Least value of E = 25 ^4 x - 50 ^2 x + 74 ^2 x is - Vidyadip

MCQ

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

A
$50$
✓
$70$
C
$75$
D
$90$

✓

Answer

Correct option: B.

$70$

b
$E=\frac{25\left(\left(\sec ^{2} x\right)^{2}-2 \sec ^{2} x+1\right)-25+74}{\tan ^{2} x}$

$=\frac{25\left(\sec ^{2} x-1\right)^{2}}{\tan ^{2} x}+\frac{49}{\tan ^{2} x}$

$\Rightarrow 25 \tan ^{2} x+49 \cot ^{2} x$

$\therefore \quad E_{\min }=2 \times 5 \times 7=70$

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