If the function f(x)=begin{cases} frac{tan a( | vedclass

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(A) For $f(x)$ to be continuous at $x=1$, $\lim_{x 	o 1^-} f(x) = \lim_{x 	o 1^+} f(x)$.$\lim_{x 	o 1^-} \frac{	an a(x-1)}{x-1} = a$.$\lim_{x 	o

Vedclass · Accepted Answer

$284$

Vedclass · Answer

(A) For $f(x)$ to be continuous at $x=1$, $\lim_{x 	o 1^-} f(x) = \lim_{x 	o 1^+} f(x)$.$\lim_{x 	o 1^-} \frac{	an a(x-1)}{x-1} = a$.$\lim_{x 	o 1^+} \frac{x^3-125}{x^2-25} = \frac{1-125}{1-25} = \frac{-124}{-24} = \frac{31}{6}$.Thus, $a = \frac{31}{6}$.For $f(x)$ to be continuous at $x=4$, $\lim_{x 	o 4^-} f(x) = \lim_{x 	o 4^+} f(x)$.$\lim_{x 	o 4^-} \frac{x^3-125}{x^2-25} = \frac{64-125}{16-25} = \frac{-61}{-9} = \frac{61}{9}$.$\lim_{x 	o 4^+} \frac{b^x-1}{x} = \frac{b^4-1}{4}$.Equating them: $\frac{61}{9} = \frac{b^4-1}{4} \Rightarrow 244 = 9b^4 - 9 \Rightarrow 9b^4 = 253$.Now, $6a + 9b^4 = 6(\frac{31}{6}) + 253 = 31 + 253 = 284$.

If the function $f(x)=\begin{cases} \frac{\tan a(x-1)}{x-1}, & \text{if } 0 < x < 1 \\ \frac{x^3-125}{x^2-25}, & \text{if } 1 \leq x \leq 4 \\ \frac{b^x-1}{x}, & \text{if } x > 4 \end{cases}$ is continuous in its domain, then $6a + 9b^4 = $

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