Question:
The value of b for which the function
$f(x)=\left\{\begin{array}{ll}5 x-4, & 0
(a) $-1$
(b) 0
(c) $\frac{13}{3}$
(d) 1
Solution:
(a) $-1$
Given: $f(x)$ is continuous at every point of its domain. So, it is continuous at $x=1$.
$\Rightarrow \lim _{x \rightarrow 1^{+}} f(x)=f(1)$
$\Rightarrow \lim _{h \rightarrow 0} f(1+h)=f(1)$
$\Rightarrow \lim _{h \rightarrow 0}\left(4(1+h)^{2}+3 b(1+h)\right)=5(1)-4$
$\Rightarrow 4+3 b=1$
$\Rightarrow-3=3 b$
$\Rightarrow b=-1$
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