Let g(x) = int_{0}^{x} f(t) dt,where f is a c | vedclass

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

Question

(A) Given that $\frac{1}{3} \leq f(t) \leq 1$ for all $t \in [0, 1]$ and $0 \leq f(t) \leq \frac{1}{2}$ for all $t \in (1, 3]$.We need to find the ran

Vedclass · Accepted Answer

$[\frac{1}{3}, 2]$

Vedclass · Answer

(A) Given that $\frac{1}{3} \leq f(t) \leq 1$ for all $t \in [0, 1]$ and $0 \leq f(t) \leq \frac{1}{2}$ for all $t \in (1, 3]$.We need to find the range of $g(3) = \int_{0}^{3} f(t) dt$.We can split the integral as $g(3) = \int_{0}^{1} f(t) dt + \int_{1}^{3} f(t) dt$.For the first part,$\int_{0}^{1} \frac{1}{3} dt \leq \int_{0}^{1} f(t) dt \leq \int_{0}^{1} 1 dt$,which gives $\frac{1}{3} \leq \int_{0}^{1} f(t) dt \leq 1$.For the second part,$\int_{1}^{3} 0 dt \leq \int_{1}^{3} f(t) dt \leq \int_{1}^{3} \frac{1}{2} dt$,which gives $0 \leq \int_{1}^{3} f(t) dt \leq \frac{1}{2} 	imes (3 - 1) = 1$.Adding these two inequalities,we get $\frac{1}{3} + 0 \leq g(3) \leq 1 + 1$,which simplifies to $\frac{1}{3} \leq g(3) \leq 2$.Thus,the interval is $[\frac{1}{3}, 2]$.

Let $g(x) = \int_{0}^{x} f(t) dt$,where $f$ is a continuous function in $[0, 3]$ such that $\frac{1}{3} \leq f(t) \leq 1$ for all $t \in [0, 1]$ and $0 \leq f(t) \leq \frac{1}{2}$ for all $t \in (1, 3]$. The largest possible interval in which $g(3)$ lies is:

Similar Questions

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \log \left(\frac{2-\sin \theta}{2+\sin \theta}\right) d \theta$ is equal to

The value of the integral $\int_{-\pi}^{\pi} \frac{\cos^2 x}{1+a^x} dx$,where $a > 0$,is

$\int_{0}^{\pi} x f(\sin x) dx = $

If $\int_0^\pi {x\,f({{\cos }^2}x + {{\tan }^4}x)\,dx} = k\int_0^{\pi /2} {f({{\cos }^2}x + {{\tan }^4}x)\,dx,}$ then the value of $k$ is

Evaluate the integral: $\int_{-1}^{1} \left[ \sqrt{1+x+x^{2}} - \sqrt{1-x+x^{2}} \right] dx$

Vedclass Test Series

Exam Paper Generator

Online Exam Module