If f(x)= 2 x 5 x -a, for x>0 =4 for x=0 =x^2+b-3, for x<0 is continuous at x=0, find a and b.

f(x) is continuous at x=0 (given) _ x 0^ + f(x)=f(0) _ x 0^ + ( 2 x 5 x -a )=4 _ x 0^ + 2 x 5 x - _ x 0^ + a=4 1 5 _ x 0^ + 2 x 2 x (2)- _ x 0^ + a=4 1 5 (1)(2)-a=4 [ x 0,2 x 0 2 5 -a=4 [ and _ 0^ + =1 ] 2 5 -4=a a=-18 5 _ x 0^ - (x^2+b-3 )=4

If f(x)= 2 x 5 x -a, for x>0 =4 for x=0 =x^2+b-3, for x<0 is cont…

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Question

$ \text {If } \mathrm{f}(\mathrm{x})=\frac{\sin 2 x}{5 x}-\mathrm{a}, \text { for } \mathrm{x}>0$
$ =4 \text { for } \mathrm{x}=0 $
$ =\mathrm{x}^2+\mathrm{b}-3, \text { for } \mathrm{x}<0$
is continuous at $x=0$, find $a$ and $b$.

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Answer

$f(x)$ is continuous at $x=0$ (given)
$ \therefore \lim _{x \rightarrow 0^{+}} \mathrm{f}(x)=\mathrm{f}(0)$
$\therefore \lim _{x \rightarrow 0^{+}}\left(\frac{\sin 2 x}{5 x}-\mathrm{a}\right)=4$
$\therefore \quad \lim _{x \rightarrow 0^{+}} \frac{\sin 2 x}{5 x}-\lim _{x \rightarrow 0^{+}} \mathrm{a}=4$
$\therefore \quad \frac{1}{5} \lim _{x \rightarrow 0^{+}} \frac{\sin 2 x}{2 x} \times(2)-\lim _{x \rightarrow 0^{+}} \mathrm{a}=4$
$\therefore \frac{1}{5}(1)(2)-\mathrm{a}=4 \quad[\because x \rightarrow 0,2 x \rightarrow 0$
$\therefore \frac{2}{5}-\mathrm{a}=4 \quad\left[ \text { and } \lim _{\theta \rightarrow 0^{+}} \frac{\sin \theta}{\theta}=1 \right]$
$\therefore \quad \frac{2}{5}-4=\mathrm{a}$
$\therefore \quad \mathrm{a}=-\frac{18}{5}$
$\therefore \lim _{x \rightarrow 0^{-}}\left(x^2+\mathrm{b}-3\right)=4 $

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