If $f (x)$ = ${x^{2} + 3 x + a, f o r x \leq 1; b x + 2, f o r x > 1}$
is everywhere differnetiable , find the values of a and b

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Solution

Given :
$f (x) = x^{2} + 3 x + a$ , for $x \leq 1$
$= b x + 2$ , for $x > 1$
The function is everywhere differentiable. So, it is also continuous.
Therefore,
$l i m_{x \to 1} f (x) = l i m_{x \to 1^{+}} f (x)$
$l i m_{x \to 1} x^{2} + 3 x + a = l i m_{x \to 1} b x + 2$
$4 + a = b + 2$
$a - b = - 2$ ......(1)
Now, $f (x)$ is differentiable at $x = 1$ , therefore,
$l i m_{x \to 1^{- 1}} \frac{f (x) - f (1)}{x - 1} = l i m_{x \to 1^{+}} \frac{f (x) - f (1)}{x - 1}$
$l i m_{x \to 1} \frac{(x^{2} + 3 x + a) - (a + 4)}{x - 1} = l i m_{x \to 1} \frac{(b x + 2) - (b + 2)}{x - 1}$
$l i m_{x \to 1} \frac{x^{2} + 3 x - 4}{x - 1} = l i m_{x \to 1} \frac{b x - b}{x - 1}$
$l i m_{x \to 1} \frac{(x + 4) (x - 1)}{x - 1} = l i m_{x \to 1} \frac{b (x - 1)}{x - 1}$
$l i m_{x \to 1} (x + 4) = l i m_{x \to 1} b$
$b = 5$
Therefore, $a = 3$

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