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JEE Advanced 2017 Paper 2 Mathematics Q11
Mathematics
Problem
Let
f
(
x
)
=
1
−
x
(
1
+
∣
1
−
x
∣
)
∣
1
−
x
∣
cos
(
1
1
−
x
)
f(x)=\dfrac{1-x(1+|1-x|)}{|1-x|}\cos\left(\dfrac{1}{1-x}\right)
f
(
x
)
=
∣1
−
x
∣
1
−
x
(
1
+
∣1
−
x
∣
)
cos
(
1
−
x
1
)
for
x
≠
1
x\ne1
x
=
1
. Then
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