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Question

Let f,g:RR be defined by f(x)=x2cosx2 and g(x)=xsinx2. Then

A
f(x)=g(x) has exactly one solution.
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B
f(x)=g(x) has exactly two solutions.
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C
f(x)g(x) as x
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D
The minimum value of f(g(x)) is 1.
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Solution

The correct option is C f(x)g(x) as x
Let h(x)=f(x)g(x)
Then h(x)=x2cosx2xsinx2
h(x)=2xxcosx2
h(x)=x2(4cosx)

h(x)>0 x>0
and h(x)<0 x<0
So, h is monotonically increasing for x>0
and h is monotonically decreasing for x<0

Also, h(x) as x±,
h(0)=12
So, by intermediate value theorem,
a value of x(,0) such that h(x)=0
and a value of x(0,) such that h(x)=0

f(g(x))=(xsinx2)212cos(xsinx2)
Since, (xsinx2)2 is a positive quantity, hence minimum value of f(g(x)) is occure when cos(xsinx2) is maximum and (xsinx2)2 is minimum.
The maximum value of cos(xsinx2) is 1 when x=nπ, nZ.
And at x=nπ, nZ, (xsinx2)2=0
minf(g(x))=12

Alternate Solution:
The graph of y=x2 is a parabola passing through origin.
Since, cosx[1,1], so at higher value of x, x2cosx2x2
Hence, the graph of y=x2cosx2 is looks like parabola, shifted from origin.

And the graph of y=xsinx2 is sinusoidal with amplitude x2.
Now, plot the graph of both the functions.
From graph, it is clear that f(x)=g(x) has exactly two solutions.

f(x)g(x)=x2cosx2xsinx2
=x2(1cosx2x2sinx2x)
So, as x±f(x)g(x)

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