Let f : [1/2, 1] rightarrow mathbb{R} be a po | vedclass

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

Question

(D) Given $f^{\prime}(x) < 2f(x)$,we have $f^{\prime}(x) - 2f(x) < 0$. Multiply by the integrating factor $e^{-2x}$: $e^{-2x} f^{\prime}(x) - 2e^{-2x}

Vedclass · Accepted Answer

$(0, (e - 1)/2)$

Vedclass · Answer

(D) Given $f^{\prime}(x) Multiply by the integrating factor $e^{-2x}$: $e^{-2x} f^{\prime}(x) - 2e^{-2x} f(x) $\frac{d}{dx} (e^{-2x} f(x)) This implies that $g(x) = e^{-2x} f(x)$ is a strictly decreasing function on $[1/2, 1]$. Since $x \ge 1/2$,we have $g(x) $e^{-2x} f(x) Thus,$f(x) Integrating both sides from $1/2$ to $1$: $\int_{1/2}^1 f(x) dx Since $f(x) > 0$,the integral is greater than $0$. Therefore,$\int_{1/2}^1 f(x) dx \in (0, (e - 1)/2)$.

Let $f : [1/2, 1] \rightarrow \mathbb{R}$ be a positive,non-constant and differentiable function such that $f^{\prime}(x) < 2f(x)$ and $f(1/2) = 1$. Then the value of $\int_{1/2}^1 f(x) dx$ lies in the interval

Similar Questions

The value of $\int_{0}^{\infty} \frac{x}{(1+x)(x^{2}+1)} dx$ is

Evaluate the definite integral $\int_{0}^{\frac{\pi}{4}} \sin 2x \,dx$.

$\int_0^{x} \frac{t^2}{\sqrt{a^2+t^2}} dt =$

The integral $\int_{0}^{2} ||x-1|-x| dx$ is equal to (in $.5$)

$\int_0^2 \frac{x^3}{(x^2 + 1)^{3/2}} \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module