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)=−x3−sinx+[x2+2a]..(ii) Also, given f(−x)=−f(x) ∴f(x)+f(−x)=0 ⇒2[x2+2a]=0 (by (i) and (ii)) ⇒0≤x2+2a<1∀−3≤x≤3
As x2+2>0, we need a>0.
Now, x2+2<a⇒a>x2+2∀x∈[−3,3] ∴a>11 (∵ Maximum value of x2+2 in −3≤x≤3 is 11)