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Question: If the function \(a = 0,b = 0\) is continuous at each point of its domain, then the value of \(a = 1...

If the function a=0,b=0a = 0,b = 0 is continuous at each point of its domain, then the value of a=1,b=−1a = 1,b = - 1 is

A

2

B

1/3

C

2/3

D

– 1/3

Answer

1/3

Explanation

Solution

f(x)=lim⁡x→0(2x−sin⁡−1x2x+tan⁡−1x)=f(0)f ( x ) = \lim _ { x \rightarrow 0 } \left( \frac { 2 x - \sin ^ { - 1 } x } { 2 x + \tan ^ { - 1 } x } \right) = f ( 0 ) ,(00\left( \frac { 0 } { 0 } \right. form ))

Applying L-Hospital’s rule, f(0)=lim⁡x→0(2−11−x2)(2+11+x2)f ( 0 ) = \lim _ { x \rightarrow 0 } \frac { \left( 2 - \frac { 1 } { \sqrt { 1 - x ^ { 2 } } } \right) } { \left( 2 + \frac { 1 } { 1 + x ^ { 2 } } \right) }

=2−12+1=13= \frac { 2 - 1 } { 2 + 1 } = \frac { 1 } { 3 }

Trick : f (0) = lim⁡x→02x−sin⁡−1x2x+tan⁡−1x\lim _ { x \rightarrow 0 } \frac { 2 x - \sin ^ { - 1 } x } { 2 x + \tan ^ { - 1 } x } ⇒lim⁡x→02−sin⁡−1xx2+tan⁡−1xx=2−12+1=13\Rightarrow \lim _ { x \rightarrow 0 } \frac { 2 - \frac { \sin ^ { - 1 } x } { x } } { 2 + \frac { \tan ^ { - 1 } x } { x } } = \frac { 2 - 1 } { 2 + 1 } = \frac { 1 } { 3 }.