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Question

If f(x)={x2(sgn [x])+{x} ,0x<2sinx+|x3| ,2x<4,
where [.] and {.} represent the greatest integer and the fractional part function, respectively. Then

A
f(x) is differentiable at x=1
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B
f(x) is continuous but non-differentiable at x=1
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C
f(x) is non-differentiable at x=2
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D
f(x) is discontinuous at x=2
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Solution

The correct options are
B f(x) is continuous but non-differentiable at x=1
C f(x) is non-differentiable at x=2
D f(x) is discontinuous at x=2
f(x)={x2(sgn [x])+{x} ,0x<2sinx+|x3| ,2x<4

sgn (x)=1 ,x>00 ,x=01 ,x<0

For continuity at x=1
lim x1+f(x)=lim x1+x2(sgn [x])+{x} =1 sgn(1)+0 =1+0=1
lim x1f(x)=lim x1x2(sgn [x])+{x} =1 sgn(0)+1 =0+1=1

Also, f(1)=1
L.H.L=R.H.L=f(1)
Hence, f(x) is continuous at x=1

Now, for differentiability,
f(1+)=limh0f(1+h)f(1)h
=limh0(1+h)2(sgn [1+h])+{1+h}1h
=limh0(1+h)2+h1h
=limh0h2+3hh
=limh0(h+3)
=3

f(1)=limh0f(1h)f(1)h
=limh0(1h)2(sgn [1h])+{1h}1h
=limh0(1h)2×0+1h1h
=limh0hh
=1
f(1+)f(1)
Hence, f(x) is not differentiable at x=1.

Now, at x=2
lim x2f(x)=lim x2x2(sgn [x])+{x} =4 sgn(1)+1 =4+1=5
lim x2+f(x)=lim x2+sinx+|x3| =sin2+1

L.H.LR.H.L
Hence, f(x) is discontinuous at x=2 and then f(x) is also non-differentiable at x=2.

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