If $$f(x) = \left\{ {\matrix{ {{1 \over {|x|}}} & {;\,|x|\, \ge 1} \cr {a{x^2} + b} & {;\,|x|\, < 1} \cr } } \right.$$ is differentiable at every point of the domain, then the values of a and b are respectively :
Solution
$$f(x) = \left\{ {\matrix{
{{1 \over {|x|}},} & {|x| \ge 1} \cr
{a{x^2} + b,} & {|x| < 1} \cr
} } \right.$$<br><br>$$ = \left\{ {\matrix{
{ - {1 \over x};} & {x \le - 1} \cr
{a{x^2} + b;} & { - 1 < x < 1} \cr
{{1 \over x};} & {x \ge 1} \cr
} } \right.$$<br><br>As f(x) is differentiable so it is also continuous,<br><br> at x = 1,<br><br>$$\mathop {\lim }\limits_{x \to {1^ - }} f(x) = \mathop {\lim }\limits_{x \to {1^ + }} f(x)$$<br><br>$\Rightarrow a + b = {1 \over 1}$<br><br>$\Rightarrow a + b = 1$ ...... (1)<br><br>As f(x) is differentiable, so at x = 1<br><br>L.H.D. = R.H.D.<br><br>$\Rightarrow 2ax = - {1 \over {{x^2}}}$<br><br>$\Rightarrow 2a = - 1$<br><br>$\Rightarrow a = - {1 \over 2}$<br><br>From (1), $b = {3 \over 2}$
About this question
Subject: Mathematics · Chapter: Limits, Continuity and Differentiability · Topic: Limits and Standard Results
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