Question

The function f(x) = $\lim_{x \rightarrow \pi/6}\left\lbrack \frac{3\sin x - \sqrt{3}\cos x}{6x - \pi} \right\rbrack$is not defined at x = π/4. The value of f(π/4) so that f is continuous at x = π/4 is –

• A. $\sqrt{3}$
• B. 1/$\frac{1}{\sqrt{3}}$
• C. 2
• D. None of these

Answer 1

Answer: (B)

1/$\frac{1}{\sqrt{3}}$

Answer 2

f$\left( \frac { \pi } { 4 } \right)$ = $\lim _ { x \rightarrow \frac { \pi } { 4 } } ( \sin 2 x ) ^ { \tan ^ { 2 } 2 x }$

= $\lim _ { x \rightarrow \frac { \pi } { 4 } } e ^ { \tan ^ { 2 } 2 x ( \sin 2 x - 1 ) }$ = $\lim _ { x \rightarrow \frac { \pi } { 4 } } e ^ { \frac { \sin 2 x - 1 } { \cot ^ { 2 } 2 x } }$ , $\frac { 0 } { 0 }$

= $\lim _ { x \rightarrow \frac { \pi } { 4 } } e ^ { \frac { 2 \cos 2 x - 0 } { 2 \cot 2 x \left( - \operatorname { cosec } ^ { 2 } 2 x \right) \cdot 2 } }$ = $\lim _ { x \rightarrow \frac { \pi } { 4 } } e ^ { \frac { - \sin 2 x } { 2 } }$ = e^–1/2