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  • Solve! - Limit , continuity and differentiability - JEE Main-18

Let S= \left \{ \left ( \lambda ,\mu \right )\epsilon \mathbf{R}\times \mathbf{R} :f\left ( t \right )=\left (\left | \lambda \right |e^{\left | t \right |} -\mu \right )\cdot \sin \left ( 2\left | t \right | \right ),t\epsilon \mathbf{R},is\; a\; differential\; function\right \}. Then S ia a subset of : 

  • Option 1)

    \mathbf{R}\times \left [0,\infty \right ]

     

     

     

  • Option 2)

    \left [ 0,\infty \right ]\times \mathbf{R}

  • Option 3)

    \mathbf{R}\times \left [-\infty ,0 \right ]

  • Option 4)

    \left [-\infty ,0 \right ]\times\mathbf{R}

 

Answers (2)

best_answer

As we learned 

@3484

@3488

@3509

 

f\left ( t \right )=\left ( \left | \lambda \right |e^{\left | t \right |} -\mu \right )\sin \left ( 2\left | t \right | \right )

=\left ( \left | \lambda \right |e^{t}-\mu \right )\sin 2t\; \; \; \; t> 0

=\left ( \left | \lambda \right |e^{-t}-\mu \right )\sin 2t\; \; \; \; t< 0

f(t)  is differentiable at t=0 

LHD = RHD

\left | \lambda \right |\sin 2\left ( 0 \right )+\left ( \left | \lambda \right |e^{\circ} -\mu \right )2\cos \infty =\left | \lambda \right |\sin 2\left ( \infty \right )-2\cos \left ( 0 \right )\left ( \lambda e^{-0} -\mu \right )

\Rightarrow 4\left ( \left | \lambda \right |-\mu \right )=0\Rightarrow \left | \lambda \right |=\mu

S\equiv \left ( \lambda ,\mu \right )=\left \{ \lambda \equiv R,\: \mu \equiv \left ( 0,\infty \right ) \right \}

S is subset of R\times \left ( 0,\infty \right )


Option 1)

\mathbf{R}\times \left [0,\infty \right ]

 

 

 

Option 2)

\left [ 0,\infty \right ]\times \mathbf{R}

Option 3)

\mathbf{R}\times \left [-\infty ,0 \right ]

Option 4)

\left [-\infty ,0 \right ]\times\mathbf{R}

Posted by

Himanshu

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