}{2^2}+1=0$
$ \Rightarrow \frac{t^2}{2}-\frac{9 t}{4}+1=0$
$\Rightarrow 2 t^2-9 t+4=0$
$\Rightarrow 2 t^2-8 t-t+4=0$
$\Rightarrow 2 t(t-4)-1(t-4)=0$
$\Rightarrow( t -4)(2 t -1)=0$
$\Rightarrow t -4=0$ or $2 t -1=0$
$\Rightarrow t =4$ or $t =\frac{1}{2}$
So, $2^x=4$ or $2^x=\frac{1}{2}$
$\Rightarrow 2^x=2^2$ or $2^x=2^{-1}$
$\Rightarrow x =2$ or $x =-1$.

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Question 25 Marks

Solve for $x:2^{2x} + 2^{x +2} - 4 \times 2^3 = 0$

Answer

$2^{2x} + 2^{x +2} - 4 x 2^3 = 0$
$\Rightarrow 2^{2x} + 2^{x+2} - 2^2 x 2^3 = 0$
$\Rightarrow 2^{2x} + 2^x . 2^2 - 2^{2+3} = 0 ......($Using $a^m x a^n= a^{m+n})$
$\Rightarrow 2^{2x}+ 2^x . 2^2 - 2^5 = 0$
$\Rightarrow 2^{2x} + 2^x . 4 - 32 = 0$
Put $2^x=t$
So, $2^{2 x }= t ^2$
$2^{2x} + 2^{x+2} - 32 = 0$
becomes $t^2 + 4t - 32 = 0$
$\Rightarrow (t + 8)(t - 4) = 0$
$\Rightarrow t + 8 = 0$ or $t - 4 = 0$
$\Rightarrow t = -8 = 0$ or $t = 4$
$\Rightarrow 2^x = -8$ or $2^x = 4$
$\Rightarrow 2x = -2^3$ or $2^x = 2^2$
Using the second equation $2^x=2^2$, we get $x=2$.

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Question 35 Marks

Solve for $x:2^{2x+3} - 9 x 2^x+ 1 = 0$

Answer

$2^{2 x+3}-9 \times 2^x+1=0$
$2^{2 x} \cdot 2^3-9 \times 2^x+1=0$
Put $2^x=t$, so, $2^{2 x}=t^2$
So, $2^{2 x} \cdot 2^3-9 \times 2^x+1=0$
becomes $8 t^2-9 t+1=0$
$\Rightarrow 8 t^2-8 t-t+1=0$
$\Rightarrow 8 t ( t -1)-( t -1)=0$
$\Rightarrow t -1=0$ or $8 t -1=0$
$\Rightarrow t =1$ or $t =\frac{1}{8}$
$\Rightarrow 2^{ x }=1$ or $2^{ x }=\frac{1}{2^3}$
$\Rightarrow 2^{ x }=2^0$ or $2^{ x }=2^{-3}$
$\Rightarrow x =0$ or $x =-3$

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Question 45 Marks

Solve for $x:2^{x+3}+2^{x+1}=320$

Answer

$2^{x+3}+2^{x+1}=320$
$\Rightarrow 2^{x+3}+2^{x+1}=2^6 \times 5$
$\Rightarrow 2^x \cdot 2^3+2^x \cdot 2^1=2^6 \times 5$
$\Rightarrow 2^x\left(2^3+2^1\right)=2^6 \times 5$
$\Rightarrow 2^x(8+2)=2^6 \times 5$
$\Rightarrow 2^x(10)=2^6 \times 5$
$\Rightarrow 2^x\left(\frac{10}{5}\right)=2^6$
$\Rightarrow 2^x \cdot 2^6$
$\Rightarrow \frac{2^x \cdot 2}{2^6}=1$
$\Rightarrow 2^{x+1-6}=1 \times 2^0$
$\Rightarrow 2^{x-5}=1 \times 2^0$
$\Rightarrow x-5=0$
$\Rightarrow x=5$

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Question 55 Marks

Simplify the following:$\left(\frac{27}{343}\right)^{\frac{2}{3}} \div \frac{1}{\left(\frac{625}{1296}\right)^{\frac{1}{4}}} \times \frac{536}{\sqrt[3]{27}}$

Answer

$\left(\frac{27}{343}\right)^{\frac{2}{3}} \div \frac{1}{\left(\frac{625}{1296}\right)^{\frac{1}{4}}} \times \frac{536}{\sqrt[3]{27}}$
$=\left(\frac{3^3}{7^3}\right)^{\frac{2}{3}} \div \frac{1}{\left(\frac{5^4}{2^4 \times 3^4}\right)^{\frac{1}{4}}} \times \frac{2^3 \times 67}{\sqrt[3]{3^3}}$
$=\left(\frac{3^3}{7^3}\right)^{\frac{2}{3}} \div \frac{1}{\left(\frac{5^4}{2^4 \times 3^4}\right)^{\frac{1}{4}}} \times \frac{2^3 \times 67}{\left(3^3\right)^{\frac{1}{3}}}$
$=\left(\frac{3^{3 \times \frac{2}{3}}}{7^{3 \times \frac{2}{3}}}\right) \div \frac{1}{\left(\frac{5^{4 \times \frac{1}{4}}}{2^{4 \times \frac{1}{4}} \times 3^4 \times \frac{1}{4}}\right)} \times \frac{2^3 \times 67}{3^{3 \times \frac{1}{3}}} \ldots($Using $\left(a^m\right)^n=a^{m n})$
$=\left(\frac{3^2}{7^2}\right) \div \frac{1}{\left(\frac{5^1}{2^1 \times 3^1}\right)} \times \frac{2^3 \times 67}{3^1}$
$=\left(\frac{3^2}{7^2}\right) \div\left(\frac{2^1 \times 3^1}{5^1}\right) \times\left(\frac{2^3 \times 67}{3^1}\right) \ldots . .($Using $a^m \times a^n=a^{m-n}$ and $a^m \div a^n=a^{m-n})$
$=\left(\frac{3^2}{7^2}\right) \div\left(\frac{5^1}{2^1 \times 3^1}\right) \times\left(\frac{2^3 \times 67}{3^1}\right)$
$=3^{2-1-1} \times 2^{3-1} \times 5^1 \times 7^2 \times 67$
$=3^0 \times 2^2 \times 5^1 \times 7^2 \times 67$
$=1 \times 4 \times 5 \times 49 \times 67 \ldots($Using $a^0=1)$
$=65660 \text {. } $

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Question 65 Marks

Simplify the following:$\left\{\left( a ^{ m }\right)^{ m -\frac{1}{ m }}\right\}^{\frac{1}{ m +1}}$

Answer

$\left\{\left(a^m\right)^{m-\frac{1}{m}}\right\}^{\frac{1}{m+1}}$
$=(a)^{m \times\left(m-\frac{1}{m}\right) \times\left(\frac{1}{m+1}\right)} \ldots($Using $a^m \div a^n=a^{m-n})$
Consider, $m \times\left(m-\frac{1}{m}\right) \times\left(\frac{1}{m}+1\right)$
$=\left(m^2-1\right) \times\left(\frac{1}{m}+1\right)$
$=m^2 \times\left(\frac{1}{m}+1\right)-1 \times\left(\frac{1}{m}+1\right)$
$=\frac{m^2}{m+1}-\frac{1}{m+1}$
$=\frac{m^2-1}{m+1}$
$=\frac{(m-1)(m+1)}{m+1}$
$=m-1(a) m \times\left(m-\frac{1}{m}\right) \times\left(\frac{1}{m}+1\right)$
$=a^{m-1} .$

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Question 75 Marks

Simplify the following:$\sqrt[3]{x^4 y^2} \div \sqrt[6]{x^5 y^{-5}}$

Answer

$ \sqrt[3]{x^4 y^2} \div \sqrt[6]{x^5 y^{-5}}$
$=\left(x^4 y^2\right)^{\frac{1}{3}} \div\left(x^5 y^{-5}\right)^{\frac{1}{6}}$
$=\left(x^{4 \times \frac{1}{3}} y^{2 \times \frac{1}{3}}\right) \div\left(x^{5 \times \frac{1}{6}} y^{-5 \times \frac{1}{6}}\right) \quad \ldots($Using $\left( a ^{ m }\right)^n= a ^{ mn })$
$=\left(x^{\frac{4}{3}} y^{\frac{2}{3}}\right) \div\left(x^{\frac{5}{6}} y^{-\frac{5}{6}}\right)$
$=\frac{x^{\frac{4}{3}} y^{\frac{2}{3}}}{x^{\frac{5}{6}} y^{-\frac{5}{6}}}$
$=x^{\frac{4}{3}-\frac{5}{6}} y^{\frac{2}{3}-\left(-\frac{5}{6}\right)} \ldots($Using $\left( a ^{ m }\right)^{ n }= a ^ {mn})$
$=x^{\frac{1}{2}} y^{\frac{3}{2}}$
$=x^{\frac{1}{2}}\left(y^3\right)^{\frac{1}{2}}$
$=\sqrt{x} \sqrt{y^3}$
$=\sqrt{x y^3} . \ldots($Using $\left.\left(a^m\right)^n=a^{m n}\right)$

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Question 85 Marks

Simplify the following:$\left(a^{\frac{1}{3}}+a^{-\frac{1}{3}}\right)\left(a^{\frac{2}{3}}-1+a^{-\frac{2}{3}}\right)$

Answer

$\left(a^{\frac{1}{3}}+a^{-\frac{1}{3}}\right)\left(a^{\frac{2}{3}}-1+a^{-\frac{2}{3}}\right)$
$=a^{\frac{1}{3}}\left(a^{\frac{2}{3}}-1+a^{-\frac{2}{3}}\right)+a^{-\frac{1}{3}}\left(a^{\frac{2}{3}}-1+a^{-\frac{2}{3}}\right)$
$=\left(a^{\frac{1}{3}} \times a^{\frac{2}{3}}-a^{\frac{1}{3}} \times 1+a^{\frac{1}{3}} \times a^{-\frac{2}{3}}\right)+\left(a^{-\frac{1}{3}} \times a^{\frac{2}{3}}-a^{-\frac{1}{3}} \times 1+a^{-\frac{1}{3}} \times a^{-\frac{2}{3}}\right)$
$=\left(a^{\frac{1}{3}+\frac{2}{3}}-a^{\frac{1}{3}} \times 1+a^{\frac{1}{3}+\frac{2}{3}}\right)+\left(a^{-\frac{1}{3}+\frac{2}{3}}-a^{-\frac{1}{3}}+a^{-\frac{1}{3}-\frac{2}{3}}\right) \ldots($Using $a^m \times a^n=a^{m+n})$
$=\left(a^1-a^{\frac{1}{3}}+a^{-\frac{1}{3}}\right)+\left(a^{\frac{1}{3}}-a^{-\frac{1}{3}}+a^{-1}\right)$
$=a-a^{\frac{1}{3}}+a^{-\frac{1}{3}}+a^{\frac{1}{3}}-a^{-\frac{1}{3}}+\frac{1}{a}$
$=a+\frac{1}{a} .$

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Question 95 Marks

Simplify the following:$\left(\frac{36 m^{-4}}{49 n^{-2}}\right)^{-\frac{3}{2}}$

Answer

$\left(\frac{36 m^{-4}}{49 n^{-2}}\right)^{-\frac{3}{2}}$
$=\left(\frac{6^2 m^{-4}}{7^2 n^{-2}}\right)^{-\frac{3}{2}}$
$=\left(\frac{6^{2 \times\left(-\frac{3}{2}\right)} m^{-4 \times\left(-\frac{3}{2}\right)}}{7^{2 \times\left(-\frac{3}{2}\right)} n^{-2} \times\left(-\frac{3}{2}\right)}\right) \ldots \ldots($Using $(a \times b)^n=a^n \times b^n$ and $\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n})$
$=\frac{6^{-3} m^6}{7^{-3} n^3}$
$=\frac{7^3 m^6}{6^3 n^3} \cdots \cdots($Using $a^{-n}=\frac{1}{a^n})$
$=\frac{343 m^6}{216 n^3} .$

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Question 105 Marks

Simplify the following:$\left(\frac{64 a^{12}}{27 b^6}\right)^{-\frac{2}{3}}$

Answer

$\left(\frac{64 a^{12}}{27 b^6}\right)^{-\frac{2}{3}}$
$=\left(\frac{2^6 a^{12}}{3^3 b^6}\right)^{-\frac{2}{3}}$
$=\left(\frac{2^{6 \times\left(-\frac{2}{3}\right)} a^{12 \times\left(-\frac{2}{3}\right)}}{3^{3 \times\left(-\frac{2}{3}\right)} b^{6 \times\left(-\frac{2}{3}\right)}}\right) \ldots \ldots($Using $\left(a \times b^n\right)=a^n \times b^n$ and $\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n})$
$=\frac{2^{-4} a^{-8}}{3^{-2} b^{-4}}$
$=\frac{3^2 b^4}{2^4 a^8} \cdots \cdot($Using $a^{-n}=\frac{1}{a^n})$
$=\frac{9 b^4}{16 a^8} .$

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Question 115 Marks

Simplify the following and express with positive index:$3p^{-2}q^3 \div 2p^3q^{-2}$

Answer

$3 p^{-2} q^3 \div 2 p^3 q^{-2}$
$=\frac{3 p^{-2} q^3}{2 p^3 q^{-2}}$
$=\frac{3}{2}\left[\frac{p^{-2}}{p^3} \times \frac{q^3}{q^{-2}}\right]$
$=\frac{3}{2}\left[\left(p^{-2} \div p^3\right) \times\left(q^3 \div q^{-2}\right)\right]$
$=\frac{3}{2}\left[\left(p^{-2-3}\right) \times\left(q^{3-(-2)}\right)\right] \ldots($Using $a^m \div a^n=a^{m-n})$
$=\frac{3}{2}\left[\left(p^{-5}\right) \times\left(q^5\right)\right]$
$=\frac{3}{2}\left[\left(\frac{1}{p^5}\right) \times\left(q^5\right)\right]$
$=\frac{3 q^5}{2 p^5} .$

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Question 125 Marks

Prove the following:$\frac{x^{ p ( q - r )}}{x^{ q ( p - r )}} \div\left(\frac{x^{ q }}{x^{ p }}\right)^{ r }=1$

Answer

$\text { L.H.S } $
$=\frac{x^{ p ( q - r )}}{x^{ q ( p - r )}} \div\left(\frac{x^{ q }}{x^{ p }}\right)^{ r } $
$=\frac{x^{ p ( q - r )}}{x^{ q ( p - r )}} \div \frac{x^{ qr }}{x^{ pr }} \ldots($Using $\left(a ^{ m }\right)^{ n }= a ^{ mn})$
$=\frac{x^{ p ( q - r )}}{x^{ q ( p - r )}} \times \frac{x^{ pr }}{x^{ qr }} $
$=\frac{x^{ pq - pr }}{x^{ pq - qr }} \times \frac{x^{ pr }}{x^{ qr }} $
$=\frac{x^{ pq - pr + pr }}{x^{ pq - qr + qr }} \ldots ($Using $a ^{ m } \times a ^{ n }= a ^{ m + n })$
$=\frac{x^{ pq }}{x^{ pq }} $
$=1 $
$=\text { R.H.S }$
Hence proved.

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Question 135 Marks

Prove the following:$\sqrt[a b]{\frac{x^{ a }}{x^{ b }}} \cdot \sqrt[ bc ]{\frac{x^{ b }}{x^{ c }}} \cdot \sqrt[ ca ]{\frac{x^{ c }}{x^{ a }}}=1$

Answer

$\text{L.H.S.}$
$ =\sqrt[ ab ]{\frac{x^{ a }}{x^{ b }}} \cdot \sqrt[ bc ]{\frac{x^{ b }}{x^{ c }}} \cdot \sqrt[ a ]{\frac{x^{ c }}{x^{ ca }}}$
$=\left(\frac{x^{ a }}{x^{ b }}\right)^{\frac{1}{ ah }} \cdot\left(\frac{x^{ b }}{x^{ c }}\right)^{\frac{1}{ bc }} \cdot\left(\frac{x^{ c }}{x^{ a }}\right)^{\frac{1}{ ca }}$
$=\frac{x^{\frac{1}{b}}}{x \frac{1}{ a }} \cdot \frac{x^{\frac{1}{ c }}}{x \frac{1}{ b }} \cdot \frac{x^{\frac{1}{ a }}}{x \frac{1}{ c }} \ldots . .($Using $\left( a ^{ m }\right)^{ n }= a ^{ mn })$
$=x^{\frac{1}{ b }-\frac{1}{ a }} \cdot x^{\frac{1}{ c }-\frac{1}{ b }} \cdot x^{\frac{1}{ a }-\frac{1}{ c }} \ldots($Using $a ^{ m } \div a ^{ n }= a ^{ m - n })$
$=x^{\frac{ a - b }{ ab }} \cdot x^{\frac{ b - c }{ bc }} \cdot x^{\frac{ c - a }{ ac }}$
$=x^{\frac{a b}{a b}+\frac{b c}{k x}+\frac{c a t}{a c}}...($Using $\left.a^m \times a^n=a^{m+n}\right)$
$=x^{\frac{a c- bc + ab - ac + bc - ab }{ abc }}$
$=x^{\frac{0}{a b c}}$
$= x ^0$
$=1\dots...($Using $\left.a^0=1\right)$
$=\text { R.H.S. }$
Hence proved.

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Question 145 Marks

Prove the following:$\left(\frac{a^m}{a^n}\right)^{m+n+1} \cdot\left(\frac{a^n}{a^1}\right)^{n+1-m} \cdot\left(\frac{a^1}{a^m}\right)^{1+m-n}$

Answer

$\text{L.H.S.}$
$=\left(\frac{a^m}{a^n}\right)^{m+n+1} \cdot\left(\frac{a^n}{a^1}\right)^{n+1-m} \cdot\left(\frac{a^1}{a^m}\right)^{1+m-n}$
$=\frac{a^{m(m+n-1)}}{a^{n(m+n-1)}} \cdot \frac{a^{n(n+1-m)}}{a^{1(n+1-m)}} \cdot \frac{a^{1(1+m-n)}}{a^{m(1+m-n)}} \ldots \ldots($Using $\left(a^m\right)^n=a^{m n})$
$=\frac{a^{m^x+m n-m}}{a^{n^x+m n-n}} \cdot \frac{a^{n^x-m n+n}}{a^{n+1-m}} \cdot \frac{a^{1+m-n}}{a^{m^x-m n+m}}$
$=a^{m^x+m n-m-\left(n^x+m n-n\right)} \cdot a^{n^x-m n-(n+1-m)} \cdot a^{1+m-n-\left(m^x-m n+m\right)} \ldots($Using $a^m \div a^n=a^{m-n})$
$=a^{m^x+m n-m-n^x-m n+n} \cdot a^{n^x-m n+n-n-1+m} \cdot a^{1+m-n-m m^x-m n+m}$
$=a^{m^x+m n-m-n^x-m n+n+n^x-m n+n-n-1+m+1+m-n-m m^x+m n-m} \ldots . ($Using $a^m \times a^n=a^{m+n})$
$=a^0$
$=1 \ldots .($Using $a^{\circ}=1)$
$=\text { R.H.S. }$
Hence proved.

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Question 155 Marks

Prove the following:$\left(\frac{x^{ a + b }}{x^{ c }}\right)^{ a - b } \cdot\left(\frac{x^{ c + a }}{x^{ b }}\right)^{ c - a } \cdot\left(\frac{x^{ b + c }}{x a }\right)^{ b - c }=1$

Answer

$\text{L.H.S.}$
$=\left(\frac{a^m}{a^n}\right)^{m+n+1} \cdot\left(\frac{a^n}{a^1}\right)^{n+1-m} \cdot\left(\frac{a^1}{a^m}\right)^{1+m-n}$
$=\frac{a^{m(m+n-1)}}{a^{n(m+n-1)}} \cdot \frac{a^{n(n+1-m)}}{a^{1(n+1-m)}} \cdot \frac{a^{1(1+m-n)}}{a^{m(1+m-n)}} \ldots . . .($Using $\left(a^m\right)^n=a^{m n})$
$=\frac{a^{m^x+m n-m}}{a^{n^x+m n-n}} \cdot \frac{a^{n^x-m n+n}}{a^{n+1-m}} \cdot \frac{a^{1+m-n}}{a^{m^x-m n+m}}$
$=a^{m^2+m n-m-\left(n^2+m n-n\right)} \cdot a^{n^x-m n-(n+1-m)} \cdot a^{1+m-n-\left(m^2-m n+m\right)}$
$($Using $\left.a^m \div a^n=a^{m-n}\right)$
$= a ^{ m ^2+ mn - m - n ^2- mn + n } \cdot a ^{ n ^2- mn + n - n -1+ m } \cdot a ^{1+ m - n - m ^2- mn + m }$
$= a ^{ m ^2+ mn - m - n ^2- mn + n + n ^2- mn + n - n -1+ m +1+ m - n - m ^2+ mn - m }$
$($Using $\left.a^m \times a^n=a^{m-n}\right)$
$=a^a$
$=1 \ldots($Using $a^{\circ}=1)$
$=\text { R.H.S. }$
Hence proved.

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Question 165 Marks

Evaluate the following:$\left(\frac{64}{216}\right)^{\frac{2}{3}} \times\left(\frac{16}{36}\right)^{-\frac{3}{2}}$

Answer

$\left(\frac{64}{216}\right)^{\frac{2}{3}} \times\left(\frac{16}{36}\right)^{-\frac{3}{2}}$
$=\left(\frac{2^6}{6^3}\right)^{\frac{2}{3}} \times\left(\frac{2^4}{6^2}\right)^{-\frac{3}{2}}$
$=\frac{\left(2^6\right)^{\frac{2}{3}}}{\left(6^3\right)^{\frac{2}{3}}} \times \frac{\left(2^4\right)^{-\frac{3}{2}}}{\left(6^2\right)^{-\frac{3}{2}}}$
$=\frac{(2)^{6 \times \frac{2}{3}}}{(6)^{3 \times \frac{2}{3}}} \times \frac{(2)^{4 \times\left(-\frac{3}{2}\right)}}{\left(6^2\right)^{2 \times\left(-\frac{3}{2}\right)}} \ldots \ldots($Using $\left(a^m\right)^n=a^{m n})$
$=\frac{(2)^{2 \times 2}}{(6)^2} \times \frac{(2)^{2 \times(-3)}}{(6)^{-3}}$
$=\frac{(2)^4}{(6)^2} \times \frac{(2)^{-6}}{(6)^{-3}}$
$=\frac{(2)^4}{(6)^2} \times \frac{(6)^3}{(2)^6} \ldots \ldots($Using $\left.a^{m+}=\frac{1}{a^m}\right)$
$=\frac{(2)^4}{(2)^6} \times \frac{(6)^3}{(6)^2}$
$=(2)^{4-6} \times(6)^{3-2} \ldots . .($Using $a^m \div a^n=a^{m-n})$
$=(2)^{-2} \times(6)^1$
$=\frac{1}{2^2} \times 6$
$=\frac{1}{4} \times 6$
$=\frac{3}{2} .$

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Question 175 Marks

Find the value of $'a\ '$ and $' b\ '$ if:$(\sqrt{243})^{ a } \div 3^{ b +1}=1$ and $27^{ b }-81^{4-\frac{ a }{2}}=0$

Answer

$ (\sqrt{243})^{ a } \div 3^{ b +1}=1$ and $27^{ b }-81^{4-\frac{ a }{2}}=0$
$\Rightarrow\left(\sqrt{3^5}\right)^{ a } \div 3^{ b +1}$ and $\left(3^3\right)^{ b }-\left(3^4\right)^{4-\frac{ a }{2}}=0$
$\Rightarrow\left(3^5\right)^{\frac{ a }{2}} \div 3^{ b +1}=1$ and $3^{3 b }-\left(3^4\right)^{4-\frac{ a }{2}}=0$
$\Rightarrow 3^{\left(\frac{5 a }{2}\right)} \div 3^{ b +1}=1$ and $3^{(3 b )}-3^{4\left(4-\frac{ a }{2}\right)}=0$
$\Rightarrow 3^{\left(\frac{5 a }{2}- b -1\right)}=1$ and $3^{(3 b )}-3^{16-2 a }=0$
$\Rightarrow 3^{\left(\frac{5 a }{2}- b -1\right)}=3^{\circ}$ and $3^{3 b }=3^{16-2 a }$
$\Rightarrow \frac{5 a }{2}- b -1=0$ and $3 b =16-2 a$
$\Rightarrow \frac{5 a }{2}- b =1$ and $2 a +3 b =16$
$\Rightarrow 5 a -2 b =2$ and $2 a +3 b =16$
Multiply the equations by $3$ and $2$ respectively.
$\Rightarrow 15 a -6 b =6$ and $4 a +6 b =32$
Adding the equations,
$19 a =38$
$\Rightarrow a =2$
Substitute the value of $a$ in $5 a-2 b=2$ to find $b$.
$5 a-2 b=2$
$\Rightarrow 5(2)-2 b=2$
$\Rightarrow 10-2 b=2$
$\Rightarrow b=4 $
Hence, $a=2$ and $b=4$.

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Question 185 Marks

Find the value of $' a\ '$ and $' b\ '$ if:$9^{2 a}=(\sqrt[3]{81})^{-\frac{6}{b}}=(\sqrt{27})^2$

Answer

$9^{2 a}=(\sqrt[3]{81})^{-\frac{6}{b}}=(\sqrt{27})^2$
$\Rightarrow 9^{2 a}=\left(\sqrt[3]{3^4}\right)^{-\frac{6}{b}}=\left(\sqrt{3^3}\right)^2$
$\Rightarrow\left(3^2\right)^{2 a}=\left(3^{4 \times \frac{1}{1}}\right)^{-\frac{6}{b}}=\left(3^{3 \times \frac{1}{2}}\right)^2$
$\Rightarrow 3^{4 a}=\left(3^1\right)^{-\frac{8}{b}}=\left(3^1\right)^3$
$\Rightarrow 3^{4 a}=\frac{-8}{b}=3$
$\Rightarrow 3^{4 a}=3$ and $\frac{-8}{b}=3$
$\Rightarrow 4 a =3$ and $b=\frac{-8}{3}$
$\Rightarrow a =\frac{3}{4}$ and $b=\frac{-8}{3} .$

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Question 195 Marks

If $a^x=b^y=c^z$ and $a b c=1$, show that $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$

Answer

$a^x=b^y=c^z$
So, $a^x=b^y$
$\Rightarrow a=b^{\frac{y}{x}}\ldots . .($Usinga $^{\frac{1}{n}}=\sqrt[n]{a})$
$b^y=c^z$
$\Rightarrow c=b^{\frac{y}{x}}\ldots . .($Using $a^{\frac{1}{n}}=\sqrt[n]{a})$
and $a b c=1$
$\Rightarrow b ^{\frac{y}{x}} \cdot b \cdot b ^{\frac{y}{x}}=1$
$\Rightarrow b ^{\frac{y}{x}} \cdot b \cdot b ^{\frac{y}{x}}=1$
$\Rightarrow b^{\frac{y}{x}+1+\frac{y}{x}}=b^{\circ}......($Using $\left.a^{\circ}=1\right)$
$\Rightarrow \frac{y}{x}+1+\frac{y}{z}=0$
Divide throughout by $y$.
$\Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$
Hence proved.

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Question 205 Marks

Find the value of $(8p)^p$ if $9^{p + 2} - 9^p = 240.$

Answer

$9 p+2-9 p=240$
$\Rightarrow 9^p\left(9^2-1\right)=240$
$\Rightarrow 9^p(80)=240$
$\Rightarrow 9^p=3$
$\Rightarrow 3^{2 p}=3$
$\Rightarrow 2 p=1$
$\Rightarrow p=\frac{1}{2}$
$(8 p)^p=\left(2^3 p\right)^p$
$=\left(2^3 \cdot \frac{1}{2}\right)^{\frac{1}{2}}$
$=\left(2^{3-1}\right)^{\frac{1}{2}}$
$=\left(2^2\right)^{\frac{1}{2}}$
$=2 .$

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Question 215 Marks

Show that: $\frac{1}{1+ a ^{ p - q }}+\frac{1}{1+ a ^{ q - p }}$

Answer

$=\frac{1}{1+a^{p-q}}+\frac{1}{1+a^{q-p}}$
$=\frac{1+a^{q-p}+1+a^{p-q}}{\left(1+a^{p-q}\right)\left(1+a^{q-p}\right)}$
$=\frac{2+a^{-(p-q)}+a^{p-q}}{\left(1+a^{p-q}\right)\left(1+a^{-(q-p)}\right)}$
$=\frac{2+a^{-(p-q)}+a^{p-q}}{1+a^{-(p-q)}+a^{p-q}+a^{p-q} \cdot a^{-(p-q)}}$
$=\frac{2+a^{-(p-q)}+a^{p-q}}{1+a^{-(p-q)}+a^{p-q}+a^{p-q-p+q}}$
$=\frac{2+a^{-(p-q)}+a^{p-q}}{1+a^{-(p-q)}+a^{p-q}+a^0}$
$=\frac{2+a^{-(p-q)}+a^{p-q}}{1+a^{-(p-q)}+a^{p-q}+1}$
$=\frac{2+a^{-(p-q)}+a^{p-q}}{2+a^{-(p-q)}+a^{p-q}}$
$=1 $
$=\text{R.H.S.}$
Hence proved.

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[5 marks sum] - MATHEMATICS STD 9 Questions - Vidyadip