Q92 / 150 · MHT CET 2026 (11 April Shift 2) · Mathematics
If $\int_{0}^{k} \dfrac{1}{1 + 4^x} dx = \log_2\left(\dfrac{4}{3}\right)$, then $k = $ _____
- A$\dfrac{2}{\sqrt{3}}$
- Correct answerB$\dfrac{3}{2}$
- C$\dfrac{\sqrt{3}}{2}$
- D$\dfrac{2}{3}$
Solution
Let $I = \int_{0}^{k} \dfrac{1}{1 + 4^x} dx$
Substitute $4^x = t \Rightarrow x \ln 4 = \ln t \Rightarrow dx = \dfrac{dt}{t \ln 4}$
When $x = 0$, $t = 1$. When $x = k$, $t = 4^k$.
$I = \int_{1}^{4^k} \dfrac{1}{1 + t} \dfrac{dt}{t \ln 4} = \dfrac{1}{\ln 4} \int_{1}^{4^k} \left( \dfrac{1}{t} - \dfrac{1}{1 + t} \right) dt$
$I = \dfrac{1}{\ln 4} \left[ \ln t - \ln(1 + t) \right]_{1}^{4^k} = \dfrac{1}{\ln 4} \left[ \ln \left( \dfrac{t}{1 + t} \right) \right]_{1}^{4^k}$
$I = \dfrac{1}{2 \ln 2} \left( \ln \left( \dfrac{4^k}{1 + 4^k} \right) - \ln \left( \dfrac{1}{2} \right) \right) = \dfrac{1}{2 \ln 2} \ln \left( \dfrac{2 \cdot 4^k}{1 + 4^k} \right)$
Given $I = \log_2\left(\dfrac{4}{3}\right) = \dfrac{\ln(4/3)}{\ln 2}$
Equating the two expressions:
$\dfrac{1}{2 \ln 2} \ln \left( \dfrac{2 \cdot 4^k}{1 + 4^k} \right) = \dfrac{\ln(4/3)}{\ln 2}$
$\ln \left( \dfrac{2 \cdot 4^k}{1 + 4^k} \right) = 2 \ln \left( \dfrac{4}{3} \right) = \ln \left( \dfrac{16}{9} \right)$
$\dfrac{2 \cdot 4^k}{1 + 4^k} = \dfrac{16}{9}$
$18 \cdot 4^k = 16 + 16 \cdot 4^k$
$2 \cdot 4^k = 16 \Rightarrow 4^k = 8$
$2^{2k} = 2^3 \Rightarrow 2k = 3 \Rightarrow k = \dfrac{3}{2}$
Answer: $\dfrac{3}{2}$
Substitute $4^x = t \Rightarrow x \ln 4 = \ln t \Rightarrow dx = \dfrac{dt}{t \ln 4}$
When $x = 0$, $t = 1$. When $x = k$, $t = 4^k$.
$I = \int_{1}^{4^k} \dfrac{1}{1 + t} \dfrac{dt}{t \ln 4} = \dfrac{1}{\ln 4} \int_{1}^{4^k} \left( \dfrac{1}{t} - \dfrac{1}{1 + t} \right) dt$
$I = \dfrac{1}{\ln 4} \left[ \ln t - \ln(1 + t) \right]_{1}^{4^k} = \dfrac{1}{\ln 4} \left[ \ln \left( \dfrac{t}{1 + t} \right) \right]_{1}^{4^k}$
$I = \dfrac{1}{2 \ln 2} \left( \ln \left( \dfrac{4^k}{1 + 4^k} \right) - \ln \left( \dfrac{1}{2} \right) \right) = \dfrac{1}{2 \ln 2} \ln \left( \dfrac{2 \cdot 4^k}{1 + 4^k} \right)$
Given $I = \log_2\left(\dfrac{4}{3}\right) = \dfrac{\ln(4/3)}{\ln 2}$
Equating the two expressions:
$\dfrac{1}{2 \ln 2} \ln \left( \dfrac{2 \cdot 4^k}{1 + 4^k} \right) = \dfrac{\ln(4/3)}{\ln 2}$
$\ln \left( \dfrac{2 \cdot 4^k}{1 + 4^k} \right) = 2 \ln \left( \dfrac{4}{3} \right) = \ln \left( \dfrac{16}{9} \right)$
$\dfrac{2 \cdot 4^k}{1 + 4^k} = \dfrac{16}{9}$
$18 \cdot 4^k = 16 + 16 \cdot 4^k$
$2 \cdot 4^k = 16 \Rightarrow 4^k = 8$
$2^{2k} = 2^3 \Rightarrow 2k = 3 \Rightarrow k = \dfrac{3}{2}$
Answer: $\dfrac{3}{2}$
Hence, the correct answer is option: 'B'.