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SSC CGL Tier II 25 Oct 2015 Paper 1

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Question : 2 of 100

Marks: +1, -0

Let

x=\frac{\sqrt{13}+\sqrt{11}}{\sqrt{13}-\sqrt{11}}

y=\frac{1}{x}

then the value of

3x^{2}-5xy+3y^{2}

is

1717
1177
1771
1171

Solution:

(1)

x=\frac{\sqrt{13}+\sqrt{11}}{\sqrt{13}-\sqrt{11}}

On rationalising the denominator

=\frac{\sqrt{13}+\sqrt{11}}{\sqrt{13}-\sqrt{11}}\times\frac{\sqrt{13}+\sqrt{11}}{\sqrt{13}+\sqrt{11}}

=\frac{(\sqrt{13}+\sqrt{11})^{2}}{(\sqrt{13})^{2}-(\sqrt{11})^{2}}

=\frac{13+11+2\sqrt{143}}{13-11}

=\frac{24+2\sqrt{143}}{2}=12+\sqrt{143}

∴

y=\frac{1}{x}=\frac{1}{12+\sqrt{143}}

=\frac{1}{12+\sqrt{143}}\times\frac{12-\sqrt{143}}{12-\sqrt{143}}

=\frac{12-\sqrt{143}}{144-143}=12-\sqrt{143}

\therefore x-y=12+\sqrt{143}-12+\sqrt{143}=2\sqrt{143}

and

xy=(12+\sqrt{143})(12-\sqrt{143})

=144-143=1

\therefore 3x^{2}-5xy+3y^{2}=3x^{2}-6xy+3y^{2}+xy

=3(x-y)^{2}+xy

=3(2\sqrt{143})^{2}+1

=3\times4\times143+1=1716+1

=1717

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