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