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STD 12 - 7.2 definite integral — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 7.2 definite integral4 Marks
MCQ
Let $f: R \rightarrow R$ be a function defined by $f(x)=\left\{\begin{array}{cc}{[x],} & x \leq 2 \\ 0, & x>2\end{array}\right.$, where $[x]$ is the greatest integer less than or equal to $x$. If $I=\int_{-1}^2 \frac{x f\left(x^2\right)}{2+f(x+1)} d x$, then the value of $(4 I-1)$ is
  • ✓
    $0$
  • B
    $1$
  • C
    $2$
  • D
    $3$
✓

Answer

Correct option: A.
$0$
a
$I =\int_{-1}^0 \frac{ x \cdot 0}{2+0} dx +\int_0^1 \frac{ x \cdot 0}{2+1} dx +\int_1^{\sqrt{2}} \frac{ x \cdot 1}{2+0} dx +0=\frac{1}{4}$

$\Rightarrow 4 I -1=0$

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