Here, f(x) is continuous at x=1 ⇒x→1−limf(x)=x→1+limf(x)
⇒a2=a⇒a=±2
Since, f(x) is continuous at x=2
⇒x→2−limf(x)=x→2+limf(x)
⇒a=222b2−4b
Put a=2
⇒2=b2−2b⇒b2−2b−2=0
⇒b=22±4+4.2=1±3
Hence, (a,b)≡(2,1±3)
Let a,b∈R,(a=0). If the function f, defined as
f(x)={\begin{matrix}\frac{2{x}^{2}}{a},0\leq x<1 \\ a,1\leq x<\sqrt{2} \\ \frac{2{b}^{2}-4b}{{x}^{3}},\sqrt{2}\leq x<8\end{matrix},is continuous in the interval [0,∞), then an ordered pair (a,b) can be
Held on 10 Apr 2016 · Verified 6 Jul 2026.
(−2,1−3)
(2,−1+3)
(2,1−3)
(−2,1+3)
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