If f(x) = begin{cases} frac{9^x - 2 cdot 3^x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) For $f(x)$ to be continuous at $x = 0$,we must have $\lim_{x 	o 0} f(x) = f(0)$.First,evaluate the limit: $\lim_{x 	o 0} \frac{9^x - 2 \cdot 3^x

Vedclass · Accepted Answer

$\frac{31}{6}$

Vedclass · Answer

(A) For $f(x)$ to be continuous at $x = 0$,we must have $\lim_{x 	o 0} f(x) = f(0)$.First,evaluate the limit: $\lim_{x 	o 0} \frac{9^x - 2 \cdot 3^x + 1}{\log(1 + 3x) \cdot 	an 2x}$.Note that $9^x - 2 \cdot 3^x + 1 = (3^x - 1)^2$.So,the limit is $\lim_{x 	o 0} \frac{(3^x - 1)^2}{\log(1 + 3x) \cdot 	an 2x}$.Using standard limits $\lim_{x 	o 0} \frac{3^x - 1}{x} = \log 3$,$\lim_{x 	o 0} \frac{\log(1 + 3x)}{3x} = 1$,and $\lim_{x 	o 0} \frac{	an 2x}{2x} = 1$:Limit $= \lim_{x 	o 0} \frac{(\frac{3^x - 1}{x})^2 \cdot x^2}{(\frac{\log(1 + 3x)}{3x} \cdot 3x) \cdot (\frac{	an 2x}{2x} \cdot 2x)} = \frac{(\log 3)^2}{3 \cdot 2} = \frac{(\log 3)^2}{6}$.Given $f(0) = a(\log b)^c$,we have $a(\log b)^c = \frac{1}{6}(\log 3)^2$.Comparing terms,we get $a = \frac{1}{6}$,$b = 3$,and $c = 2$.Therefore,$a + b + c = \frac{1}{6} + 3 + 2 = \frac{1}{6} + 5 = \frac{31}{6}$.

If $f(x) = \begin{cases} \frac{9^x - 2 \cdot 3^x + 1}{\log(1 + 3x) \cdot \tan 2x} & , x \neq 0 \\ a(\log b)^c & , x = 0 \end{cases}$ is continuous at $x = 0$,then $a + b + c =$

Similar Questions

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 = $

If $f(x) = \begin{cases} 1 + x^2, & \text{when } 0 \le x \le 1 \\ 1 - x, & \text{when } x > 1 \end{cases}$,then

Show that the function $f$ defined by $f(x) = |1 - x + |x||$,where $x$ is any real number,is a continuous function.

$f(x) = \begin{cases} [x] + [-x], & \text{when } x \neq 2 \\ \lambda, & \text{when } x = 2 \end{cases}$
If $f(x)$ is continuous at $x = 2$,the value of $\lambda$ will be

The function $f(x) = \frac{2x^2 + 7}{x^3 + 3x^2 - x - 3}$ is discontinuous for

Vedclass Test Series

Exam Paper Generator

Online Exam Module