JEE Main 2014 · 6 Apr · Shift 1 · Mathematics Q62

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MathematicsSingle correct+4 / -1Question 62 of 90

If $z$ is a complex number such that $|z|\ge2$ , then the minimum value of $\left|z+\dfrac12\right|$ :

is strictly greater than

\dfrac52

is strictly greater than

\dfrac32

but less than

\dfrac52

is equal to

\dfrac52

lies in the interval

(1,2)

Answer and worked solution

Official answer: D

By the triangle inequality,
$\left|z+\tfrac12\right|\ge|z|-\tfrac12\ge\tfrac32,$
with equality at $z=-2$ . The minimum is $\frac32$ , which lies in $(1,2)$ .

Topics:Modulus and argument

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JEE Main 2014 · 6 Apr · Shift 1, Mathematics question 62