If f(x)= {x}{≤ft(x²+4)≤ft(x²+9)} d x and f(0)=1/5 ≤ft(2/3), then f(1)=

Correct Answer is : −110log⁡(2)-1/10 (2)−101log(2)

Correct Answer is : −110log⁡(2)-1/10 (2)−101log(2)

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MHTCET 20 Apr 2026 Shift 1 Paper

Question : 127 of 150

Marks: +1, -0

f(x)=\int \frac{x}{\left(x^{2}+4\right)\left(x^{2}+9\right)} d x

f(0)=\frac{1}{5} \log \left(\frac{2}{3}\right)

f(1)=

Solution:

Concept:

Use substitution and partial fractions to integrate, then apply the given initial condition to find the constant.

Explanation:

Let

t = x^2

, so

dt = 2x\,dx

and

x\,dx = \frac{dt}{2}

The integral becomes:

f(x) = \frac{1}{2} \int \frac{dt}{(t+4)(t+9)}

Using partial fractions:

\frac{1}{(t+4)(t+9)} = \frac{1}{5}\left(\frac{1}{t+4} - \frac{1}{t+9}\right)

So,

f(x) = \frac{1}{10} \log\left(\frac{t+4}{t+9}\right) + C

Substitute back

t = x^2

f(x) = \frac{1}{10} \log\left(\frac{x^2+4}{x^2+9}\right) + C

Apply

f(0) = \frac{1}{5}\log\left(\frac{2}{3}\right)

f(0) = \frac{1}{10}\log\left(\frac{4}{9}\right) + C = \frac{1}{5}\log\left(\frac{2}{3}\right) + C

Therefore,

C = 0

Now evaluate

f(1)

f(1) = \frac{1}{10} \log\left(\frac{1+4}{1+9}\right) = \frac{1}{10} \log\left(\frac{5}{10}\right)

f(1) = \frac{1}{10} \log\left(\frac{1}{2}\right) = -\frac{1}{10} \log 2

Answer:

Option D:

-\frac{1}{10} \log(2)

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