The value of a for which the function

(d) $-1$

Given: $f(x)=\left\{\begin{array}{l}5 x-4, \text { if } 0<\mathrm{x} \leq 1 \\ 4 x^{2}+3 a x, \text { if } 1<\mathrm{x}<2\end{array}\right.$

If $f(x)$ is continuous in its domain, then it will be continuous at $x=1$.

Now,

$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0} f(1-h)=\lim _{h \rightarrow 0}[5(1-h)-4]=5-4=1$

$\lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} f(1+h)=\lim _{h \rightarrow 0}\left[4(1+h)^{2}+3 a(1+h)\right]=4+3 a$

Since f(x) is continuous at x = 1,

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

$\Rightarrow 4+3 a=1$

$\Rightarrow 3 a=-3$

$\Rightarrow a=-1$