If f:R→R defined by f(x)=a2cos2x+x2sin2x, x≤0 eax+b, x>0 is continuous then

Ltx→0−f(x)=Ltx→0+f(x)Ltx→0−(a2cos2x+x2sin2x)=Ltx→0+eax+ba2=eb2loga=bloge∴b=2logea

If f:R→R defined by f(x)=a2cos2x+x2sin2x, x≤0 eax+b, x>0 is continuous then

Ltx→0−f(x)=Ltx→0+f(x)Ltx→0−(a2cos2x+x2sin2x)=Ltx→0+eax+ba2=eb2loga=bloge∴b=2logea

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If $f : R \to R$ defined by $f (x) = \{\begin{cases} a^{2} \cos^{2} x + x^{2} \sin^{2} x, x \leq 0 \\ e^{a x + b}, x > 0 \end{cases}$ is continuous then

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$2 b = \log | a |$

$b = 2 \log | a |$

$b^{2} = \log | a |$

$b = \log | 2 a |$

answer is A.

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$\underset{x \to 0^{-}}{L t} f (x) = \underset{x \to 0^{+}}{L t} f (x)$

$\underset{x \to 0^{-}}{L t} (a^{2} \cos^{2} x + x^{2} \sin^{2} x) = \underset{x \to 0^{+}}{L t} e^{a x + b}$

$a^{2} = e^{b}$

$2 \log a = b \log e$

$∴ b = 2 \log_{e} a$

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If f:R→R defined by f(x)=a2cos2x+x2sin2x, x≤0 eax+b, x>0 is continuous then

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If $f : R \to R$ defined by $f (x) = \{\begin{cases} a^{2} \cos^{2} x + x^{2} \sin^{2} x, x \leq 0 \\ e^{a x + b}, x > 0 \end{cases}$ is continuous then