Tag: maths

Questions Related to maths

The sum of roots of the equation $(1.25)^{1-x^2} = (0.4096)^{1+x}$

  1. Infinite

  2. $1$

  3. $2$

  4. $4$


Correct Option: D
Explanation:
$\left ( 1.25 \right )^{1-x^{2}}=\left ( 0.4096 \right )^{1+x}$

$\left ( \dfrac{125}{100} \right )^{1-x^{2}}=\left ( \dfrac{4096}{10000} \right )^{1+x}$

$\left ( \dfrac{5}{4} \right )^{1-x^{2}}=\left ( \left ( \dfrac{8}{10} \right )^{4} \right )^{1+x}$

$\left ( \dfrac{5}{4} \right )^{1-x^{2}}=\left ( \dfrac{4}{5} \right )^{4+4x}$

$\left ( \dfrac{5}{4} \right )^{1-x^{2}}=\left ( \dfrac{5}{4} \right )^{-4-4x}$

$\Rightarrow 1-x^{2}=-4-4x$

$x^{2}-4x-5=0$

$x^{2}-5x+x-5=0$
$x(x-5)+1(x-5)=0$
$(x+1)(x-5)=0$
$x=-1,5$

Therefore, Sum of the roots of equation  is $4$

Find:$\dfrac{\sqrt[3]{108}\times \sqrt[6]{4}}{\sqrt[4]{81}}$

  1. $ 2$

  2. $\frac{\sqrt[6]{4}}{\sqrt[2]{3}}$

  3. $ \sqrt[4]{6}$

  4. $\frac{\sqrt[4]{2}}{\sqrt[2]{3}}$


Correct Option: A
Explanation:

$(\frac{\sqrt[3]{180}\times \sqrt[6]{4}}{\sqrt[4]{81}})\\ ((\frac{27\times4)^{(\frac{1}{4})}\times 2^{(\frac{2}{6})}}{(3^4)^{(\frac{1}{4})}}))\\ =(\frac{3\times 2^{(\frac{2}{3}) }\times 2^{(\frac{1}{3})}}{3})\\= 2^{{(\frac{2}{3})}+{(\frac{1}{3})}}\\=2$

If $(25) _{n}\times (31) _{n}=(1015) _{n}$ then the value of $(13) _{n}\times (25) _{n}$ is $n>0$ :

  1. $(626) _{n}$

  2. $(462) _{n}$

  3. $(716) _{n}$

  4. $(676) _{n}$


Correct Option: A

If $ p= {2} ^{ \tfrac {2} {3}} + {2} ^{ \tfrac {1} {3}} $,then 

  1. ${p}^{3}-6p+6=0 $

  2. ${p}^{3}-3p-6=0 $

  3. ${p}^{3}-6p-6=0 $

  4. ${p}^{3}-3p+6=0 $


Correct Option: C
Explanation:
Given,

$p=2^{\frac{2}{3}}+2^{\frac{1}{3}}$

cubing both sides

$p^3=(2^{\frac{2}{3}}+2^{\frac{1}{3}})^3$

$p^3=\left(2^{\frac{2}{3}}\right)^3+3\left(2^{\frac{2}{3}}\right)^2\cdot \:2^{\frac{1}{3}}+3\cdot \:2^{\frac{2}{3}}\left(2^{\frac{1}{3}}\right)^2+\left(2^{\frac{1}{3}}\right)^3$

$p^3=6+6\cdot \:2^{\frac{2}{3}}+6\cdot \:2^{\frac{1}{3}}\quad $

$p^3=6+6\left ( 2^{\frac{2}{3}}+2^{\frac{1}{3}} \right )$

$p^3=6+6p$

$\therefore p^3-6-6p=0$

$\dfrac{(625)^{6.25} \times (25)^{2.6}}{(625)^{6.75} \times (5)^{1.2}} = ?$

  1. $5$

  2. $10$

  3. $15$

  4. $25$


Correct Option: D
Explanation:
Given,

$\dfrac{625^{6.25}\cdot \:25^{2.6}}{625^{6.75}\cdot \:5^{1.2}}$

$=\dfrac{625^{\tfrac{25}{4}}\cdot \:25^{\tfrac{13}{5}}}{625^{\tfrac{27}{4}}\cdot \:5^{\tfrac{6}{5}}}$

$=\dfrac{(5^4)^{\tfrac{25}{4}}\cdot \:(5^2)^{\tfrac{13}{5}}}{(5^4)^{\tfrac{27}{4}}\cdot \:5^{\tfrac{6}{5}}}$

$=\dfrac{(5^{25})\cdot \:(5)^{\tfrac{26}{5}}}{(5)^{27}\cdot \:5^{\tfrac{6}{5}}}$

$=\dfrac{5^{25+\tfrac{26}{5}}}{5^{27+\tfrac{6}{5}}}$

$=5^{25+\tfrac{26}{5}-27-\tfrac{6}{5}}$

$=5^{-2+\tfrac{20}{5}}$

$=5^{-2+4}$

$=5^2=25$

The value of $\left(\dfrac{1}{64}\right)^{-5/6}$ will be

  1. $8$

  2. $16$

  3. $36$

  4. $32$


Correct Option: D
Explanation:

Given, 

$(\dfrac{1}{64})^{\tfrac{-5}{6}}$

$= {64}^{\tfrac{5}{6}}$


$= ({2^6})^{\frac{5}{6}}$

$= ({2^5})$

$= 32$

Find $x:[3+\left { 2+(1+x^{2}) \right }^{2}]^{2}=144$

  1. $1$

  2. $0$

  3. $5$

  4. $6$


Correct Option: B
Explanation:
Given,

$\left [ 3+\left\{2+\left(1+x^2\right)\right\}^2 \right ]^2=144$

taking square root on both sides, we get,

$\left [ 3+\left\{2+\left(1+x^2\right)\right\}^2 \right ]=12$

$\left\{2+\left(1+x^2\right)\right\}^2=12-3=9$

again taking square root on both sides, we get,

$2+(1+x^2)=3$

$1+x^2=3-2=1$

$x^2=1-1=0$

$\therefore x=0$

THe value of $\dfrac{8^3 + 6^3}{8^2 - 8 \times6 + 6^2}$ is

  1. $10$

  2. $14$

  3. $2$

  4. $15$


Correct Option: B
Explanation:
Given,

$\dfrac{8^3+6^3}{8^2-8 \times 6+6^2}$

$=\dfrac{512+216}{64-48+36}$

$=\dfrac{728}{52}$

$=14$

The product $(32)(32)^{1/6}(32)^{1/36}......$ to $\infty$ is 

  1. $16$

  2. $32$

  3. $64$

  4. $0$


Correct Option: A
Explanation:

$(32)(32)^{1/6}(32)^{1/36}......$ to $\infty = (32)^{1+\dfrac{1}{6}.....\infty}$

sum of infinite GP = a=1  ,  r=$\dfrac{1}{6}$
      $\dfrac{a}{1-r}=\dfrac{1}{1-6}=\dfrac{6}{5}$
$(32)^{\dfrac{6}{5}}=16$

Simplicity
$\left[ \left{ \left( 625 \right) ^{ -\dfrac { 1 }{ 2 } } \right} ^{ -\dfrac { 1 }{ 4 } } \right] $

  1. $\dfrac{1}{\sqrt5}$

  2. $\sqrt5$

  3. 5

  4. None of these


Correct Option: B
Explanation:

$\begin{array}{l}\left[ {{{\left( {{{\left( {625} \right)}^{\frac{{ - 1}}{2}}}} \right)}^{\frac{{ - 1}}{4}}}} \right] = \left[ {{{\left( {{{\left( {{{25}^2}} \right)}^{\frac{{ - 1}}{2}}}} \right)}^{\frac{{ - 1}}{4}}}} \right]\ = \left[ {{{\left( {{{25}^{ - 1}}} \right)}^{\frac{{ - 1}}{4}}}} \right]\ = {25^{\frac{1}{4}}}\ = {5^{2 \times \frac{1}{4}}}\ = {5^{\frac{1}{2}}}\ = \sqrt 5 \end{array}$