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JEE Mains · Maths · STD 12 - 7.2 definite integral

અહી \(g ( x )=\int_{0}^{ x } f( t ) dt \) કે જ્યાં \(f\) એ \([0,3]\) પર સતત છે કે જેથી દરેક \(t \in[0,1]\) માટે \(\frac{1}{3} \leq f(t) \leq 1\) અને \(t \in(1,3]\)  માટે \(0 \leq f( t ) \leq \frac{1}{2}\) થાય છે. તો  \(g (3)\) ને સમાવતો મહતમ અંતરાલ મેળવો.

  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, \(g (3)=\int_{0}^{3} f( t ) dt =\int_{0}^{1} f( t ) dt +\int_{1}^{3} f( t ) dt\)…
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