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MHTCET 11 Apr 2026 Shift 2 Paper

Question : 132 of 150

Marks: +1, -0

\int_{0}^{k} \frac{1}{1+4^{x}} d x=\log _{2}\left(\frac{4}{3}\right)

k=

\_\_\_\_

Solution:

Concept:

Use algebraic manipulation of the integrand and compare logarithms in the same base.

Explanation:

Rewrite the integrand as

\frac{1}{1+4^{x}}=1-\frac{4^{x}}{1+4^{x}}

\int \frac{1}{1+4^{x}}\,dx=\int 1\,dx-\int \frac{4^{x}}{1+4^{x}}\,dx

Now

\int \frac{4^{x}}{1+4^{x}}\,dx=\log_{4}\left(1+4^{x}\right)

Therefore,

\int \frac{1}{1+4^{x}}\,dx=x-\log_{4}\left(1+4^{x}\right)

Applying the limits

0

k

\int_{0}^{k}\frac{1}{1+4^{x}}\,dx=\left[x-\log_{4}\left(1+4^{x}\right)\right]_{0}^{k}

=k-\log_{4}\left(1+4^{k}\right)+\log_{4}2

Write

k=\log_{4}\left(4^{k}\right)

and

\log_{4}2=\frac{1}{2}

=\log_{4}\left(4^{k}\right)-\log_{4}\left(1+4^{k}\right)+\log_{4}2

=\log_{4}\left(\frac{2\cdot 4^{k}}{1+4^{k}}\right)

Convert the given value to base

4

\log_{2}\left(\frac{4}{3}\right)=\log_{4}\left(\frac{16}{9}\right)

Hence,

\frac{2\cdot 4^{k}}{1+4^{k}}=\frac{16}{9}

Let

4^{k}=y

. Then

\frac{2y}{1+y}=\frac{16}{9}

Cross-multiplying:

18y=16+16y ⇒ 2y=16 ⇒ y=8

Thus

4^{k}=8

. Write both sides as powers of

2

2^{2k}=2^{3}

2k=3

, giving

k=\frac{3}{2}

Answer:

k=\frac{3}{2}

, i.e. Option B.

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