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Question

Let f:[3,3]R where f(x)=x3+sinx+[x2+2a] be an odd function then value of a is (where [.] represent greatest integer function and a is positive)

A
less than 11
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B
11
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C
greater than 11
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D
none of these
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Solution

The correct option is B greater than 11
Given f(x)=x3+sinx+[x2+2a]..(i)
f(x)=x3sinx+[x2+2a]..(ii)
Also, given f(x)=f(x)
f(x)+f(x)=0
2[x2+2a]=0 (by (i) and (ii))
0x2+2a<13x3
As x2+2>0, we need a>0.
Now, x2+2<aa>x2+2x[3,3]
a>11 ( Maximum value of x2+2 in 3x3 is 11)

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