ExamBro
ExamBro
JEE Mains · Maths · STD 12 - 7.2 definite integral

Let \(g ( x )=\int_{0}^{ x } f( t ) dt ,\) where \(f\) is 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 :

  1. A \(\left[-1,-\frac{1}{2}\right]\)
  2. B \(\left[-\frac{3}{2},-1\right]\)
  3. C \(\left[\frac{1}{3}, 2\right]\)
  4. D \([1,3]\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\left[\frac{1}{3}, 2\right]\)

Step-by-step Solution

Detailed explanation

\(\frac{1}{3} \leq f( t ) \leq 1 \forall t \in[0,1]\) \(0 \leq f( t ) \leq \frac{1}{2} \forall t \in(1,3]\) Now, \(g (3)=\int_{0}^{3} f( t ) dt =\int_{0}^{1} f( t ) dt +\int_{1}^{3} f( t ) dt\)…
From JEE Mains
Explore more questions on app