The least value of E = frac{25 sec^4 x - 50 s | vedclass

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Question

(B) Given expression is $E = \frac{25 \sec^4 x - 50 \sec^2 x + 74}{	an^2 x}$.We know that $\sec^2 x = 1 + 	an^2 x$.Substituting this,we get $E = \fr

Vedclass · Accepted Answer

$70$

Vedclass · Answer

(B) Given expression is $E = \frac{25 \sec^4 x - 50 \sec^2 x + 74}{	an^2 x}$.We know that $\sec^2 x = 1 + 	an^2 x$.Substituting this,we get $E = \frac{25(1 + 	an^2 x)^2 - 50(1 + 	an^2 x) + 74}{	an^2 x}$.Let $t = 	an^2 x$. Then $E = \frac{25(1 + t)^2 - 50(1 + t) + 74}{t}$.$E = \frac{25(1 + 2t + t^2) - 50 - 50t + 74}{t} = \frac{25 + 50t + 25t^2 - 50 - 50t + 74}{t}$.$E = \frac{25t^2 + 49}{t} = 25t + \frac{49}{t}$.Using the Arithmetic Mean-Geometric Mean inequality $(AM \ge GM)$,we have $\frac{25t + \frac{49}{t}}{2} \ge \sqrt{25t \cdot \frac{49}{t}}$.$25t + \frac{49}{t} \ge 2 \sqrt{25 \cdot 49} = 2 \cdot 5 \cdot 7 = 70$.Thus,the minimum value is $70$.

The least value of $E = \frac{25 \sec^4 x - 50 \sec^2 x + 74}{\tan^2 x}$ is

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