Q26 / 60 · JEE Advanced 2015 (Paper 1) · Mathematics

Let $f : R \to R$ be a function defined by $f (x) = \{\begin{matrix} [x], & x \leq 2 \\ 0, & x > 2 \end{matrix}$ , where $[x]$ is the greatest integer less than or equal to $x$ . If $I = \int_{- 1}^{2} \frac{x f (x^{2})}{2 + f (x + 1)} d x,$ then the value of $(4 I - 1)$ is

Correct answer: 0

Solution

f (x + 1) = \{\begin{matrix} [x + 1], & x + 1 \leq 2 \\ 0, & x + 1 > 2 \end{matrix} = \{\begin{matrix} [x] + 1, & x \leq 1 \\ 0, & x > 1 \end{matrix}

= \{\begin{matrix} 0, & - 1 \leq x < 0 \\ 1, & 0 \leq x < 1 \\ 0, & 1 < x < 2 \end{matrix}

f (x^{2}) = \{\begin{matrix} [x^{2}] . & x^{2} \leq 2 \\ 0, & x^{2} > 2 \end{matrix}

= \{\begin{matrix} [x^{2}] . & | x | \leq \sqrt{2} \\ 0, & | x | > \sqrt{2} \end{matrix}

= {\begin{matrix} 0, & - 1 \leq x < 0 \\ 0, & 0 \leq x < 1 \\ 1, & 1 \leq x < \sqrt{2} \\ 0, & \sqrt{2} < x < 2 \end{matrix}

l = \int_{- 1}^{0} 0 d x + \int_{0}^{1} 0 d x + \int_{1}^{\sqrt{2}} \frac{x d x}{2 + 0} + \int_{\sqrt{2}}^{2} 0 d x

= \frac{1}{4} {[x^{2}]}_{1}^{\sqrt{2}} = \frac{1}{4} [2 - 1] = \frac{1}{4}

4 l = 1

4 l - 1 = 0

Hence, the correct answer is 0.