Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section A

Chapter

Applications of Derivatives

Question:

The minimum value of $64 \sec x + 27\, cosec\, x, 0 < x <$ is

Options:

91

25

125

None of these

Correct Answer:

125

Explanation:

Let $y = 64 \sec x + 27\, cosec\, x$

$= 64 \sec x tan x – 27\, cosec\, x \cot x$

$= 64 \sec^3 x + 64 \sec x \tan^2x + 27 cosec^3x + cosec\, x \cot^2x$

Now = 0 $64 \sec x \tan x = 27\, cosec\, x \cot x$

$\tan^3x = 27/64$

$\tan x = 3/4$

Also then > 0 $( 0 < x < 1/2)$

So y is minimum when $x = tan^{–1} (3/4)$ and its min. value = $64 ( 5/4) + 27 (5/3) = 125$