If $$f(x)=x2+$$α for x≥$$02x2+1+$$β for x<$$0$$ is continuous at $$x=0$$ and $$f12=2$$ then α$$2+$$β$$2=$$

R.H.L $$=$$α, L. H.$$L=2+$$β$$f(0)=$$α ∴α$$=2+$$β$$f12=2$$⇒$$14+$$α$$=2$$⇒α$$=2$$−$$14=74$$⇒β$$=$$α−$$2=74$$−$$2=$$−$$14$$∴α$$2+$$β$$2=5016=258=3.125$$

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$If f (x) = \{\begin{array}{l} x^{2} + α for x \geq 0 \\ 2 \sqrt{x^{2} + 1} + β for x < 0 \end{array} is continuous at x = 0 and f (\frac{1}{2}) = 2 then α^{2} + β^{2} =$

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Solution

$R.H.L = α, L. H . L = 2 + β$
$f (0) = α ∴ α = 2 + β$
$\begin{array}{l} f (\frac{1}{2}) = 2 \\ \Rightarrow \frac{1}{4} + α = 2 \Rightarrow α = 2 - \frac{1}{4} = \frac{7}{4} \end{array}$
$\Rightarrow β = α - 2 = \frac{7}{4} - 2 = \frac{- 1}{4}$
$∴ α^{2} + β^{2} = \frac{50}{16} = \frac{25}{8} = 3.125$

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Continuity

Remember concepts with our Masterclasses.

If f(x)=x2+α for x≥02x2+1+β for x<0 is continuous at x=0 and f12=2 then α2+β2=

R.H.L =α, L. H.L=2+βf(0)=α ∴α=2+βf12=2⇒14+α=2⇒α=2−14=74⇒β=α−2=74−2=−14∴α2+β2=5016=258=3.125

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$If f (x) = \{\begin{array}{l} x^{2} + α for x \geq 0 \\ 2 \sqrt{x^{2} + 1} + β for x < 0 \end{array} is continuous at x = 0 and f (\frac{1}{2}) = 2 then α^{2} + β^{2} =$