Question

Statement I If $f$ is a constant function i.e., $f(x)=\lambda$ for some real number $\lambda$ and $g(x)$ is continuous function, then the function $(\lambda \cdot g)$ defined by $(\lambda \cdot g)(x)=\lambda \cdot g(x)$ is also continuous. In particular, if $\lambda=-1$, then continuity of $f$ implies continuity of $-f$
Statement II If $f$ is a constant function, $f(x)=\lambda$ and $g(x)$ is continuous function, then the function $\frac{\lambda}{g}$ defined by $\frac{\lambda}{g}(x)=\frac{\lambda}{g(x)}$ is also continuous wherever $g(x) \neq 0$. In particular, the continuity of $g$ implies continuity of $\frac{1}{g}$.

• A. Statement I is true; statement II is true; statement II is a correct explanation for statement I
• B. Statement I is true; statement II is true; statement II is not a correct explanation for statement I
• C. Statement I is true; statement II is false
• D. Statement I is false; statement II is true

Answer 1

Answer: (B)

(i) $\displaystyle\lim _{x \rightarrow C}(\lambda \cdot g)(x)=\displaystyle\lim _{x \rightarrow c} \lambda g(x) $
$ =\lambda\displaystyle \lim _{x \rightarrow c} g(x) (\because \lambda \text { is constant }) $
$=\lambda g(c) {\left[\because \displaystyle\lim _{x \rightarrow c} g(x)=g(c)\right]} $
$ =(\lambda g)(c)$
$\Rightarrow(\lambda \cdot g)(x)$ is continuous at $x=c$.
Thus, for $\lambda=-1$, the continuity of $f$ implies continuity of $-f$.
(ii) $\displaystyle\lim _{x \rightarrow c}\left(\frac{\lambda}{g}\right)(x)=\displaystyle\lim _{x \rightarrow c} \frac{\lambda}{g(x)} $
$ =\lambda \displaystyle\lim _{x \rightarrow c} \frac{1}{g(x)}=\lambda \cdot \frac{1}{\displaystyle\lim _{x \rightarrow c} g(x)} $
$=\lambda \cdot \frac{1}{g(c)} \left[\because \displaystyle\lim _{x \rightarrow c} g(x)=g(c)\right] $
$ =\frac{\lambda}{g(c)} [\because g(c) \neq 0]$
$\Rightarrow\left(\frac{\lambda}{g}\right)(x)=\frac{\lambda}{g(x)}$ is continuous.
In particular, for $\lambda=1$, the continuity of $g$ implies continuity of $\frac{1}{g}$.