- A${y'^2} + x = {y^2}$
- ✓$y'y'' + y = \sin x$
- C$y''' + y'' + y = 0$
- D$y' = y$
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$f( x )=\frac{ x ^2-3 x -6}{ x ^2+2 x +4} \text {. }$
Then which of the following statements is (are) $TRUE$ ?
$(A)$ $f$ is decreasing in the interval $(-2,-1)$
$(B)$ $f$ is increasing in the interval $(1,2)$
$(C)$ $f$ is onto
$(D)$ Range of $f$ is $\left[-\frac{3}{2}, 2\right]$
$M =\left\{( x , y ) \in R \times R : x ^2+ y ^2 \leq r ^2\right\},$
where $r >0$. Consider the geometric progression $a _{ n }=\frac{1}{2^{ n -1}}, n =1,2,3, \ldots$. Let $S _0=0$ and, for $n \geq 1$, let $S _{ n }$ denote the sum of the first $n$ terms of this progression. For $n \geq 1$, let $C_n$ denote the circle with center $\left(S_{n-1}, 0\right)$ and radius $a _{ n }$, and $D _{ n }$ denote the circle with center $\left( S _{ n -1}, S _{ n -1}\right)$ and radius $a _{ n }$.
($1$) Consider M with $r =\frac{1025}{513}$. Let $k$ be the number of all those circles $C _{ n }$ that are inside $M$. Let $l$ be the maximum possible number of circles among these $k$ circles such that no two circles intersect. Then
$(A)$ $k +2 l=22$ $(B)$ $2 k +l=26$ $(C)$ $2 k +3 l=34$ $(D)$ $3 k +2 l=40$
($2$) Consider $M$ with $r =\frac{\left(2^{199}-1\right) \sqrt{2}}{2^{158}}$. The number of all those circles $D _{ a }$ that are inside $M$ is
$(A) 198$ $(B) 199$ $(C) 200$ $(D) 201$
Give the answer or qution ($1$) and ($2$)