InterviewSolution
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1). a < b2). a > b3). a ≤ b4). a ≥ b |
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Answer» $(\begin{array}{l}{\RM{I}}.{\rm{\;}}\frac{{286{a^2}}}{{15}} - 30a =- {\rm{\;}}\frac{{14{a^2}}}{{15}} - 18 + 9A\\ \Rightarrow \frac{{286{a^2}}}{{15}} + \frac{{14{a^2}}}{{15}} - 30a - 9a + 18 = 0\end{array})$ ⇒ 20a2 - 39a + 18 = 0 ⇒ 4A(5a - 6) - 3(5a - 6) = 0 ⇒ (4a - 3)(5a - 6)=0 Then, a = +3/4 = + 0.75 or a = +6/5 = +1.2 $(\begin{array}{l}{\rm{II}}.{b^2} - {\rm{\;}}\frac{{158b}}{{63}} =- {\rm{\;}}\frac{{11}}{7}\\ \Rightarrow {b^2} - {\rm{\;}}\frac{{158b}}{{63}} =- {\rm{\;}}\frac{{99}}{{63}}\end{array})$ ⇒ 63b2 - 158b + 99 = 0 ⇒ 63b2 - 81b - 77b + 99 = 0 ⇒ 9b(7b - 9) - 11(7b - 9) = 0 ⇒ (9b - 11)(7b - 9) = 0 Then, b = + 11/9 = +1.222 or b = +9/7 = +1.286 So, when a = +0.75, a < b for b = +1.222 and a < b for b = +1.286 And when a = +1.2, a < b for b = +1.222 and a < b for b = +1.286 ∴ So, we can observe that a < b. |
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