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(B) The intersection point of the curves $y = x^2$ and $y = \frac{1}{x}$ is found by setting $x^2 = \frac{1}{x}$,which gives $x^3 = 1$,so $x = 1$. Thu

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$e^{2/3}$

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(B) The intersection point of the curves $y = x^2$ and $y = \frac{1}{x}$ is found by setting $x^2 = \frac{1}{x}$,which gives $x^3 = 1$,so $x = 1$. Thus,the intersection point is $(1, 1)$.The area of the region bounded by the curves $y = x^2$,$y = \frac{1}{x}$,the $x$-axis $(y = 0)$,and the line $x = t$ $(t > 1)$ is given by the sum of two integrals:$	ext{Area} = \int_0^1 x^2 \, dx + \int_1^t \frac{1}{x} \, dx$Evaluating the integrals:$	ext{Area} = \left[ \frac{x^3}{3} ight]_0^1 + \left[ \ln(x) ight]_1^t$$	ext{Area} = \left( \frac{1}{3} - 0 ight) + (\ln(t) - \ln(1))$Since $\ln(1) = 0$,we have:$	ext{Area} = \frac{1}{3} + \ln(t)$Given that the area is $1 \, 	ext{sq. unit}$:$\frac{1}{3} + \ln(t) = 1$$\ln(t) = 1 - \frac{1}{3} = \frac{2}{3}$Taking the exponential of both sides:$t = e^{2/3}$

If the area of the region bounded by the curves $y = x^2$,$y = \frac{1}{x}$ and the lines $y = 0$ and $x = t$ $(t > 1)$ is $1 \, \text{sq. unit}$,then $t$ is equal to

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