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In the given question, two equations numbered l and II are given. Solve both the equations and mark the appropriate answerI. \({x^{3/2}} - \;\frac{{81}}{{\sqrt x }}\; = 0\)II. 20y2 – 119y + 176 = 01. x > y2. x < y3. x ≥ y4. x ≤ y5. x = y or the relationship between x and y cannot be established |
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Answer» Correct Answer - Option 5 : x = y or the relationship between x and y cannot be established I. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaWG4bWdamaaCaaaleqabaWdbiaaiodacaGGVaGaaGOmaaaakiab % gkHiTiaacckadaWcaaWdaeaapeGaaGioaiaaigdaa8aabaWdbmaaka % aapaqaa8qacaWG4baaleqaaaaakiaacckacqGH9aqpcaaIWaaaaa!4194! {x^{3/2}} - \;\frac{{81}}{{√ x }}\; = 0\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaWG4bWdamaaCaaaleqabaWdbiaaiodacaGGVaGaaGOmaaaakiab % gkHiTiaacckadaWcaaWdaeaapeGaaGioaiaaigdaa8aabaWdbmaaka % aapaqaa8qacaWG4baaleqaaaaakiaacckacqGH9aqpcaaIWaaaaa!4194! {x^{3/2}} - \;\frac{{81}}{{√ x }}\; = 0\)\({x^{3/2}} - \;\frac{{81}}{{√ x }}\; = 0\) ⇒ \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qadaWcaaWdaeaapeGaaiiOaiaadIhapaWaaWbaaSqabeaapeWaaSaa % a8aabaWdbiaaiodaa8aabaWdbiaaikdaaaaaaOGaaiiOaiabgEna0k % aacckadaGcaaWdaeaapeGaamiEaaWcbeaakiaacckacaGGGcGaeyOe % I0IaaGioaiaaigdacaGGGcaapaqaa8qacaGGGcGaeyOgIyTaamiEaa % aacqGH9aqpcaaIWaaaaa!4B95! \frac{{\;{x^{\frac{3}{2}}}\; \times \;√ x \;\; - 81\;}}{{\;\surd x}} = 0\)\(\frac{{\;{x^{\frac{3}{2}}}\; \times \;√ x \;\; - 81\;}}{{\;\surd x}} = 0\) ⇒ x3/2 × x1/2 – 81 = 0 ⇒ x2 = 81 ⇒ x = ± 9 II. 20y2 – 119y + 176 = 0 ⇒ 20y2 – 64y – 55y + 176 = 0 ⇒ 4y(5y – 16) – 11(5y – 16) = 0 ⇒ (5y – 16)(4y – 11) = 0 ⇒ \(y = \;\frac{{16}}{5}\;,\;\frac{{11}}{4}\)
∴ x = y or the relationship between x and y cannot be established We can not take the negative value of x as in the question there is √x is positive x2 is both positive and negative |
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