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MHTCET 25 Apr 2025 Shift 1 Paper

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Question : 108 of 150

Marks: +1, -0

\int_{1}^{3} \frac{\log x^{2}}{\log \left(16 x^{2}-8 x^{3}+x^{4}\right)} d x=

1
3
$\quad \log 2$
$\frac{1}{2}$

Solution:

Concept:

Use the property

\int_a^b \frac{f(x)}{f(x)+f(a+b-x)} dx = \frac{b-a}{2}

.

Explanation:

Let

I = \int_{1}^{3} \frac{\log x^{2}}{\log(16x^{2}-8x^{3}+x^{4})} dx

.

Write the numerator as

2\log x

.

Factor the denominator:

16x^{2}-8x^{3}+x^{4} = x^{2}(x^{2}-8x+16) = x^{2}(4-x)^{2}

.

So the integrand becomes

\frac{2\log x}{\log x^{2} + \log(4-x)^{2}} = \frac{2\log x}{2\log x + 2\log(4-x)} = \frac{\log x}{\log x + \log(4-x)}

.

Here

f(x)=\log x

, and for

a=1

,

b=3

, we have

a+b-x = 4-x

.

Therefore,

I = \int_{1}^{3} \frac{f(x)}{f(x)+f(4-x)} dx = \frac{1}{2}(3-1) = 1

.

Answer:

Option A:

1

.

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