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  • How to solve this problem- - Limit , continuity and differentiability - JEE Main-3

Let f:R\rightarrow R   be defined by

f(x)=\left\{\begin{matrix} k-2x, & i\! fx\leq -1\\ 2x+3, & i\! fx> -1 \end{matrix}\right.

If f has a local minimum at x=-1, then a possible value of k is

  • Option 1)

    1

  • Option 2)

    0

  • Option 3)

    –1/2

  • Option 4)

    –1

 

Answers (1)

As we learnt in

Differentiability -

Let  f(x) be a real valued function defined on an open interval (a, b) and  x\epsilon (a, b).Then  the function  f(x) is said to be differentiable at   x_{\circ }   if

\lim_{h\rightarrow 0}\:\frac{f(x_{0}+h)-f(x_{0})}{(x_{0}+h)-x_{0}}


or\:\:\:\lim_{h\rightarrow 0}\:\frac{f(x)-f(x_{0})}{x-x_{0}}

-

 

 f (x) = \left\{\begin{matrix} k-2x & x\leqslant -1\\ 2x+3 & x\leqslant -1 \end{matrix}\right.

\\ \therefore k -2 (-1) = 2 (-1)+3\\ k+2 = -2+3 \\k+2=1\\k=-1

Since f (x) is differential so it must be continuous.


Option 1)

1

This option is incorrect.

Option 2)

0

This option is incorrect.

Option 3)

–1/2

This option is incorrect.

Option 4)

–1

This option is correct.

Posted by

Sabhrant Ambastha

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