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KCET 2026 Mathematics Solved Paper

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Question : 38 of 60

Marks: +1, -0

\int e^{-x \log 2} \cdot 2^{x} d x=

$\log x+C$
$x+C$
$\frac{1}{x}+C$
$\frac{x^{2}}{2}+C$

Solution:

Concept:

Rewrite

2^x

in terms of

e

so the integrand simplifies to a constant.

Explanation:

Use the identity

2^x = e^{x \log 2}

.

Substitute into the integrand:

e^{-x \log 2} \cdot 2^x = e^{-x \log 2} \cdot e^{x \log 2}

Apply the exponent rule

e^a \cdot e^b = e^{a+b}

:

e^{-x \log 2 + x \log 2} = e^0 = 1

The integral reduces to

\int 1 \, dx

.

Since the derivative of

x

is

1

, we get

x + C

.

This corresponds to option B.

Answer:

x + C

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