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Question

Let f(x)=xa2+x2dxb2+(dx)2, xR,

where a,b and d are non-zero real constants, Then :

A
f is an increasing function of x
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B
f is neither increasing nor decreasing function of x
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C
f is a decreasing function of x
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D
f is not a continuous function of x
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Solution

The correct option is A f is an increasing function of x
f(x)=xa2+x2(dx)b2+(dx)2

f(x)=a2+x2x12a2+x22xa2+x2 +b2+(dx)2(dx)2(dx)2b2+(dx)2b2+(dx)2

=a2+x2x2a2+x2a2+x2 +b2+(dx)2(dx)2b2+(dx)2b2+(dx)2


=a2(a2+x2)3/2+b2(b2+(dx)2)3/2>0

Hence, f is an increasing function.

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