If f(x) = begin{cases} e^x + ax, & x

Question

(B) Given that $f(x)$ is differentiable at $x = 0$. Since every differentiable function is continuous,$f(x)$ must be continuous at $x = 0$. Thus,$\lim

Vedclass · Accepted Answer

$(-3, 1)$

Vedclass · Answer

(B) Given that $f(x)$ is differentiable at $x = 0$. Since every differentiable function is continuous,$f(x)$ must be continuous at $x = 0$. Thus,$\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^+} f(x) = f(0)$. $\lim_{x 	o 0^-} (e^x + ax) = e^0 + a(0) = 1$. $\lim_{x 	o 0^+} b(x - 1)^2 = b(0 - 1)^2 = b$. Equating these,we get $b = 1$. Since $f(x)$ is differentiable at $x = 0$,the left-hand derivative $(LHD)$ must equal the right-hand derivative $(RHD)$ at $x = 0$. $LHD = \frac{d}{dx}(e^x + ax) = e^x + a$. At $x = 0$,$LHD = e^0 + a = 1 + a$. $RHD = \frac{d}{dx}(b(x - 1)^2) = 2b(x - 1)$. At $x = 0$,$RHD = 2b(0 - 1) = -2b$. Setting $LHD = RHD$,we have $1 + a = -2b$. Substituting $b = 1$,we get $1 + a = -2(1) \implies 1 + a = -2 \implies a = -3$. Therefore,$(a, b) = (-3, 1)$.

If $f(x) = \begin{cases} e^x + ax, & x < 0 \\ b(x - 1)^2, & x \ge 0 \end{cases}$ is differentiable at $x = 0$,then $(a, b)$ is

Similar Questions

Suppose $f(x) = \begin{cases} [\cos \pi x], & x \leq 1 \\ 2\{x\} - 1, & x > 1 \end{cases}$, where $[\cdot]$ and $\{\cdot\}$ denote the greatest integer function and the fractional part of $x$ respectively, then at $x = 1$:

The function $f(x) = \begin{cases} 2x + 1, & x \in \mathbb{Q} \\ x^2 - 2x + 5, & x \notin \mathbb{Q} \end{cases}$ is

Let $S$ be the set of all points in $(-\pi, \pi)$ at which the function $f(x) = \min\{\sin x, \cos x\}$ is non-differentiable. Then $S$ is a subset of which of the following?

Let $f(x) = \begin{cases} a \sin(x + b) & x \ge 0 \\ 6x^7 - x + 1 & x < 0 \end{cases}$ be differentiable for all real $x$. If $a \in \mathbb{R}$ and $b \in [0, 2\pi]$,then the number of ordered pairs $(a, b)$ is:

If $f(x)=|x-3|$,then $f^{\prime}(3)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module