JEE Advanced 2015 · Paper 1 · Mathematics Q46

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MathematicsInteger+4Question 46 of 60

Let $f:\mathbb R\to\mathbb R$ be a function defined by $f(x)=\begin{cases}[x],&x\le2\\0,&x>2\end{cases}$ , where $[x]$ is the greatest integer less than or equal to $x$ . If $I=\displaystyle\int_{-1}^{2}\frac{xf(x^2)}{2+f(x+1)}\,dx$ , then the value of $(4I-1)$ is

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Answer and worked solution

Official answer: 0

Numerator. $f(x^2)=[x^2]$ when $x^2\le2$ , and 0 when $x^2>2$ . On $[-1,2]$ it is non-zero only for $1\le x\le\sqrt2$ , where it equals 1 (the single point $x=-1$ contributes nothing).

Denominator. There $x+1\in[2,1+\sqrt2]$ , and $x+1>2$ almost everywhere, so $f(x+1)=0$ and the denominator is 2.

Integral.
$I=\int_1^{\sqrt2}\frac x2\,dx=\frac14\implies4I-1=0.$

Answer: 0

Topics:Properties of definite integrals

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JEE Advanced 2015 · Paper 1, Mathematics question 46