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Question

# Let g:[−2,2]→R where g(x)=x3+tanx+[x2+1P] be an odd function , then the value of the parameter P satisfies(Note : [a] denotes the greatest integer less than or equal to a)

A
5<P<5
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B
P<5
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C
5<P<0
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D
P>5
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Solution

## The correct option is D P>5Let h(x)=[x2+1P].Since, x3 and tanx are odd functions g(x) is odd if and only if h(x) is an odd function or h(x)=0 for all x. Now, −2≤x≤2 implies that 1≤x2+1≤5.Hence, for P≤5, the function h(x) will assume some nonzero integral values dependant on x. But then h(x) can not be an odd function.For P>5 the function h(x)=0 for all x and hence g(x) is an odd function.

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