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Q.

If fx=cosxx12sinxx22xtanxx1 then limx→0f'(x)x=

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a

1

b

-1

c

2

d

-2

answer is D.

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Detailed Solution

fx=cosxx2-2x2-x[2sinx-2xtanx]+12xsinx-x2tanx=-x2cosx-2xsinx+2x2tanx+2xsinx-x2tanx=-x2cosx+x2tanxf1(x)=-2xcosx+x2sinx+2xtanx+x2sec2xlimx→0f1(x)x=limx→0-2xcosx+x2sinx+2xtanx+x2sec2xx=limx→0-2cosx+xsinx+2tanx+xsec2x=-2+0+0+0=-2
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If fx=cosxx12sinxx22xtanxx1 then limx→0f'(x)x=