Let a and b be real numbers such that the function

$g(x)=\left\{\begin{array}{c} -3 a x^2-2, x<1 \\ b x+a^2, x \geq 1 \end{array}\right.$  is differentiable for all x ∈ R

Then the possible value(s) of a is (are)

1, 2

It is given that f(x) is differentiable for all x ∈ R. So, it has to be differentiable and hence continuous at x = 1 also.

Continuity of f(x) at x = 1 : If f(x) is continuous at x = 1, then

$\lim\limits_{x \rightarrow 1^{-}} f(x)=\lim\limits_{x \rightarrow 1^{+}} f(x)=f(1)$

$\Rightarrow \lim\limits_{x \rightarrow 1^{-}}\left(-3 a x^2-2\right)=\lim\limits_{x \rightarrow 1^{+}}\left(b x+a^2\right)=b \times 1+a^2$

$\Rightarrow -3 a-2=b+a^2$

$\Rightarrow a^2+3 a+2+b=0$             .......(i)

Differentiability of f(x) at x = 1 : If f(x) is differentiable at x = 1, then

LHD of f(x) at x = 1) = (RHD of f(x) at x = 1)

$\Rightarrow \left\{\frac{d}{d x}\left(-3 a x^2-2\right)\right\}_{x=1}=\left\{\frac{d}{d x}\left(b x+a^2\right)\right\}_{x=1}$

⇒ - 6a = b               ......(ii)

Solving (i) and (ii), we get: a = 1, b = -6 or, a = 2, b = -12.