JEE Mains · Maths · STD 12 - 7.2 definite integral
Let the function \(f :[0,2] \rightarrow R\) be defined as \(f(x)=\left\{\begin{array}{cc}e^{\min \left[x^2, x-[x]\right\}}, & x \in[0,1) \\e^{\left[x-\log _e x\right]}, & x \in[1,2]\end{array}\right.\) where [t] denotes the greatest integer less than or equal to \(t\). Then the value of the integral \(\int \limits_0^2 x f(x) d x\) is
- A \(2 e -1\)
- B \(1+\frac{3 e }{2}\)
- C \(2 e -\frac{1}{2}\)
- D \((e-1)\left(e^2+\frac{1}{2}\right)\)
Answer & Solution
Correct Answer
(C) \(2 e -\frac{1}{2}\)
Step-by-step Solution
Detailed explanation
Minimum \(m \left\{ x ^2,\{ x \}\right\}= x ^2 ; x \in[0,1)\) \({\left[ x -\log _{ e } x \right]=1 ; x \in[1,2)}\) \(\therefore f ( x )=\left\{\begin{array}{l} e ^{ x ^2} ; x \in[0,1) \\ e ; x \in[1,2)\end{array}\right.\)…
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