Q177 / 240 · KCET 2026 · Mathematics
$\int e^{-x \log 2} \cdot 2^x\,dx = $
- A$\log x + C$
- Correct answerB$x + C$
- C$\dfrac{1}{x} + C$
- D$\dfrac{x^2}{2} + C$
Solution
The given integral is $\int e^{-x \log 2} \cdot 2^x \, dx$
Using the property $e^{a \log b} = b^a$, we can write $e^{-x \log 2} = e^{\log(2^{-x})} = 2^{-x}$
Substituting this into the integral:
$\int 2^{-x} \cdot 2^x \, dx$
$\int 2^{-x + x} \, dx$
$\int 2^0 \, dx$
$\int 1 \, dx$
$x + C$
Answer: $x + C$
Using the property $e^{a \log b} = b^a$, we can write $e^{-x \log 2} = e^{\log(2^{-x})} = 2^{-x}$
Substituting this into the integral:
$\int 2^{-x} \cdot 2^x \, dx$
$\int 2^{-x + x} \, dx$
$\int 2^0 \, dx$
$\int 1 \, dx$
$x + C$
Answer: $x + C$
Hence, the correct answer is option: 'B'.