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BITSAT 2024Paper 2
Exam
BITSAT
Year
2024
Paper
Paper 2
Subject
Mathematics
Chapter
Definite Integration

BITSAT 2024 · Paper 2 · Mathematics · Definite Integration

∫_5^9 ( 3 x^2)/( 3 x^2+ (588-84 x+3 x^2)) d x is equal to

∫59log⁡3x2log⁡3x2+log⁡(588−84x+3x2)dx\int_5^9 \frac{\log 3 x^2}{\log 3 x^2+\log \left(588-84 x+3 x^2\right)} d x is equal to

Options

  1. A22
  2. B11
  3. C12\frac{1}{2}
  4. D44

Correct answer

A. 22

Step-by-step solution

Let I=∫59log⁡3x2dxlog⁡3x2+log⁡(588−84x+3x2)I=\int_5^9 \frac{\log 3 x^2 d x}{\log 3 x^2+\log \left(588-84 x+3 x^2\right)}...(i) =∫59log⁡3x2dxlog⁡3x2+log⁡3(14−x)2=∫59log⁡3(14−x)2dxlog⁡3(14−x)2+log⁡3(14−(14−x))2[∫abf(x)dx=∫abf(a+b−x)dx]\begin{aligned} & =\int_5^9 \frac{\log 3 x^2 d x}{\log 3 x^2+\log 3(14-x)^2} \\ & =\int_5^9 \frac{\log 3(14-x)^2 d x}{\log 3(14-x)^2+\log 3(14-(14-x))^2} \\ & \quad\left[\int_a^b f(x) d x=\int_a^b f(a+b-x) d x\right]\end{aligned}I=\int_5^9 \frac{\log 3(14-x)^2 d x}{\log 3(14-x)^2+\log 3 x^2}...(ii)AddingEqs.(i)and(ii),weget...(ii) Adding Eqs. (i) and (ii), we get2I=∫59log⁡3x2+log⁡3(14−x)2log⁡3(14−x)2+log⁡3x2dx2I=∫59dx=9−5=4⇒I=2\begin{aligned} & 2 I=\int_5^9 \frac{\log 3 x^2+\log 3(14-x)^2}{\log 3(14-x)^2+\log 3 x^2} d x \\ & 2 \mathrm{I}=\int_5^9 d x=9-5=4 \Rightarrow I=2 \end{aligned}$
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