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II. y3 – \(\frac{{\left( {3\;\times \;8} \right)\frac{9}{2}}}{{y\sqrt y }}\) = 01). if x > y2). if x ≥ y3). if x < y4). if x ≤ y |
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Answer» I. $(\frac{{17}}{{\sqrt X }} = \sqrt x - \frac{7}{{\sqrt x }} = 0)$ ⇒ $(\frac{{17}}{{\sqrt x }})$ + $(\frac{7}{{\sqrt x }} = \sqrt x)$ ⇒ $(\frac{{17\; + \;7}}{{\sqrt x }})$ = √x ⇒ 24 = x II. y3 – $(\frac{{\left( {3\; \times \;8} \right)\frac{9}{2}}}{{y\sqrt y }})$ = 0 ⇒ y9/2 = 249/2 ⇒ y = 24 When x = 24, x = y for y = 24 ∴ x = y |
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