If the point (k-1)/k,(k-2)/k lies on the locus of z satisfying the inequality (z+3i)/(3z+i)<1, then the interval in which k lies is

Correct answer: (D) (-∞,1)∪(5,∞). (z+3i)/(3z+i)⇒(x+iy+3i)/(3x+3iy+i) 0 Now (k-1)/k,(k-2)/k lie above the curve So k-1k^2+k-2k^2-1>0 ⇒k^2-2k+1+k^2-4k+4-k^2>0 ⇒k^2-6k+5>0 k∈(-∞,1) ∪(5,∞)

Q47 / 160 · TS EAMCET 2019 (03 May Shift 2) · Mathematics

If the point $(\frac{k - 1}{k}, \frac{k - 2}{k})$ lies on the locus of $z$ satisfying the inequality $|\frac{z + 3 i}{3 z + i}| < 1,$ then the interval in which $k$ lies is

A
$(- \infty, 2) \cup (3, \infty)$
B
$[2, 3]$
C
$[1, 5]$
Correct answer
D
$(- \infty, 1) \cup (5, \infty)$

Solution

$|\frac{z + 3 i}{3 z + i}| \Rightarrow |\frac{x + i y + 3 i}{3 x + 3 i y + i}| < 1$
$\frac{\sqrt{x^{2} + (3 + y)^{2}}}{\sqrt{9 x^{2} + (3 y + 1)^{2}}} < 1$
$\Rightarrow$ $x^{2} + 9 + y^{2} + 6 y < 9 x^{2} + 9 y^{2} + 6 y + 1$
$\Rightarrow - 8 x^{2} - 8 y^{2} + 8 < 0$
$\Rightarrow x^{2} + y^{2} - 1 > 0$
Now $(\frac{k - 1}{k}, \frac{k - 2}{k})$ lie above the curve
So ${[\frac{k - 1}{k}]}^{2} + {[\frac{k - 2}{k}]}^{2} - 1 > 0$
$\Rightarrow [k^{2} - 2 k + 1] + [k^{2} - 4 k + 4] - k^{2} > 0$
$\Rightarrow k^{2} - 6 k + 5 > 0$
$k \in (- \infty, 1) \cup (5, \infty)$

Hence, the correct answer is option: 'D'.

If the point k-1k,k-2k lies on the locus of z satisfying the inequality z+3i3z+i<1, then the interval in which k lies is

Solution

Relevant Questions

If the point $(\frac{k - 1}{k}, \frac{k - 2}{k})$ lies on the locus of $z$ satisfying the inequality $|\frac{z + 3 i}{3 z + i}| < 1,$ then the interval in which $k$ lies is