Question

lf the point $(k+1,k)$ lies inside the region bounded by the curve $x=\sqrt{25-y^2}$ and $y$-axis, then $k$ belongs to the interval

• A. $(-1,3)$
• B. $(-4,3)$
• C. $(-\infty ,-4)\cup (3,\infty )$
• D. $(-4,-3)$

Answer 1

Answer: (A)

$(-1,3)$

$x=\sqrt{25-y^2}$
$x^2+y^2=25$ is a equation of hemicircle with $x⩾0$
$\therefore (k+1{)}^2+k^2-25<0$
$2k^2+2k-24<0$
$k^2+k-12<0$
$(k+4)(k-3)<0$
$kϵ(-4,3)$ ---(1)
and as known $x⩾0$
$\Rightarrow k+1>0$
$k>-1$ ---(2)
from (1) & (2),
$kϵ(-1,3)$