JEE Main 2016 · 9 Apr · Shift 1 · Mathematics Q73

MathematicsSingle correct+4 / -1Question 73 of 90

The minimum distance of a point on the curve $y=x^2-4$ from the origin is :

\dfrac{\sqrt{19}}2

\sqrt{\dfrac{15}2}

\dfrac{\sqrt{15}}2

\sqrt{\dfrac{19}2}

Answer and worked solution

Official answer: C

$d^2=x^2+(x^2-4)^2$ . With $u=x^2$ , the minimum is at $u=\frac72$ , giving $d^2=\frac72+\frac14=\frac{15}4$ and $d=\frac{\sqrt{15}}2$ .

JEE Main 2016 · 9 Apr · Shift 1, Mathematics question 73