Let $a, b \in(a
eq 0)$. If the function $f$ is defined as $f(x)=\left\{\begin{array}{cc}\frac{2 x^2}{\mathrm{a}} & , 0 \leq x \lt 1 \ \mathrm{a} & , 1 \leq x \lt \sqrt{2} \ \frac{2 \mathrm{~b}^2-4 \mathrm{~b}}{x} & , \sqrt{2} \leq x \lt \infty\end{array} ight.$ is continuous in the interval…

Question

Let $a, b \in(a 
eq 0)$. If the function $f$ is defined as $f(x)=\left\{\begin{array}{cc}\frac{2 x^2}{\mathrm{a}} & , 0 \leq x \lt 1 \ \mathrm{a} & , 1 \leq x \lt \sqrt{2} \ \frac{2 \mathrm{~b}^2-4 \mathrm{~b}}{x} & , \sqrt{2} \leq x \lt \infty\end{array}ight.$ is continuous in the interval $[0, \infty)$, then an ordered pair $(\mathrm{a}, \mathrm{b})$ is

Aicharya · Accepted Answer

Function: $f(x)= \begin{cases}\frac{2 x^2}{a}, & 0 \leq x \lt 1 \\ a, & 1 \leq x \lt \sqrt{2} \\ \frac{2 b^2-4 b}{x}, & \sqrt{2} \leq x \lt \infty\end{cases}$
Condition: Function $f(x)$ must be continuous in the interval $[0, \infty)$. This implies: $\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x), \quad \text { and } \quad \lim _{x \rightarrow \sqrt{2}^{-}} f(x)=\lim _{x \rightarrow \sqrt{2}^{+}} f(x) .$
Step 1: Continuity at $x=1$ : $\lim _{x \rightarrow 1^{-}} f(x)=\frac{2(1)^2}{a}=\frac{2}{a}, \quad \lim _{x \rightarrow 1^{+}} f(x)=a .$
Equating these: $\frac{2}{a}=a \quad \Rightarrow \quad a^2=2 \quad \Rightarrow \quad a= \pm \sqrt{2}$
Step 2: Continuity at $x=\sqrt{2}$ : $\lim _{x \rightarrow \sqrt{2}^{-}} f(x)=a, \quad \lim _{x \rightarrow \sqrt{2}^{+}} f(x)=\frac{2 b^2-4 b}{\sqrt{2}}$
Equating these: $a=\frac{2 b^2-4 b}{\sqrt{2}} \Rightarrow a \sqrt{2}=2 b^2-4 b .$
Substitute $a=\sqrt{2}$ : $(\sqrt{2})(\sqrt{2})=2 b^2-4 b \quad \Rightarrow \quad 2=2 b^2-4 b$
Simplify: $b^2-2 b-1=0$
Solve for $b$ : $b=\frac{2 \pm \sqrt{4+4}}{2}=1 \pm \sqrt{3} .$
Final Values: $a=\sqrt{2}, b=1-\sqrt{3}$. Answer: $(\sqrt{2}, 1-\sqrt{3})$, Option 3.

Let $a, b \in(a \neq 0)$. If the function $f$ is defined as $f(x)=\left\{\begin{array}{cc}\frac{2…

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