If $f : [- 2, 2] \to R$ defined by $f (x) = x^{3} + tan x + [\frac{x^{2} + 1}{p}]$ is an odd function, then the least value of $[p]$ is
( $[.]$ represents the greatest integer function)

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Solution

$f (x) = x^{3} + tan x + [\frac{x^{2} + 1}{p}]$
$f$ is an odd function.
$∴ f (- x) = - f (x) \Rightarrow - x^{3} - tan x + [\frac{x^{2} + 1}{p}] = - x^{3} - tan x - [\frac{x^{2} + 1}{p}] \Rightarrow [\frac{x^{2} + 1}{p}] = 0$

Now, $x \in [- 2, 2]$
$∴ x^{2} + 1 \in [1, 5]$
So, for $f$ to be an odd function, $p \in (5, \infty)$
So, the least value of $[p]$ is $5$ .

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If f:[−2,2]→R defined by f(x)=x3+tanx+[x2+1p] is an odd function, then the least value of [p] is ([.] represents the greatest integer function)

f(x)=x3+tanx+[x2+1p] f is an odd function. ∴f(−x)=−f(x)⇒−x3−tanx+[x2+1p]=−x3−tanx−[x2+1p]⇒[x2+1p]=0 Now, x∈[−2,2] ∴x2+1∈[1,5] So, for f to be an odd function, p∈(5,∞) So, the least value of [p] is 5.

If $f : [- 2, 2] \to R$ defined by $f (x) = x^{3} + tan x + [\frac{x^{2} + 1}{p}]$ is an odd function, then the least value of $[p]$ is
( $[.]$ represents the greatest integer function)