If f(x) = intlimits_0^x {{e^t}{{sin }^{ - 1}} | vedclass

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(B) Given $f(x) = \int\limits_0^x {{e^t}{{\sin }^{ - 1}}(t - 1)\ln t\,dt}$.By the Fundamental Theorem of Calculus,$f^\prime(x) = e^x \sin^{-1}(x-1) \l

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$f(x)$ is an increasing function

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(B) Given $f(x) = \int\limits_0^x {{e^t}{{\sin }^{ - 1}}(t - 1)\ln t\,dt}$.By the Fundamental Theorem of Calculus,$f^\prime(x) = e^x \sin^{-1}(x-1) \ln x$.The domain of $f(x)$ is determined by the domain of the integrand: $x > 0$ and $-1 \leq x-1 \leq 1$,which implies $x \in (0, 2]$.Now,analyze the sign of $f^\prime(x)$ in the interval $(0, 2]$:$1$. $e^x > 0$ for all $x$.$2$. $\ln x  0$ for $x \in (1, 2]$.$3$. $\sin^{-1}(x-1)  0$ for $x \in (1, 2]$.For $x \in (0, 1)$: $f^\prime(x) = (+) \cdot (-) \cdot (-) = (+) > 0$.For $x \in (1, 2]$: $f^\prime(x) = (+) \cdot (+) \cdot (+) = (+) > 0$.At $x = 1$,$f^\prime(1) = 0$. Since $f^\prime(x) \geq 0$ for all $x \in (0, 2]$,the function $f(x)$ is an increasing function.

If $f(x) = \int\limits_0^x {{e^t}{{\sin }^{ - 1}}(t - 1)\ln t\,dt}$ for $x > 0$,then:

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