Tag: maths

Questions Related to maths

If ${x}^{2}+\cfrac{1}{{x}^{2}}=102$, then $x-\cfrac{1}{x}$ is equal to

  1. $8$

  2. $10$

  3. $12$

  4. $13$


Correct Option: B
Explanation:
Given,

$\Rightarrow x^2+\dfrac{1}{x^2}=102$

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

$\Rightarrow \left ( x-\dfrac{1}{x} \right )^2=102-2=100$

$\Rightarrow x-\dfrac{1}{x} =10$

If ${x}^{3}+\cfrac{1}{{x}^{3}}=110$, then $x+\cfrac{1}{x}$ is equal to

  1. $5$

  2. $10$

  3. $15$

  4. None of these


Correct Option: A
Explanation:

$\left(x+\dfrac{1}{x}\right)^3=\left(x^3+\dfrac{1}{x^3}\right)+3\left(x+\dfrac{1}{x}\right)$


$\Rightarrow$  $\left(x+\dfrac{1}{x}\right)^3=110+3\left(x+\dfrac{1}{x}\right)$

Let $y=x+\dfrac{1}{x}$

$\Rightarrow$  $y^3=110+3y$

$\Rightarrow$  $y^3-3y-110=0$

$\Rightarrow$  $y^3-5y^2+5y^2-3y-110=0$

$\Rightarrow$  $y^2(y-5)+5y^2-25y+22y-110=0$

$\Rightarrow$  $y^2(y-5)+5y(y-5)+22(y-5)=0$

$\Rightarrow$  $(y-5)(y^2+5y+22)=0$

$\Rightarrow$  $y=5$ or $y=\dfrac{-5\pm3\sqrt{7}i}{2}$

$\therefore$  $x+\dfrac{1}{x}=5$

If $x+\cfrac{1}{x}=4$, then ${x}^{4}+\cfrac{1}{{x}^{4}}$ is equal to

  1. $196$

  2. $194$

  3. $192$

  4. $190$


Correct Option: B
Explanation:
Given,

$x+\dfrac{1}{x}=4$

squaring on both sides, we get,

$x^2+\dfrac{1}{x^2}+2=16$

$x^2+\dfrac{1}{x^2}=16-2=14$

squaring on both sides, we get,

$x^4+\dfrac{1}{x^4}+2=196$

$x^4+\dfrac{1}{x^4}=196-2=194$

Evaluate $\displaystyle \left ( \frac{2x}{7} - \frac{7y}{4} \right )^{2}$

  1. $\displaystyle \frac{x^{2}}{49} + \frac{17y^{2}}{16} - xy$

  2. $\displaystyle \frac{4x^{2}}{49} + \frac{49y^{2}}{16} - xy$

  3. $\displaystyle \frac{4x^{2}}{9} + \frac{49y^{2}}{4} - xy$

  4. $\displaystyle \frac{x^{2}}{13} + \frac{49y^{2}}{13} - xy$


Correct Option: B
Explanation:

$\left ( \dfrac{2x}{7} - \dfrac{7y}{4} \right ) ^{2}$


Using,

$(a-b)^2=a^2-2ab+b^2$

$=\left( \dfrac{2x}{7}\right)^2-2\left( \dfrac{2x}{7}\right)\left(\dfrac{7y}{4} \right)+\left(\dfrac{7y}{4} \right)^2$

$=\dfrac{4x^2}{49}-xy+\dfrac{49y^2}{16}$

Evaluate $\displaystyle \left ( \frac{7}{8}x + \frac{4}{5}y \right ) ^{2}$

  1. $\displaystyle \frac{49}{64}x^{2} + \frac{16}{25}y^{3} + \frac{7}{5}xy$

  2. $\displaystyle \frac{78}{32}x^{2} + \frac{16}{25}y^{2} + \frac{1}{5}xy$

  3. $\displaystyle \frac{49}{64}x^{2} + \frac{16}{25}y^{2} + \frac{7}{5}xy$

  4. $\displaystyle \frac{78}{32}x^{2} + \frac{16}{25}y^{2} + \frac{1}{6}xy$


Correct Option: C
Explanation:

$\left ( \frac{7}{8}x + \frac{4}{5}y \right ) ^{2}$
Using,
$(a+b)^2=a^2+2ab+b^2$
$=( \frac{7}{8}x)^2+2( \frac{7}{8}x)(\frac{4}{5}y )+(\frac{4}{5}y )^2$
$=\frac{49}{64}x^2+\frac{7}{5}xy+\frac{16}{25}y^2$

Find square of the following expression

$\displaystyle 3a - 4b$

  1. $\displaystyle a^{2} - 24ab + 48b^{2}$

  2. $\displaystyle 9a^{2} - 4ab + 48b^{2}$

  3. $\displaystyle 9a^{2} - 24ab + 16b^{2}$

  4. $\displaystyle a^{2} - 24ab + 16b^{2}$


Correct Option: C
Explanation:

Squaring,
$(3a - 4b)^2$
Using, $(a-b)^2=a^2-2ab+b^2$
$=(3a)^2-2(3a)(4b)+(4b)^2$
$=9a^2-24ab+16b^2$

Find square of the following expression

$\displaystyle \frac{3a}{2b} - \dfrac{2b}{3a}$

  1. $\displaystyle \dfrac{9a^{2}}{4b^{2}} - 2 + \dfrac{b^{2}}{a^{2}}$

  2. $\displaystyle \dfrac{9a^{2}}{4b^{2}} + 2 + \dfrac{4b^{2}}{9a^{2}}$

  3. $\displaystyle \dfrac{a^{2}}{4b^{2}} - 2 + \dfrac{4b^{2}}{9a^{2}}$

  4. $\displaystyle \dfrac{9a^{2}}{4b^{2}} - 2 + \dfrac{4b^{2}}{9a^{2}}$


Correct Option: D
Explanation:

Squaring,
$(\dfrac{3a}{2b} - \dfrac{2b}{3a})^2$
Using, $(a-b)^2=a^2-2ab+b^2$
$=(\dfrac{3a}{2b})^2-2(\dfrac{3a}{2b})(\dfrac{2b}{3a})+(\dfrac{2b}{3a})^2$
$=\dfrac{9a^2}{4b^2}-2+\dfrac{4b^2}{9a^2}$

Find the square of  $\displaystyle a + 2b + c$

  1. $\displaystyle a^{2} + b^{2} + c^{2} + ab + bc + ac$

  2. $\displaystyle a^{2} + 4b^{2} + c^{2} + 4ab + 4bc + 2ac$

  3. $\displaystyle a^{3} + 4b^{3} + c^{3} + 8ab + 8bc + 8ac$

  4. $\displaystyle a^{3} + b^{3} + c^{3} + 4ab + 4bc + 2ac$


Correct Option: B
Explanation:

Squaring,
$(a + 2b + c)^2$
Using $(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ac$
$=a^2+(2b)^2+c^2+2(a)(2b)+2(2b)(c)+2(a)(c)$
$=a^2+4b^2+c^2+4ab+4bc+2ac$

Find the square of $\displaystyle 2a - b - 3c$

  1. $\displaystyle 4a^{2} + b^{2} + 9c^{2} - 4ab + 6bc - 12ca$

  2. $\displaystyle a^{3} + b^{3} + c^{2} - 4ab + 6bc - ca$

  3. $\displaystyle a^{2} + b^{2} - c^{2} - 4ab + 6bc - 2ca$

  4. $\displaystyle 4a^{3} + b^{3} + 9c^{2} - 4ab - 6bc - 12ca$


Correct Option: A
Explanation:

   $2a-b-3c$
Squaring, we get
   $(2a-b-3c)^2$
Using $(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ac$
$=(2a)^2+(-b)^2+(-3c)^2+2(2a)(-b)+2(-b)(-3c)+2(2a)(-3c)$
$=4a^{2} + b^{2} + 9c^{2} - 4ab + 6bc - 12ca$

Evaluate :

$\displaystyle \left ( \frac{2x}{5} + \frac{3y}{4} - \frac{4z}{7} \right )^{2}$

  1. $\displaystyle \frac{x^{2}}{25} + \frac{9y^{2}}{16} + \frac{z^{2}}{49} + \frac{3}{5} xy - \frac{6}{7}yz - \frac{16}{35}zx$

  2. $\displaystyle \frac{4x^{2}}{25} + \frac{9y^{2}}{16} + \frac{16z^{2}}{49} + \frac{1}{5} xy - \frac{6}{3}yz - \frac{36}{35}zx$

  3. $\displaystyle \frac{x^{2}}{25} + \frac{9y^{2}}{16} + \frac{5z^{2}}{49} + \frac{3}{5} xy - \frac{6}{7}yz - \frac{16}{35}zx$

  4. $\displaystyle \frac{4x^{2}}{25} + \frac{9y^{2}}{16} + \frac{16z^{2}}{49} + \frac{3}{5} xy - \frac{6}{7}yz - \frac{16}{35}zx$


Correct Option: D
Explanation:

    $ \left ( \dfrac{2x}{5} + \dfrac{3y}{4} - \dfrac{4z}{7} \right )^{2}$
Squaring, we get
Using $(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ac$
$=( \dfrac{2x}{5})^2+(\dfrac{3y}{4})^2+(- \dfrac{4z}{7})^2+2( \dfrac{2x}{5})(\dfrac{3y}{4})+2(\dfrac{3y}{4})(- \dfrac{4z}{7})+2( \dfrac{2x}{5})(- \dfrac{4z}{7})$
$=\dfrac{4x^{2}}{25} + \dfrac{9y^{2}}{16} + \dfrac{16z^{2}}{49} + \dfrac{3}{5} xy - \dfrac{6}{7}yz - \dfrac{16}{35}zx$