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TS EAMCET 2017 Code A Solved Paper

Section: Mathematics

Question : 4 of 160

Marks: +1, -0

	List I	List II
I.	$\int\limits_{-1}^{1}x\|x\|dx$	a) $\frac{\pi}{2}$
II.	$\int\limits_{0}^{\frac{\pi}{2}}\left(1+\log\left(\frac{4+3\sin x}{4+3\cos x}\right)\right)dx$	b) $\int\limits_{0}^{\frac{a}{2}}f(x)dx$
III.	$\int\limits_{0}^{a}f(x)dx$	c) $\int\limits_{0}^{a}[f(x)+f(-x)]dx$
IV.	$\int\limits_{-a}^{a}f(x)dx$	d) 0
		e) $\int\limits_{0}^{a}f(a-x)dx$

	I	II	III	IV
(1) (2) (3) (4)	D D D A	A A C D	C C A B	C B E C

Solution:

(I) As

x|x|

is an odd function therefore,

\int\limits_{-1}^{1}x|x|dx=0

(II) Let us consider,

I=\int\limits_{0}^{\;\;\frac{\pi}{2}}\left[1+\left(\log\;\;\frac{4+3\sin x}{4+3\sin x}\right)\right]dx

=\int\limits_{0}^{\;\;\frac{\pi}{2}}\left[1+\left(\log\;\;\frac{4+3\sin\left(\;\;\frac{\pi}{2}-x\right)}{4+3\cos\left(\;\;\frac{\pi}{2}-x\right)}\right)\right]dx

=\int\limits_{0}^{\;\;\frac{\pi}{2}}\left[1+\left(\log\;\;\frac{4+3\cos x}{4+3\sin x}\right)\right]dx

Now,

2I=\int\limits_{0}^{\;\;\frac{\pi}{2}}\left[2+\log\;\;\frac{4+3\sin x}{4+3\cos x}+\log\;\;\frac{4+3\cos x}{4+3\sin x}\right]dx

2I=\int\limits_{0}^{\;\;\frac{\pi}{2}}\left(2+\log\;\;\frac{(4+3\sin x)(4+3\cos x)}{(4+3\cos x)(4+3\sin x)}\right)dx

2I=\int\limits_{0}^{\;\;\frac{\pi}{2}}\left(2+\log 1\right)dx

I=\;\;\frac{\pi}{2}

(III) from the properties of integration,

\int\limits_{0}^{a}f(x)dx=\int\limits_{0}^{a}f(a-x)dx

(IV) From the properties of integration,

\int\limits_{-a}^{a}f(x)dx=\int\limits_{0}^{a}f(x)dx+\int\limits_{0}^{a}f(-x)dx

=\int\limits_{0}^{a}[f(x)+f(-x)]dx

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