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(A) For the function $f(x)$ to be continuous at $x = 0$,the left-hand limit,right-hand limit,and the value of the function at $x = 0$ must be equal. $

Vedclass · Accepted Answer

$b = 2 \log |a|$

Vedclass · Answer

(A) For the function $f(x)$ to be continuous at $x = 0$,the left-hand limit,right-hand limit,and the value of the function at $x = 0$ must be equal. $1$. Left-hand limit $(LHL)$ at $x = 0$: $\lim_{x \rightarrow 0^-} f(x) = \lim_{h \rightarrow 0} f(0-h) = \lim_{h \rightarrow 0} (a^2 \cos^2 h + b^2 \sin^2 h) = a^2(1)^2 + b^2(0)^2 = a^2$. $2$. Right-hand limit $(RHL)$ at $x = 0$: $\lim_{x \rightarrow 0^+} f(x) = \lim_{h \rightarrow 0} f(0+h) = \lim_{h \rightarrow 0} e^{a(h)+b} = e^b$. $3$. Value of the function at $x = 0$: $f(0) = a^2 \cos^2(0) + b^2 \sin^2(0) = a^2$. Since the function is continuous,$LHL = RHL = f(0)$,therefore: $a^2 = e^b$. Taking the natural logarithm on both sides: $\ln(a^2) = \ln(e^b)$ $2 \ln |a| = b$ Thus,$b = 2 \log |a|$.

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

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