MCQ
If $f(x) = \left\{ {\begin{array}{*{20}{c}}{{e^x} + ax,}&{x < 0}\\{b{{(x - 1)}^2},}&{x \ge 0}\end{array}} \right.$ is differentiable at $x = 0,$ then $(a,\,b)$ is
- A$( - 3,\, - 1)$
- ✓$( - 3,\,\,1)$
- C$(3,\,\,1)$
- D$(3,\, - 1)$
Hence, $f(x)$ will be continuous at $x = 0$.
$\mathop {{\rm{lim}}}\limits_{x \to {0^ - }} ({e^x} + ax) = \mathop {{\rm{lim}}}\limits_{x \to {0^ + }} b{(x - 1)^2}$
==> ${e^0} + a \times 0 = b{(0 - 1)^2}$ ==> $b = 1$….. $(i)$
But $f(x)$ is differentiable at $x = 0$, then
$Lf'(x) = Rf'(x)$ ==> $\frac{d}{{dx}}({e^x} + ax) = \frac{d}{{dx}}b{(x - 1)^2}$
==> ${e^x} + a = 2b(x - 1)$
At $x = 0,$ ${e^0} + a = - 2b$ ==> $a + 1 = - 2b$ ==> $a = - 3$
==> $(a,\,\,b) = ( - 3,\,\,1)$.
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