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  • Answer please! - Limit , continuity and differentiability - JEE Main-15

The function f\left ( x \right )= \tan ^{-1}\left ( \sin x+\cos x \right ) is an increasing function in

  • Option 1)

    \left ( 0,\frac{\pi }{2} \right )

  • Option 2)

    \left ( -\frac{\pi }{2},\frac{\pi }{2} \right )

  • Option 3)

    \left (\frac{\pi }{4},\frac{\pi }{2} \right )

  • Option 4)

    \left (-\frac{\pi }{2},\frac{\pi }{4} \right )

 

Answers (1)

best_answer

As we have learned

Strictly increasing functions -

A function f(x) is called strictly increasing in an interval l.

if\:for\:\:\:x_{1}<x_{2}\Rightarrow f(x_{1})<f(x_{2})
 

or\:if\:for\:\:\:x_{1}>x_{2}\Rightarrow f(x_{1})>f(x_{2})


\therefore \:\:\frac{dy}{dx}=f'(x)>0\:\:\:x\epsilon (a,b)

 

- wherein

 

 f'(x) = \frac{1}{(1+(\sin x + \cos x)^2)}\times ( \cos x - \sin x) \: \: and \: \: f'(x)> 0 \\\Rightarrow \cos x - \sin x > 0 \\ \Rightarrow \cos x > \sin x

 

 

 

 

 


Option 1)

\left ( 0,\frac{\pi }{2} \right )

Option 2)

\left ( -\frac{\pi }{2},\frac{\pi }{2} \right )

Option 3)

\left (\frac{\pi }{4},\frac{\pi }{2} \right )

Option 4)

\left (-\frac{\pi }{2},\frac{\pi }{4} \right )

Posted by

Himanshu

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