4 Marks Questions

Question 14 Marks

Simplify : $\left[1-\left\{1-\left(1-x^2\right)^{-1}\right\}^{-1}\right]^{-1 / 2}$

Answer

$\left[1-\left\{1-\left(1-x^2\right)^{-1}\right\}^{-1}\right]^{-1 / 2}$
$=\left[1-\left\{1-\left(\frac{1}{1-x^2}\right)\right\}^{-1}\right]^{-1 / 2} \quad\left[\right.$ Using $\left.a^{-n}=\frac{1}{a^n}\right]$
$=\left[1-\left\{\frac{1-x^2-1}{1-x^2}\right\}^{-1}\right]^{-1 / 2}$
$=\left[1-\left[\frac{-x^2}{1-x^2}\right]^{-1}\right]^{-1 / 2}$
$=\left[1+\frac{1-x^2}{x^2}\right]^{-1 / 2}$
$=\left[\frac{x^2+\left(1-x^2\right)}{x^2}\right]^{-1 / 2}$
$=\left[\frac{1}{x^2}\right]^{-1 / 2}$
$=\left(x^{-2}\right)^{-1 / 2}$
$\begin{array}{l}=x^{-2 \times(-1 / 2)}=x \\ \quad\left[ U \operatorname{sing}\left(a^m\right)^n=a^{n t r}\right]\end{array}$

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Question 24 Marks

Suppose that certain amount $P$ is invested at an annual rate of $6.5 \%$, compounded annually. How long will it take for the amount to triple?
[use, $\log _e 3=1.0986$ ]

Answer

We use the formula,
$
\begin{array}{c}
A=Pe^{r t} \\
\text { Given, } A=3 P, r=6.5 \%=\frac{6.5}{100}=0.065
\end{array}
$
$\begin{aligned} \text { Therefore, } & 3 P =P e^{0.065 t} \\ \Rightarrow 3 =e^{0.065 t}\end{aligned}$
Taking log both sides, we get
$
\begin{array}{ll}
\\
\Rightarrow \quad & \log _e 3=\log _e e^{0.065 t} \\
\Rightarrow \quad & \log _e 3=0.065 t \log _e e \\ \Rightarrow \quad & \log _e 3 =0.065 t \quad[\because \log e=1]
\end{array}
$
$\begin{array}{ll}\Rightarrow & t=\frac{\log _e 3}{0.065} \\ \Rightarrow & t=\frac{1.0986}{0.065}=16.9015 \cong 16.9\end{array}$
Therefore, it would take 16.9 years for any initial investment $P$ to triple. Under the given condition.

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Question 34 Marks

Oskie-946 has a decay rate of $13.5 \%$. If the original sample was 50 gms , how long will it take for only 10 gms of the sample to remain ?
[Given, $\log _e 0.2=-1.60943$ ]

Answer

We use the formula of exponential decay i.e.,
$\begin{aligned}
m(t) & =m_0 e^{-r t} \\
Given,\ r & =13.5 \%=\frac{13.5}{100}=0.135 \\
m(t) & =10 \text { and } m_0=50
\end{aligned}
$
Therefore,
$10=50 e^{-0.135 t}$
$\Rightarrow \quad 0.2=e^{-0.135 t}$
Taking log both sides, we get
$\log (0.2)=\log \left(e^{-0.135 t}\right)$
$\begin{array}{rlrl}\Rightarrow & & \log (0.2) & =-0.135 t \log e \\ \Rightarrow & & \log (0.2) & =-0.135 t \\ \Rightarrow & & t & =-\frac{\log (0.2)}{0.135} \\ & && =-\frac{(-1.60943)}{0.135} \\ && & =11.9217\end{array}$
Hence, sample will take approximately 12 years, for remaining 10 gms .

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Question 44 Marks

If $a=\log _{24} 12, b=\log _{36} 24$ and $c=\log _{48} 36$, then prove that $1+a b c=2 b c$.

Answer

$
\begin{aligned}
\text { L.H.S. }= & 1+a b c \\
= & 1+\log _{24} 12 \times \log _{36} 24 \times \log _{48} 36 \\
= & 1+\frac{\log _{10} 12}{\log _{10} 24} \times \frac{\log _{10} 24}{\log _{10} 36} \times \frac{\log _{10} 36}{\log _{10} 48} \quad 1 \\
& \quad\left[\text { Using } \log _b x=\frac{\log _a x}{\log _a b}\right]
\end{aligned}
$
$\begin{array}{l}=1+\frac{\log _{10} 12}{\log _{10} 48} \\ =\frac{\log _{10} 48+\log _{10} 12}{\log _{10} 48} \\ =\frac{\log _{10}(48 \times 12)}{\log _{10} 48}\end{array}$
$\begin{array}{l}{\left[\text { Applying rule } \log _a(m n)=\log _a m+\log _a n\right]} \\ =\frac{\log _{10} 576}{\log _{10} 48}\end{array}$
$
\begin{array}{l}
=\frac{\log _{10}(24)^2}{\log _{10} 48} \\
=\frac{2 \log _{10} 24}{\log _{10} 48}
\end{array}
$
[Applying rule $\log _a m^n=n \log _a m$ ]
$\begin{array}{l}=2 \times\left(\frac{\log _{10} 24}{\log _{10} 36}\right) \times\left(\frac{\log _{10} 36}{\log _{10} 48}\right) \\ =2 \times\left(\log _{36} 24\right) \times\left(\log _{48} 36\right) \\ =2 \times b \times c \\ =2 b c=\text { R.H.S. } \quad \text { Hence Proved. }\end{array}$

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Question 54 Marks

Prove that : $\frac{\log _3 8}{\log _9 16 \cdot \log _4 10}=3 \log _{10} 2$

Answer

Change all the logarithms on L.H.S. to the base 10 by using the formula.
$\log _b x=\frac{\log _a x}{\log _a b}$
We may write,
$
\begin{array}{l}
\log _3 8=\frac{\log _{10} 8}{\log _{10} 3}=\frac{\log _{10} 2^3}{\log _{10} 3}=\frac{3 \log _{10} 2}{\log _{10} 3} \\
\log _9 16=\frac{\log _{10} 16}{\log _{10} 9}=\frac{\log _{10} 2^4}{\log _{10} 3^2}=\frac{4 \log _{10} 2}{2 \log _{10} 3} \\
\log _4 10=\frac{\log _{10} 10}{\log _{10} 4}=\frac{1}{\log _{10} 2^2}=\frac{1}{2 \log _{10} 2}
\end{array}
$
[Since, $\log _a m^n=n \log _a m$ and $\left.\log _{10} 10=1\right]$
Now,
$
\begin{aligned}
\text { L.H.S. } & =\frac{3 \log _{10} 2}{\log _{10} 3} \times \frac{2 \log _{10} 3}{4 \log _{10} 2} \times \frac{2 \log _{10} 2}{1} \\
& =\frac{3 \times 2 \times 2}{4} \cdot \frac{\log _{10} 2 \times \log _{10} 3 \times \log _{10} 2}{\log _{10} 3 \times \log _{10} 2} \\
& =3 \log _{10} 2=\text { R.H.S. } \quad \text { Hence Proved. }
\end{aligned}
$

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Question 64 Marks

(i) $(0.04)^{-1.5}=$ ?
(ii) $(25)^{7.5} \times(5)^{2.5} \div(125)^{1.5}=5^?$
(iii) If $\left(\frac{a}{b}\right)^{x-1}=\left(\frac{b}{a}\right)^{x-3}$, then find the value of $x$.

Answer

(i)
$
\begin{aligned}
(0.04)^{-1.5} & =\left(\frac{4}{100}\right)^{-1.5} \\
& =\left(\frac{1}{25}\right)^{-1.5}
\end{aligned}
$
$\begin{array}{l}=\left(\frac{1}{25}\right)^{-3 / 2} \\ =(25)^{3 / 2} \\ =\left(5^2\right)^{3 / 2}=5^3=125 \\ \quad\left[ Using\left(a^m\right)^n=a^{m n}\right]\end{array}$

$\begin{array}{l}\text { (ii) Let }(25)^{7.5} \times(5)^{2.5} \div(125)^{1.5}=5^x \\ \text { Then, }\left(5^2\right)^{7.5} \times(5)^{2.5} \div\left(5^3\right)^{1.5}=5^x \\ \left(5^{2 \times 7.5} \times 5^{2.5}\right) \div 5^{3 \times 1.5}=5^x \quad\left[\text { Using }\left(a^{m 1}\right)^n=a^{n m}\right]\end{array}$
$\Rightarrow \quad\left(5^{15} \times 5^{2.5}\right) \div 5^{4.5}=5^x$
$\Rightarrow \quad \frac{5^{15+2.5}}{5^{4.5}}=5^x \quad\left[\right.$ Using $\left.a^m \cdot a^n=a^{m+n}\right]$
$\Rightarrow \quad 5^{15+2.5-4.5}=5^x$
$\Rightarrow 5^x=5^{13}$
Since, same base has equal powers.
$
\therefore \quad x=13$

(iii) Given,$\left(\frac{a}{b}\right)^{x-1}=\left(\frac{b}{a}\right)^{x-3}$
$\Rightarrow \quad\left(\frac{a}{b}\right)^{x-1}=\left(\frac{a}{b}\right)^{-(x-3)}$
$\Rightarrow \quad\left(\frac{a}{b}\right)^{x-1}=\left(\frac{a}{b}\right)^{3-x}$
We know that, if base is equal, then index is also equal$
\begin{aligned}
\Rightarrow & & x-1 & =3-x \\
\Rightarrow & & 2 x & =4 \Rightarrow x=2 .
\end{aligned}
$

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Question 74 Marks

If $2^x=3^y=6^{-z}$, then find the value of $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$.

Answer

Let$
2^x=3^y=6^{-z}=k
$
$
\Rightarrow \quad 2^x=k \Rightarrow(k)^{1 / x}=2
$
Similarly,$3^y=k \Rightarrow(k)^{\frac{1}{y}}=3$
and$6^{-z}=k \Rightarrow(k)^{\frac{-1}{z}}=6$
Since,$6=2 \times 3$
Therefore,$
(k)^{\frac{-1}{z}}=(k)^{\frac{1}{x}} \times(k)^{\frac{1}{y}}
$
$
\begin{array}{l}
\Rightarrow \quad(k)^{-\frac{1}{z}}=(k)^{\frac{1}{x}+\frac{1}{y}} \\
{\left[\text { Using } a^m \cdot a^n=a^{m+n}\right] 1 }
\end{array}
$
We know that, if base is equal, then index is also equal.
$\begin{aligned} \therefore & \frac{-1}{z} =\frac{1}{x}+\frac{1}{y} \\ \Rightarrow & \frac{1}{x}+\frac{1}{y}+\frac{1}{z} & =0 .\end{aligned}$

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Question 84 Marks

Simplify :
\[\frac{1}{1+a^{m-n}+a^{m-p}}+\frac{1}{1+a^{n-m}+a^{n-p}}+\frac{1}{1+a^{p-m}+a^{p-n}} .\]

Answer

Let $T_1=\frac{1}{1+a^{m-n}+a^{m-p}}$
$\Rightarrow \quad T _1=\frac{1}{1+\frac{a^m}{a^n}+\frac{a^m}{a^p}}$
[Using $\left.\frac{a^m}{a^n}=a^{m-n}\right]$
$\begin{array}{l} \Rightarrow \quad T _1=\frac{a^{n+p}}{a^{n+p}+a^{m+p}+a^{m+n}} \\ {\left[\text { Using } a^m \cdot a^n=a^{m+n}\right] }\end{array}$
Similarly, let $T_2=\frac{1}{1+a^{n-m}+a^{n-p}}$
$\Rightarrow \quad T _2=\frac{1}{1+\frac{a^n}{a^m}+\frac{a^n}{a^p}}$
$\Rightarrow \quad T _2=\frac{a^{m+p}}{a^{m+p}+a^{m+n}+a^{n+p}}$
and let $T_3=\frac{1}{1+a^{p-m}+a^{p-n}}$
$\Rightarrow \quad T _3=\frac{1}{1+\frac{a^p}{a^m}+\frac{a^p}{a^n}}$
$\Rightarrow \quad T _3=\frac{a^{m+n}}{a^{m+n}+a^{m+p}+a^{n+p}}$
Now, $\quad T _1+ T _2+ T _3=\frac{1}{1+a^{m-n}+a^{m-p}}$$+\frac{1}{1+a^{n-m}+a^{n-p}}+\frac{1}{1+a^{p-m}+a^{p-n}}$
$=\frac{a^{n+p}}{a^{n+p}+a^{m+p}+a^{m+n}}+\frac{a^{m+p}}{a^{m+p}+a^{m+n}+a^{n+p}}$$+\frac{a^{m+n}}{a^{m+n}+a^{m+p}+a^{n+p}}$
$=\frac{a^{n+p}+a^{m+p}+a^{m+n}}{a^{n+p}+a^{m+p}+a^{m+n}}$
$=1$
Thus,
$\frac{1}{1+a^{m-n}+a^{m-p}}+\frac{1}{1+a^{n-m}+a^{n-p}}+\frac{1}{1+a^{p-m}+a^{p-n}}=1$

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Question 94 Marks

If $x=3^{\frac{1}{3}}+3^{-\frac{1}{3}}$, then find the value of $3 x^3-9 x$.

Answer

Given, $\quad x=3^{\frac{1}{3}}+3^{-\frac{1}{3}}$
On cubing both sides, we get
$
x^3=\left[3^{\frac{1}{3}}+3^{\frac{-1}{3}}\right]^3
$
$\Rightarrow \quad x^3=\left(3^{\frac{1}{3}}\right)^3+\left(3^{-\frac{1}{3}}\right)^3+3.3^{\frac{1}{3}} 3^{\frac{-2}{3}}+3.3^{\frac{2}{3}} \cdot 3^{\frac{-1}{3}}$
$\left[\right.$ Using $\left.(a+b)^3=a^3+b^3+3 a^2 b+3 a b^2\right]$
$\begin{array}{l} \Rightarrow \quad x^3=3^1+3^{-1}+3 \cdot 3^{\frac{-1}{3}}+3.3^{\frac{1}{3}} \\ {\left[\text { Using }\left(a^n\right)^n=a^{m n}\right] }\end{array}$
$\Rightarrow \quad x^3=3+\frac{1}{3}+3\left(3^{\frac{1}{3}}+3^{\frac{-1}{3}}\right)$
$\left[\right.$ Using $a^{-n}=\frac{1}{a^n}$ ]
$\Rightarrow \quad x^3=3+\frac{1}{3}+3 x \quad[$ From eq (i) $]$
Now, substituting the value of $x^3$ in $3 x^3-9 x$, we get
$
\begin{aligned}
3 x^3-9 x & =3\left(3+\frac{1}{3}+3 x\right)-9 x \\
& =9+1+9 x-9 x \\
& =10
\end{aligned}
$
Hence value of $3 x^3-9 x$ is 10 .

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