x=a−−√+1a−−√,x=a+1a,x = \sqrt a + \frac{1}{{\sqrt a }}{\text{,}} y=a−1a," role="presentation">y=a−−√−1a−−√,y=a−1a,y = \sqrt a - \frac{1}{{\sqrt a }}{\text{,}} (a>0)" role="presentation">(a>0)(a>0)\left( {a > 0} \right) then the value of x4 + y4 - 2x2y2 is?" /> x=a−−√+1a−−√,x=a+1a,x = \sqrt a + \frac{1}{{\sqrt a }}{\text{,}} y=a−1a," role="presentation">y=a−−√−1a−−√,y=a−1a,y = \sqrt a - \frac{1}{{\sqrt a }}{\text{,}} (a>0)" role="presentation">(a>0)(a>0)\left( {a > 0} \right) then the value of x4 + y4 - 2x2y2 is?" /> x=a−−√+1a−−√,x=a+1a,x = \sqrt a + \frac{1}{{\sqrt a }}{\text{,}} y=a−1a," role="presentation">y=a−−√−1a−−√,y=a−1a,y = \sqrt a - \frac{1}{{\sqrt a }}{\text{,}} (a>0)" role="presentation">(a>0)(a>0)\left( {a > 0} \right) then the value of x4 + y4 - 2x2y2 is?" />

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If $x = \sqrt{a} + \frac{1}{\sqrt{a}},$ $x = \sqrt a + \frac{1}{{\sqrt a }}{\text{,}}$ $y = \sqrt{a} - \frac{1}{\sqrt{a}},$ $y = \sqrt a - \frac{1}{{\sqrt a }}{\text{,}}$ $(a > 0)$ $\left( {a > 0} \right)$ then the value of x⁴ + y⁴ - 2x²y² is?

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} x = \sqrt{a} + \frac{1}{\sqrt{a}} \\ y = \sqrt{a} - \frac{1}{\sqrt{a}} \\ Put a = 4 \\ x = 2 + \frac{1}{2} = \frac{5}{2} \\ y = 2 - \frac{1}{2} = \frac{3}{2} \\ Then, x^{4} + y^{4} - 2 x^{2} y^{2} \\ = {(x^{2} - y^{2})}^{2} \\ = {(\frac{25}{4} - \frac{9}{4})}^{2} \\ = 16 \end{aligned}

$\eqalign{ & {\text{ }}x = \sqrt a + \frac{1}{{\sqrt a }} \cr & {\text{ }}y = \sqrt a - \frac{1}{{\sqrt a }} \cr & {\text{Put }}a = 4 \cr & x = 2 + \frac{1}{2} = \frac{5}{2} \cr & y = 2 - \frac{1}{2} = \frac{3}{2} \cr & {\text{Then, }}{x^4} + {y^4} - 2{x^2}{y^2}{\text{ }} \cr & = {\left( {{x^2} - {y^2}} \right)^2} \cr & = {\left( {\frac{{25}}{4} - \frac{9}{4}} \right)^2} \cr & = {\text{ 16}} \cr}$

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y = \sqrt{a} - \frac{1}{\sqrt{a}},

$y = \sqrt a - \frac{1}{{\sqrt a }}{\text{,}}$

(a > 0)

$\left( {a > 0} \right)$ then the value of x⁴ + y⁴ - 2x²y² is?", "text": "If

x = \sqrt{a} + \frac{1}{\sqrt{a}},

$x = \sqrt a + \frac{1}{{\sqrt a }}{\text{,}}$

y = \sqrt{a} - \frac{1}{\sqrt{a}},

$y = \sqrt a - \frac{1}{{\sqrt a }}{\text{,}}$

(a > 0)

$\left( {a > 0} \right)$ then the value of x⁴ + y⁴ - 2x²y² is?", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" }, "answerCount": "4", "acceptedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} x = \sqrt{a} + \frac{1}{\sqrt{a}} \\ y = \sqrt{a} - \frac{1}{\sqrt{a}} \\ Put a = 4 \\ x = 2 + \frac{1}{2} = \frac{5}{2} \\ y = 2 - \frac{1}{2} = \frac{3}{2} \\ Then, x^{4} + y^{4} - 2 x^{2} y^{2} \\ = {(x^{2} - y^{2})}^{2} \\ = {(\frac{25}{4} - \frac{9}{4})}^{2} \\ = 16 \end{aligned}

$\eqalign{ & {\text{ }}x = \sqrt a + \frac{1}{{\sqrt a }} \cr & {\text{ }}y = \sqrt a - \frac{1}{{\sqrt a }} \cr & {\text{Put }}a = 4 \cr & x = 2 + \frac{1}{2} = \frac{5}{2} \cr & y = 2 - \frac{1}{2} = \frac{3}{2} \cr & {\text{Then, }}{x^4} + {y^4} - 2{x^2}{y^2}{\text{ }} \cr & = {\left( {{x^2} - {y^2}} \right)^2} \cr & = {\left( {\frac{{25}}{4} - \frac{9}{4}} \right)^2} \cr & = {\text{ 16}} \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} x = \sqrt{a} + \frac{1}{\sqrt{a}} \\ y = \sqrt{a} - \frac{1}{\sqrt{a}} \\ Put a = 4 \\ x = 2 + \frac{1}{2} = \frac{5}{2} \\ y = 2 - \frac{1}{2} = \frac{3}{2} \\ Then, x^{4} + y^{4} - 2 x^{2} y^{2} \\ = {(x^{2} - y^{2})}^{2} \\ = {(\frac{25}{4} - \frac{9}{4})}^{2} \\ = 16 \end{aligned}

$\eqalign{ & {\text{ }}x = \sqrt a + \frac{1}{{\sqrt a }} \cr & {\text{ }}y = \sqrt a - \frac{1}{{\sqrt a }} \cr & {\text{Put }}a = 4 \cr & x = 2 + \frac{1}{2} = \frac{5}{2} \cr & y = 2 - \frac{1}{2} = \frac{3}{2} \cr & {\text{Then, }}{x^4} + {y^4} - 2{x^2}{y^2}{\text{ }} \cr & = {\left( {{x^2} - {y^2}} \right)^2} \cr & = {\left( {\frac{{25}}{4} - \frac{9}{4}} \right)^2} \cr & = {\text{ 16}} \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }

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