Question

The value of a for which the function
$f\left(x\right)=\left\{\begin{array}{ll}5x-4,& \mathrm{if}0<x\le 1\\ 4x^2+3ax,& \mathrm{if}1<x<2\end{array}\right.$is continuous at every point of its domain, is

(a) $\frac{13}{3}$
(b) 1
(c) 0
(d) −1

Answer 1

(d) −1

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

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

Now,
$$\begin{gathered} \underset{x\to 1^{-}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(1-h\right)=\underset{h\to 0}{\lim }\left[5\left(1-h\right)-4\right]=5-4=1 \\ \underset{x\to 1^{+}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(1+h\right)=\underset{h\to 0}{\lim }\left[4{\left(1+h\right)}^2+3a\left(1+h\right)\right]=4+3a \end{gathered}$$

Since f(x) is continuous at x = 1,
$\underset{x\to 1^{-}}{\lim }f\left(x\right)=\underset{x\to 1^{+}}{\lim }f\left(x\right)$

$$\begin{gathered} \Rightarrow 4+3a=1 \\ \Rightarrow 3a=-3 \\ \Rightarrow a=-1 \end{gathered}$$