Question

If ƒ(x) = $\frac{1}{4}f(x) = \left\{ \begin{aligned} & \begin{matrix} x\sin\frac{1}{x}, & x \neq 0 \end{matrix} \\ & \begin{matrix} 0, & x = 0 \end{matrix} \end{aligned} \right.\$ for x ∈R, then ƒ(x) is

continuous in –

• A. (–∞, ∞)
• B. (–∞, ∞) – $\lim_{x \rightarrow 0}f(x)$
• C. (–∞, ∞) – $f(x) = \left\{ \begin{aligned} & \begin{matrix} \frac{\sin\lbrack x\rbrack}{\lbrack x\rbrack}, & \lbrack x\rbrack \neq 0 \end{matrix} \\ & \begin{matrix} 0, & \lbrack x\rbrack = 0 \end{matrix} \end{aligned} \right.\$
• D. None of these

Answer 1

Answer: (B)

(–∞, ∞) – $\lim_{x \rightarrow 0}f(x)$

Answer 2

If |2 sin x| < 1, ƒ(x)

= $\lim _ { n \rightarrow \infty }$ $\frac { x } { ( 2 \sin x ) ^ { 2 n } + 1 }$ = $\frac { \mathrm { x } } { 0 + 1 }$ = x.

If | 2 sin x | = 1, ƒ(x) = $\lim _ { n \rightarrow \infty }$ $\frac { x } { ( 2 \sin x ) ^ { 2 n } + 1 }$ = $\frac { \mathrm { x } } { 1 + 1 }$ = $\frac { 1 } { 2 }$x..

If | 2 sin x | > 1, ƒ(x) = $\lim _ { n \rightarrow \infty }$ $\frac { x } { ( 2 \sin x ) ^ { 2 n } + 1 }$

= $\frac { x } { \infty + 1 }$ = 0.

Also, | 2 sin x | < 1, ⇒ | sin x | < $\frac { 1 } { 2 }$

⇒ $\frac { - 1 } { 2 }$ < sin x < $\frac { 1 } { 2 }$ ⇒ nπ – $\frac { \pi } { 6 }$ < x < nπ +$\frac { \pi } { 6 }$,

| 2 sin x | = 1 ⇒ | sin x | = $\frac { 1 } { 2 }$

⇒ sin x = ±$\frac { 1 } { 2 }$ ⇒ x = nπ ± $\frac { \pi } { 6 }$.

and |2 sin x| > 1⇒ | sin x | > $\frac { 1 } { 2 }$

⇒ sin x > $\frac { 1 } { 2 }$ or sin x < $\frac { - 1 } { 2 }$

⇒ x = nπ ± $\frac { \pi } { 6 }$.

Thus, we have

ƒ(x) = The only doubtful points are x = nπ ± $\frac { \pi } { 6 }$.

Clearly, ƒ(x) is not continuous at x = nπ ±$\frac { \pi } { 6 }$.

∴ ƒ(x) is continuous in (–∞, ∞)

– $\left\{ \mathrm { x } \left\lvert \, \mathrm { x } = \mathrm { n } \pi \pm \frac { \pi } { 6 } \right. , \mathrm { n } \in \mathrm { I } \right\}$