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JEE Advanced 2015 Paper 1 Mathematics Q6
Mathematics
Problem
Let
f
:
R
→
R
f:\mathbb R\to\mathbb R
f
:
R
→
R
be a function defined by
f
(
x
)
=
{
⌊
x
⌋
,
x
≤
2
,
0
,
x
>
2
,
f(x)=\begin{cases}\lfloor x\rfloor,&x\le2,\\0,&x>2,\end{cases}
f
(
x
)
=
{
⌊
x
⌋
,
0
,
x
≤
2
,
x
>
2
,
where
⌊
x
⌋
\lfloor x\rfloor
⌊
x
⌋
is the greatest integer less than or equal to
x
x
x
. If
I
=
∫
−
1
2
x
f
(
x
2
)
2
+
f
(
x
+
1
)
d
x
I=\displaystyle\int_{-1}^{2}\frac{x f(x^2)}{2+f(x+1)}\,dx
I
=
∫
−
1
2
2
+
f
(
x
+
1
)
x
f
(
x
2
)
d
x
, then the value of
(
4
I
−
1
)
(4I-1)
(
4
I
−
1
)
is
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