If f(x)=x−4|x−4|+a,x<4a+b, x=4x−4|x−4|+b,x>4 Then, f(x) is continuous at x=4 when a2+b2=

If f(x)=x−4|x−4|+a,x4 Then, f(x) is continuous at x=4 when a2+b2=

Solution:

We have,

$\begin{aligned} lim_{x \to 4^{-}} f (x) = lim_{x \to 4^{-}} \frac{x - 4}{| x - 4 |} + a = lim_{x \to 4^{-}} \frac{x - 4}{- (x - 4)} + a = a - 1 \\ lim_{x \to 4^{+}} f (x) = lim_{x \to 4^{+}} \frac{x - 4}{| x - 4 |} + b = lim_{x \to 4^{+}} \frac{x - 4}{x - 4} + b = b + 1 \end{aligned}$

and, $f (4) = a + b$

If f(x) is continuous at x = 4, then

$\begin{array}{l} lim_{x \to 4^{-}} f (x) = f (4) = lim_{x \to 4^{+}} f (x) \\ a - 1 = a + b = b + 1 \Rightarrow b = - 1 and a = 1 \Rightarrow a^{2} + b^{2} = 2 \end{array}$

$\Rightarrow f (| x |) = \{\begin{array}{r} - 2 x + 1, x < 0 \\ 2 x + 1, x \geq 0 \end{array}$

We find that f(x) is neither continuous nor differentiable at $x = 0 .$ . At all other points $f (x)$ is continuous and differentiable. $f (| x |)$ is every where continuous but not differentiable at $x = 0 .$

If $f (x) = \{\begin{array}{cc} \frac{x - 4}{| x - 4 |} + a, & x < 4 \\ a + b & , x = 4 \\ \frac{x - 4}{| x - 4 |} + b, & x > 4 \end{array}$ Then, f(x) is continuous at $x = 4 when a^{2} + b^{2} =$

Solution:

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