If the function f(x)=⎧⎪
⎪
⎪
⎪
⎪⎨⎪
⎪
⎪
⎪
⎪⎩1−cos2pxx2,x<0q2,x=01−√1−xx,x>0 is continuous at x=0, then the value(s) of 2(p+q2) can be
A
0
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B
1
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C
2
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D
4
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Solution
The correct option is C2 For function to be continuous at x=0 R.H.L.=L.H.L=f(0)
Now, R.H.L.=limx→0+1−√1−xx×1+√1−x1+√1−x =limx→0+xx(1+√1−x) ∴R.H.L.=12=q2
L.H.L.=limx→0−1−cos2pxx2 =limx→0−2sin2pxx2=2p2 ⇒2p2=12 ⇒p=±12 ∴2(p+q2)=±1+1=0 or 2