If the area bounded by curves y = 2^kx, x = 0, x = 2 in first quadrant is 3/( 2), then k is (k ≠ 0) Option 1: 1/2 Option 2: 1 Option 3: -1 Option 4: 2

Answer: 1. Solution: Area =_0^2 2^kxdx = (2^2k-1)/(k 2). Equating to 3/( 2) gives (2^2k-1)/k=3. Testing the options, k=1 gives (4-1)/1=3, which satisfies the equation.

Calculus

Application of Integrals

Medium

core

If the area bounded by curves $y = 2^{kx}$ , $x = 0$ , $x = 2$ in first quadrant is $\frac{3}{\log 2}$ , then $k$ is ( $k \neq 0$ )

1/2

-1

✅ Correct Option: 2

Area $=\int_0^2 2^{kx}dx = \frac{2^{2k}-1}{k\log 2}$ .

Equating to $\frac{3}{\log 2}$ gives $\frac{2^{2k}-1}{k}=3$ .

Testing the options, $k=1$ gives $\frac{4-1}{1}=3$ , which satisfies the equation.

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If the area bounded by curves $y = 2^{kx}$ , $x = 0$ , $x = 2$ in first quadrant is $\frac{3}{\log 2}$ , then $k$ is ( $k \neq 0$ )