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AP EAMCET Engineering 22 Apr 2019 Shift 1 Paper

Section: Mathematics

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Question : 75 of 160

Marks: +1, -0

If

A_n

\int\limits_{\pi/2}^{\infty} e^{-x} , \cos^n x

\frac{A_4 - A_6}{A_4}

=

$\frac{3}{2}$
$\frac{7}{37}$
$\frac{5}{37}$
$\frac{2}{7}$

Solution:

From the given expression,

A_{6} = \int\limits_{\frac{\pi}{2}}^{\infty} e^{-x} \cos^{6} x \, dx

= \left[ -e^{-x} \cos^{6} x \right]_{\frac{\pi}{2}}^{\infty} - \int\limits_{\frac{\pi}{2}}^{\infty} (-e^{-x})

6 \cos^{5} x (-\sin x) \, dx

= 0 - 6 \int\limits_{\frac{\pi}{2}}^{\infty} e^{-x} \cos^{5} x \sin x \, dx

This gives,

\frac{A_6}{-6} = \int\limits_{\frac{\pi}{2}}^{\infty} e^{-x} \cos^{5} x \sin x \, dx

-\frac{1}{6} A_6 = ( \left[ -e^{-x} \cos^{5} x \sin x \right]_{\frac{\pi}{2}}^{\infty}

- \int\limits_{\frac{\pi}{2}}^{\infty} (-e^{-x}) \left[ 5 \cos^{4} x (-\sin x) \sin x + \cos^{5} x \cdot \cos x \right] \, dx )

= 0 + \int\limits_{\frac{\pi}{2}}^{\infty} (e^{-x}) [ -5 \cos^{4} x (1-\cos^{2} x)

+ \cos^{6} x ] \, dx

= -5 \int\limits_{\frac{\pi}{2}}^{\infty} (e^{-x}) [\cos^{4} x] \, dx + 6 \int\limits_{\frac{\pi}{2}}^{\infty} (e^{-x}) [\cos^{6} x] \, dx

This implies,

-\frac{1}{6} A_6 = -5 A_4 + 6 A_6

5 A_4 = \frac{37}{6} A_6

A_6 = \frac{30}{37} A_4

Now,

\frac{A_4 - A_6}{A_4} = \frac{A_4 - \frac{30}{37} A_4}{A_4}

= \frac{7}{37}

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