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TS EAMCET 4-Aug-2021 Shift 1 Question Paper

Section: Mathematics

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Question : 78 of 160

Marks: +1, -0

\int\limits_{5}^{9} \frac{\log 3x^2}{\log 3x^2 + \log (588-84x+3x^2)}

dx

=

$2$
$1$
$\frac{1}{2}$
$4$

Solution: 👈: Video Solution

Let

I \; = \int\limits_{5}^{9} \frac{\log 3 x^{2} d x}{\log 3 x^{2}+\log (588-84 x+3 x^{2})} \dots \; \text{ (i) } \;

\; = \int\limits_{5}^{9} \frac{\log 3 x^{2} d x}{\log 3 x^{2}+\log 3(14-x)^{2}}

\; = \int\limits_{5}^{9} \frac{\log 3(14-x)^{2} d x}{\log 3(14-x)^{2}+\log 3(14-(14-x))^{2}}

\; \left[ \int\limits_{a}^{b} f(x) d x = \int\limits_{a}^{b} f(a+b-x) d x \right]

I \; = \int\limits_{5}^{9} \frac{\log 3(14-x)^{2} d x}{\log 3(14-x)^{2}+\log 3 x^{2}}

. . . (ii)

Adding Eqs. (i) and (ii), we get

2 I \; = \int\limits_{5}^{9} \frac{\log 3 x^{2}+\log 3(14-x)^{2}}{\log 3(14-x)^{2}+\log 3 x^{2}} d x

\; = \int\limits_{5}^{9} d x = 9-5 = 4

I \; = 2

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