IIFT 2019 Question 2

Question 2

If $$x=8-\sqrt{32}$$ and $$y=2+\sqrt{2}$$, then $$\left(x+\frac{1}{y}\right)^2$$ is given by:


$$x=8-\sqrt{32}$$ and $$y=2+\sqrt{2}$$,

We have to find the value of $$\left(x+\frac{1}{y}\right)^2$$

$$\left(8-\sqrt{\ 32}+\ \frac{\ 1}{2+\sqrt{\ 2}}\right)^2$$

$$(\ \frac{\ 8\left(2+\sqrt{\ 2}\right)-\sqrt{\ 32}\left(2+\sqrt{\ 2}\right)+\ \ 1}{2+\sqrt{\ 2}})^2$$

$$\left(\ \frac{\ 9}{2+\sqrt{\ 2}}\right)^2$$

$$\ \frac{\ 81}{6+2\sqrt{\ 2}}$$

$$\left(\ \frac{\ 1}{y}\right)^2=\ \left(\frac{\ 1}{2+\sqrt{\ 2}}\right)^2$$ = $$6+2\sqrt{\ 2}$$

=$$\left(\ \frac{\ 2-\sqrt{\ 2}}{2}\right)^{^2}$$

=$$\ \frac{\ 6-4\sqrt{2\ }}{4}$$

=$$\ \frac{\ 3-2\sqrt{2\ }}{2}$$

$$x^2=64+32-64\sqrt{\ 2}$$

=$$96-64\sqrt{\ 2}$$

=32($$3-2\sqrt{\ 2}$$) = 32*$$2y^2$$

we get, $$x^{2\ }=\ 64y^2$$

$$\ \frac{\ 81}{y^2}=\ \frac{\ 81}{64}x^2$$

D is the correct answer.

Alternative solution,

$$xy=\left(8-\sqrt{\ 32}\right)\left(2+\sqrt{\ 2}\right)=4\sqrt{\ 2}\left(\sqrt{\ 2}-1\right)\times\ \sqrt{\ 2}\left(\sqrt{\ 2}+1\right)=8\left(2-1\right)=8$$

(As $$xy=8\ \longrightarrow\ y=\dfrac{x}{8}$$)

$$\left(x+\frac{1}{y}\right)^2=\left(\frac{\left(xy+1\right)}{y}\right)^2=\left(\frac{\left(xy+1\right)\times\ x}{8}\right)^2=\left(\frac{\left(8+1\right)\times\ x}{8}\right)^2=\frac{81}{64}\times\ x^2$$

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Prabhat Bhatt

2 years, 5 months ago

y=2+sqrt2=2/(2-sqrt2)=>1/y=(2-sqrt2)/2; =>y=x/8;
Hence proved;

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