Question

Let $p$ be the number of positive integral values which are not contained in the set of values of ' $a$ '. Such that function $f:[-3,3]-\left\{\frac{\pi }{2}\right\}\to$ Rdefined by $f(x)=\tan (\sin (\sin x))+\left[\frac{(x-2{)}^2}{a}\right]$ is an odd function. Then the value of $\left[\frac{p-3}{7}\right]$, is
[Note : $[k]$ denotes greatest integer less than or equal to $k$.]

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (C)

$f(x)=-f(-x)$
$\Rightarrow \left[\frac{(x-2{)}^2}{a}\right]=0\forall x\in [-3,3]$
$\Rightarrow (x-2{)}^2\in [0,25]$
$\text{ for }\left[\frac{(x-2{)}^2}{a}\right]=0\forall x\in [-3,3],a\in (25,\infty )$
$\Rightarrow p=25$
$\Rightarrow \left[\frac{25-3}{7}\right]=\left[\frac{22}{7}\right]=3$