Tag: maths

Questions Related to maths

The value of $\left (\dfrac {32}{243}\right )^{-3/5}$ is _____.

maths indices negative indices law of indices laws of indices

  1. $\dfrac {27}{8}$

  2. $\dfrac {8}{27}$

  3. $\dfrac {16}{27}$

  4. $\dfrac {27}{16}$

Show answer
Correct Option: A
Explanation:
We need to find value of $\left (\dfrac {32}{243}\right )^{-3/5}$
It can be written as,
$\left(\dfrac {2^5}{3^5}\right)^{-3/5}$
$\Rightarrow \left (\dfrac {3}{2}\right)^{3} = \dfrac {27}{8}$

$\left {\left (\dfrac {1}{3}\right )^{-3} -\left (\dfrac {1}{2}\right )^{-3}\right } \div \left (\dfrac {1}{4}\right )^{-3} = ?$

maths indices negative indices law of indices laws of indices

  1. $\dfrac {19}{64}$

  2. $\dfrac {64}{19}$

  3. $\dfrac {27}{16}$

  4. $\dfrac {16}{27}$

Show answer
Correct Option: A
Explanation:
We need to find value of $\left \{\left (\dfrac {1}{3}\right )^{-3} -\left (\dfrac {1}{2}\right )^{-3}\right \} \div \left (\dfrac {1}{4}\right )^{-3} = ?$
$\left (\dfrac{1}{3}\right)^{-3}=3^{3}$
$\left (\dfrac{1}{2}\right)^{-3}=2^{3}$
$\left (\dfrac{1}{4}\right)^{-3}=4^{3}$
So, given expression becomes,
$\dfrac {3^{3}-2^{3}}{4^{3}}$ $=\dfrac {19}{64}$. 
Hence, A is the right option.

$\dfrac{1}{5}$ can also be expressed as:

maths indices negative indices law of indices laws of indices

  1. $5^{-1}$

  2. $\dfrac{1}{25}$

  3. $\dfrac{-1}{25}$

  4. None of these

Show answer
Correct Option: A
Explanation:

$\dfrac 15 = \dfrac {1}{5^1}=5^{-1}$

$-2^{-3}$ can also be expressed as:

maths indices negative indices law of indices laws of indices

  1. $\dfrac{1}{8}$

  2. $-\dfrac{1}{8}$

  3. $8$

  4. $-8$

Show answer
Correct Option: B
Explanation:

$-2^{-3} = \dfrac {1}{-2 \times -2 \times -2} = -\dfrac 18$


So, option B is correct.

The inverse of the function $f : R \rightarrow {x \in R : x < 1}$ given by $f(x)=\dfrac{e^{x}-e^{-x}}{e^{x}+e^{-x}},$ is 

maths indices negative indices law of indices laws of indices

  1. $\dfrac{1}{2} \log \dfrac{1+x}{1-x}$

  2. $\dfrac{1}{2} \log \dfrac{2+x}{2-x}$

  3. $\dfrac{1}{2} \log \dfrac{1-x}{1+x}$

  4. $none\ of\ these$

Show answer
Correct Option: A
Explanation:

Given $f(x)=\cfrac{e^{x}-e^{-x}}{e^{x}+e^{-x}}$

So $y=\cfrac{e^x-e^{-x}}{e^{x}+e^{-x}}$
$\implies y=\cfrac{e^{2 x}-1}{e^{2 x}+1}$
$\implies y(e^{2 x}+1)=e^{2 x}-1$
$\implies e^{2 x}(y-1)=-y-1$
$\implies e^{2 x}=\cfrac{1+y}{1-y}$
$\implies 2 x=\log \cfrac{1+y}{1-y}$
$\implies x=\cfrac{1}{2}\log \cfrac{1+y}{1-y}$
$f^{-1}(x)=\cfrac{1}{2}\log\cfrac{1+x}{1-x}$

If $\sqrt [ 3 ]{ a+\sqrt { b }  } =7+4\sqrt { 3 } $, then $\sqrt [ 3 ]{ { a }^{ 2 }-b } =$

maths indices negative indices law of indices laws of indices

  1. $0$

  2. $1$

  3. $-1$

  4. $7$

Show answer
Correct Option: B
Explanation:

We have.

$ \sqrt[3]{a+\sqrt{b}}=7+4\sqrt{3} $

$ {{(a+\sqrt{b})}^{1/3}}=(7+4\sqrt{3}) $

$ a+\sqrt{b}={{(7+4\sqrt{3})}^{3}}\ \ ......\ \ (1) $


$ \text{Similarly,} $

$ a-\sqrt{b}={{(7-4\sqrt{3})}^{3}}\ \ ......\ \ (2) $


On multiplying (1) and (2) to. We get,

$ (a+\sqrt{b})(a-\sqrt{b})={{(7+4\sqrt{3})}^{3}}{{(7-4\sqrt{3})}^{3}} $

$ {{a}^{2}}-b={{\left[ (7+4\sqrt{3})(7-4\sqrt{3}) \right]}^{3}} $

$ {{a}^{2}}-b={{\left[ 49-16\times 3 \right]}^{3}} $

$ {{({{a}^{2}}-b)}^{1/3}}=(1) $

$ \sqrt[3]{{{a}^{2}}-b}=1 $


Hence, this is the answer

$(x+y)^{-1}(x^{-1}+y^{-1})$ is the reciprocal of

maths indices negative indices law of indices laws of indices

  1. $x+y$

  2. $xy$

  3. $\displaystyle\frac{1}{xy}$

  4. $\displaystyle\frac{x}{y}$

Show answer
Correct Option: B
Explanation:

$\displaystyle(x+y)^{-1}(x^{-1}+y^{-1})=\frac{1}{x+y}\left(\frac{1}{x}+\frac{1}{y}\right)=\frac{1}{x+y}\times\frac{x+y}{xy}=\frac{1}{xy}$

The value of ${1-[1-(1-n)^{-1}]^{-1}}^{-1}$ is

maths indices negative indices law of indices laws of indices

  1. $0$

  2. $1$

  3. $n$

  4. $\displaystyle\frac{1}{n}$

Show answer
Correct Option: C
Explanation:

$\displaystyle{1-[1-(1-n)^{-1}]^{-1}}^{-1}=\left[1-\left(1-\frac{1}{1-n}\right)^{-1}\right]^{-1}$

$\displaystyle=\left[1-\left(\frac{1-n-1}{1-n}\right)^{-1}\right]^{-1}=\left[1+\left(\frac{n}{1-n}\right)^{-1}\right]^{-1}=\left(1+\frac{1-n}{n}\right)^{-1}$

$\displaystyle=\left(\frac{n+1-n}{n}\right)^{-1}=\left(\frac{1}{n}\right)^{-1}=n$

The value of ${ \left[ { \left( { 3 }^{ 2 } \right)  }^{ 2 } \right]  }^{ -1 }$ is-

maths indices negative indices law of indices laws of indices

  1. $81$

  2. $-81$

  3. $-0.0123$

  4. $0.0123$

Show answer
Correct Option: D
Explanation:
$\cfrac { 1 }{ { \left( { 3 }^{ 2 } \right)  }^{ 2 } } $        $\because (a^m)^n=a^{mn}$
$=\cfrac { 1 }{ { 3 }^{ 4 } }$
$ =\cfrac { 1 }{ 81 }$
$ =0.0123$

9798745995 is written with periods in International system is __________.

maths numbers in indian and international systems international place value system international system of numeration use of commas and comparison and number systems

  1. 97.98,74,59,95

  2. 9,79,87.45,995

  3. 9,798,745,995

  4. 979,874,599,5

Show answer
Correct Option: C
Explanation:

correct option is C..