Let the maximum value of expression y = x 4 - x 2 x 6 + 2 x 3 - 1 for x 1 is p q , where ' p ' and ' q ' are relatively prime natural numbers, then p + q =

Question

Let the maximum value of expression $y = \frac{x^{4} - x^{2}}{x^{6} + 2 x^{3} - 1}$ for $x > 1$ is $\frac{p}{q}$ , where $' p'$ and $' q'$ are relatively prime natural numbers, then $p + q =$

Smart Achievers · Accepted Answer

(7)

$y = \frac{x - \frac{1}{x}}{x^{3} - \frac{1}{x^{3}} + 2},$

$Let t = x - \frac{1}{x} > 0 for x > 1,$

$y = \frac{t}{t (t^{2} + 3) + 2},$

$x^{3} - \frac{1}{x^{3}} = t (t^{2} + 3)$

$= \frac{1}{t^{2} + \frac{2}{t} + 3}$ $(\begin{matrix} t^{2} + \frac{2}{t} = t^{2} + \frac{1}{t} + \frac{1}{t} \geq 3 \\ ∴ t^{2} + \frac{2}{t} + 3 \geq 6 (AM \geq GM) \end{matrix},$

$y_{\max} = \frac{1}{{(t^{2} + \frac{2}{t} + 3)}_{\min}} = \frac{1}{6},$ $p = 1, q = 6$

Solution

Q.
Let the maximum value of expression $y = \frac{x^{4} - x^{2}}{x^{6} + 2 x^{3} - 1}$ for $x > 1$ is $\frac{p}{q}$ , where $' p'$ and $' q'$ are relatively prime natural numbers, then $p + q =$

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Solution

Q. Let the maximum value of expression y=x4-x2x6+2x3-1 for x>1 is pq, where 'p' and 'q' are relatively prime natural numbers, then p+q=

Ans. (7) y=x-1xx3-1x3+2, Let t=x-1x>0 for x>1, y=tt(t2+3)+2, x3-1x3=t(t2+3) =1t2+2t+3 (t2+2t=t2+1t+1t≥3∴ t2+2t+3≥6 (AM≥GM), ymax=1(t2+2t+3)min=16, p=1, q=6

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Q.
Let the maximum value of expression $y = \frac{x^{4} - x^{2}}{x^{6} + 2 x^{3} - 1}$ for $x > 1$ is $\frac{p}{q}$ , where $' p'$ and $' q'$ are relatively prime natural numbers, then $p + q =$