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  • Can someone explain - Limit , continuity and differentiability - JEE Main-16

Let a function f(x) is such that f'(x) = x^3(x-2)^2 then number of points of inflexion of f(x) when f(x) is defined on \mathbb R will be ?

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  • Option 2)

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Answers (1)

best_answer

As we have learned

Condition for inflexion -

1.  f''(x) changes sign x passes through the point  a.

2.  only those values of  x  for  f''(x)  change signs are points of inflexion.

ex:\:\:\:f''(x)=x^{2}(x-2),\:\:\:f''(x)=0

at\:\:\:x=0\:\:and\:\:x=2\:\:\:but\:only\:\:\:x=2  point\:of\:inflexion\:because\:at\:x=2\:\:\:f''(x)\:\:changes\:sign\:only

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 \because f'(x)= x^{3}(x-2)^{2}

\Rightarrow f''(x)= x^{3}\cdot 2(x-2)+ (x-2)^{2}\cdot 3x^{2}

\Rightarrow f''(x)= x^{2}\cdot (x-2)(2x+3x-6)

\Rightarrow f''(x)=x^{2}(x-2)(5x-6)

This f'(x) will change sign only about x=2 and x=6/5 so only two points of inflextion

 

 

 

 

 

 

 


Option 1)

0

Option 2)

1

Option 3)

2

Option 4)

3

Posted by

Himanshu

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