JEE Main 2016 · 10 Apr · Shift 1 · Mathematics Q71

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MathematicsSingle correct+4 / -1Question 71 of 90

Let a, b ∈ R, (a ≠ 0). If the function f defined as
$f(x)=\begin{cases}\dfrac{2x^2}a, & 0\le x<1\\ a, & 1\le x<\sqrt2\\ \dfrac{2b^2-4b}{x^3}, & \sqrt2\le x<\infty\end{cases}$
is continuous in the interval [0, ∞), then an ordered pair (a, b) is :

(\sqrt2,\ 1-\sqrt3)

(-\sqrt2,\ 1+\sqrt3)

(\sqrt2,\ -1+\sqrt3)

(-\sqrt2,\ 1-\sqrt3)

Answer and worked solution

Official answer: A

At x = 1: $\frac2a=a$ , so $a=\pm\sqrt2$ . At $x=\sqrt2$ : $a=\frac{2b^2-4b}{2\sqrt2}$ . For $a=\sqrt2$ , $b^2-2b-2=0$ and $b=1\pm\sqrt3$ ; for $a=-\sqrt2$ there is no real b. So $(\sqrt2,1-\sqrt3)$ works.

Topics:Continuity

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JEE Main 2016 · 10 Apr · Shift 1, Mathematics question 71