Let f : R rightarrow R be a continuous odd fu | vedclass

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

Question

(B) Since $f$ is an odd function,$f(-t) = -f(t)$.$F(1) = \int_{-1}^1 f(t) dt = 0$ because the integral of an odd function over a symmetric interval $[

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(B) Since $f$ is an odd function,$f(-t) = -f(t)$.$F(1) = \int_{-1}^1 f(t) dt = 0$ because the integral of an odd function over a symmetric interval $[-a, a]$ is $0$.$G(1) = \int_{-1}^1 t|f(f(t))| dt = 0$ because the integrand $h(t) = t|f(f(t))|$ is an odd function $(h(-t) = -t|f(f(-t))| = -t|f(-f(t))| = -t|f(f(t))| = -h(t))$.Using $L'H\hat{o}pital's$ rule for the limit $\lim_{x \rightarrow 1} \frac{F(x)}{G(x)}$:$\lim_{x \rightarrow 1} \frac{F'(x)}{G'(x)} = \lim_{x \rightarrow 1} \frac{f(x)}{x|f(f(x))|} = \frac{f(1)}{1 \cdot |f(f(1))|} = \frac{1/2}{|f(1/2)|} = \frac{1}{14}$.Therefore,$|f(1/2)| = 7$. Since $f$ is a continuous odd function vanishing at only one point (which must be $x=0$) and $f(1) = 1/2 > 0$,$f(x)$ must be positive for $x > 0$. Thus,$f(1/2) = 7$.

Similar Questions

The minimum value of the twice differentiable function $f(x) = \int_{0}^{x} e^{x-t} f'(t) dt - (x^2 - x + 1) e^x, x \in R$,is.

The value of $\mathop {\lim }\limits_{x \to 0} \frac{{\int_0^x {\cos {t^2}dt} }}{x}$ is

Let $f(x) = \int_{1}^{x} \sqrt{2 - t^2} dt$. Then the real roots of the equation $x^2 - f'(x) = 0$ are

The value of $x$ that maximises the value of the integral $\int\limits_x^{x + 3} {t(5 - t)\,dt}$ is

Let $f$ be a differentiable function satisfying $f(x) = \frac{2}{\sqrt{3}} \int_{0}^{\sqrt{3}} f \left(\frac{\lambda^{2} x}{3}\right) d\lambda$ for $x > 0$ and $f(1) = \sqrt{3}$. If $y = f(x)$ passes through the point $(\alpha, 6)$,then $\alpha$ is equal to $.........$

Vedclass Test Series

Exam Paper Generator

Online Exam Module