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  • I need help with - Limit , continuity and differentiability - JEE Main-4

f\left ( x \right )= \begin{Bmatrix} \left ( x-1 \right )^{\frac{1}{2-x}}, x> 1,x\neq 2 \\ k, x=2 \end{Bmatrix}

The value of k for which f is continuous at x=2 is :

  • Option 1)

    1

  • Option 2)

    e

  • Option 3)

    e^{-1}

  • Option 4)

    e^{-2}

 

Answers (2)

best_answer

As we learned,

 

Rule for continuous -

A function is continuous at  x = a if and only if 

        L=R=V

L.H.L    R.H.L   value at  x = a.

- wherein

Where 

L=\lim_{x\rightarrow a^{-}}\:f(x)

R=\lim_{x\rightarrow a^{+}}\:f(x)

V_{I}=\lim_{x\rightarrow a}\:f(x)

 

 

Let f\left ( x \right )= \begin{Bmatrix} \left ( x-1 \right )^{\frac{1}{2-x}}, x> 1,x\neq 2 \\ k, x=2 \end{Bmatrix}

                               

 

for x=2^{+}

Limit

\lim_{x\rightarrow 2}\left ( x-1 \right )^{\frac{1}{2-x}}=\lim_{x\rightarrow 2}\left ( 1+x-2 \right )^{\frac{1}{2-x}}

\lim_{x\rightarrow 2}=e^{\frac{x-2}{2-x}}=e^{-1}

Thus k=e^{-1}


Option 1)

1

Option 2)

e

Option 3)

e^{-1}

Option 4)

e^{-2}

Posted by

Himanshu

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