CAMANISHPARMAR wrote:
Which of the following are pairs of numbers that are reciprocals of each other?
(I) \(\frac{-5}{12}\) and \(\frac{12}{5}\)
(II) \(\frac{\sqrt{7}}{2}\) and \(\frac{2\sqrt{7}}{7}\)
(III) \(\sqrt{17}\) - 4 and \(\sqrt{17}\) + 4
(A) II only
(B) III only
(C) I and II
(D) I and III
(E) II and III
One number is the reciprocal of another if their product equals 1
(I) \(\frac{-5}{12}\) and \(\frac{12}{5}\)
\((\frac{-5}{12}*\frac{12}{5})=\frac{-60}{60}=-1\)
NO. The product is \(-1\)
(II) \(\frac{\sqrt{7}}{2}\) and \(\frac{2\sqrt{7}}{7}\)
\((\frac{\sqrt{7}}{2}* \frac{2\sqrt{7}}{7})=\frac{2*\sqrt{7}*\sqrt{7}}{2*7}=\frac{2*7}{2*7}=1\)
That works.
(III) \(\sqrt{17}-4\) and \(\sqrt{17}+4\)
• Product = 1?
Difference of squares:\((a+b)(a-b)=a^2-b^2\)
\((\sqrt{17}-4)(\sqrt{17}+4)=(17-16)=1\)
• OR: \((n * \frac{1}{n})=1\)?
\((\sqrt{17}+4)*\frac{1}{(\sqrt{17}+4)}=\)
Rationalize the denominator of the second term
\((\sqrt{17}+4)*(\frac{1}{(\sqrt{17}+4)}*\frac{\sqrt{17}-4}{\sqrt{17}-4})=\)
\((\sqrt{17}+4)*(\frac{\sqrt{17}-4}{17-16})=\)
\(\frac{(\sqrt{17}-4)(\sqrt{17}+4)}{1}=(\sqrt{17}-4)(\sqrt{17}+4)=(17-16)=1\)
That works.
Roman numerals II and III are reciprocal pairs.
Answer E
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