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Question: \(\frac{1}{1 - x}\) is equal to...

11−x\frac{1}{1 - x} is equal to

A

[x2+12]+cos⁡−1[x2−12]\left\lbrack x^{2} + \frac{1}{2} \right\rbrack + \cos^{- 1}\left\lbrack x^{2} - \frac{1}{2} \right\rbrack

B

{π2,π}\left\{ \frac{\pi}{2},\pi \right\}

C

{π2}\left\{ \frac{\pi}{2} \right\}

D

x2+x+1\sqrt{x^{2} + x + 1}

Answer

{π2,π}\left\{ \frac{\pi}{2},\pi \right\}

Explanation

Solution

lim⁡x→π3sin⁡(π3−x)2cos⁡x−1\lim _ { x \rightarrow \frac { \pi } { 3 } } \frac { \sin \left( \frac { \pi } { 3 } - x \right) } { 2 \cos x - 1 } (form 00\frac { 0 } { 0 } )

By L′ Hospital’s rule

lim⁡x→π3−cos⁡(π3−x)−2sin⁡x=−cos⁡0−2sin⁡π3=12⋅32=13\lim _ { x \rightarrow \frac { \pi } { 3 } } \frac { - \cos \left( \frac { \pi } { 3 } - x \right) } { - 2 \sin x } = \frac { - \cos 0 } { - 2 \sin \frac { \pi } { 3 } } = \frac { 1 } { 2 \cdot \frac { \sqrt { 3 } } { 2 } } = \frac { 1 } { \sqrt { 3 } } Alternative solution:

lim⁡x→π/3sin⁡(π3−x)2cos⁡x−1=lim⁡x→π/32sin⁡(π3−x)2cos⁡(π3−x)22(cos⁡x−cos⁡π3)\lim _ { x \rightarrow \pi / 3 } \frac { \sin \left( \frac { \pi } { 3 } - x \right) } { 2 \cos x - 1 } = \lim _ { x \rightarrow \pi / 3 } \frac { 2 \sin \frac { \left( \frac { \pi } { 3 } - x \right) } { 2 } \cos \frac { \left( \frac { \pi } { 3 } - x \right) } { 2 } } { 2 \left( \cos x - \cos \frac { \pi } { 3 } \right) }

= lim⁡x→π/3sin⁡(π3−x)2cos⁡(π3−x)22sin⁡(π3−x)2sin⁡(π3+x)2=lim⁡x→π/3cos⁡(π3−x)22sin⁡(π3+x)2\lim _ { x \rightarrow \pi / 3 } \frac { \sin \frac { \left( \frac { \pi } { 3 } - x \right) } { 2 } \cos \frac { \left( \frac { \pi } { 3 } - x \right) } { 2 } } { 2 \sin \frac { \left( \frac { \pi } { 3 } - x \right) } { 2 } \sin \frac { \left( \frac { \pi } { 3 } + x \right) } { 2 } } = \lim _ { x \rightarrow \pi / 3 } \frac { \cos \frac { \left( \frac { \pi } { 3 } - x \right) } { 2 } } { 2 \sin \frac { \left( \frac { \pi } { 3 } + x \right) } { 2 } }

= 12⋅32=13\frac { 1 } { 2 \cdot \frac { \sqrt { 3 } } { 2 } } = \frac { 1 } { \sqrt { 3 } }