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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$f(x)$ is an increasing function

Vedclass · Answer

(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:

Similar Questions

The function $f(x) = \sin^4 x + \cos^4 x$ increases,if

Let $(2, 3)$ be the largest open interval in which the function $f(x) = 2 \log_e(x-2) - x^2 + ax + 1$ is strictly increasing and $(b, c)$ be the largest open interval in which the function $g(x) = (x-1)^3(x+2-a)^2$ is strictly decreasing. Then $100(a+b-c)$ is equal to:

The function $f(x) = \cot^{-1} x + x$ is increasing in the interval.

If $f(x) = x^3 - x^2 + 100x + 1001$,then:

Given the function $f(x) = \left( \frac{e^{2x} - 1}{e^{2x} + 1} \right)$,the function is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module