The value of b for which the function

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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