If f(x)={a2cos2x+b2sin2x,x≤0eax+b,x>0} f(x) is continuous at x=0 then

limx→0−a2cos2x+b2sin2x=a2limx→0+eax+b=ebLtx→0−f(x)=Ltx→0+f(x)a2=eb2log|a|=bloge2log|a|=b

If f(x)={a2cos2x+b2sin2x,x≤0eax+b,x>0} f(x) is continuous at x=0 then

limx→0−a2cos2x+b2sin2x=a2limx→0+eax+b=ebLtx→0−f(x)=Ltx→0+f(x)a2=eb2log|a|=bloge2log|a|=b

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If $f (x) = {\begin{matrix} a^{2} \cos^{2} x + b^{2} \sin^{2} x, & x \leq 0 \\ e^{a x + b}, & x > 0 \end{matrix}}$ $f (x)$ is continuous at $x = 0$ then

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$\log a = 2 \log | b |$

$2 \log | b | = e$

$a = b$

$2 \log | a | = b$

answer is A.

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$\begin{array}{l} \lim_{x \to 0^{-}} a^{2} \cos^{2} x + b^{2} \sin^{2} x = a^{2} \\ \lim_{x \to 0^{+}} e^{a x + b} = e^{b} \end{array}$

$\begin{array}{l} \underset{x \to 0^{-}}{L t} f (x) = \underset{x \to 0^{+}}{L t} f (x) \\ a^{2} = e^{b} \\ 2 \log | a | = b \log e \\ 2 \log | a | = b \end{array}$

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If f(x)={a2cos2x+b2sin2x,x≤0eax+b,x>0} f(x) is continuous at x=0 then

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If $f (x) = {\begin{matrix} a^{2} \cos^{2} x + b^{2} \sin^{2} x, & x \leq 0 \\ e^{a x + b}, & x > 0 \end{matrix}}$ $f (x)$ is continuous at $x = 0$ then