If the function g(x) = begin{cases} ae^x, & x | vedclass

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

Question

(B) For the function $g(x)$ to be differentiable at $x = 0$,it must first be continuous at $x = 0$.Continuity at $x = 0$ implies $\lim_{x 	o 0^-} g(x

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) For the function $g(x)$ to be differentiable at $x = 0$,it must first be continuous at $x = 0$.Continuity at $x = 0$ implies $\lim_{x 	o 0^-} g(x) = \lim_{x 	o 0^+} g(x) = g(0)$.$\lim_{x 	o 0^-} ae^x = ae^0 = a$.$\lim_{x 	o 0^+} (b\cos x + x) = b\cos(0) + 0 = b$.Thus,$a = b$  .......$(1)$For differentiability,the left-hand derivative $(LHD)$ must equal the right-hand derivative $(RHD)$ at $x = 0$.$LHD = \frac{d}{dx}(ae^x) = ae^x$. At $x = 0$,$LHD = a$.$RHD = \frac{d}{dx}(b\cos x + x) = -b\sin x + 1$. At $x = 0$,$RHD = -b\sin(0) + 1 = 1$.Equating $LHD = RHD$,we get $a = 1$.Since $a = b$,we have $b = 1$.Therefore,$a^2 + b^2 = 1^2 + 1^2 = 1 + 1 = 2$.

If the function $g(x) = \begin{cases} ae^x, & x \le 0 \\ b\cos x + x, & x > 0 \end{cases}$ is differentiable,then the value of $a^2 + b^2$ is

Similar Questions

If $f(x) = \begin{cases} \frac{x-1}{2x^2-7x+5}, & \text{for } x \neq 1 \\ -\frac{1}{3}, & \text{for } x=1 \end{cases}$,then $f^{\prime}(1)$ is equal to:

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

The number of points,where the function $f: R \rightarrow R, f(x) = |x-1| \cos |x-2| \sin |x-1| + (x-3)|x^2-5x+4|$ is $NOT$ differentiable,is:

If $f(x + y) = f(x) + f(y) + |x|y + xy^2$,$\forall x, y \in R$ and $f'(0) = 0$,then

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module