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  • Please,please help me - The maximum value of the function on the set{ } is : - Limit , continuity and differentiability - JEE Main

 

The maximum value of the function f(x)=3x^{3}-18x^{2}+27x-40  on the set  S={   {{x\epsilon R:x^{2}+30\leq 11x}}  }  is :

  • Option 1)

     

    -122

  • Option 2)

     

    -222

  • Option 3)

     

    222

  • Option 4)

     

    122

Answers (1)

best_answer

 

Method for maxima or minima -

By second derivative method :

Step\:1.\:\:find\:values\:of\:x\:for\:\frac{dy}{dx}=0

Step\:\:2.\:\:\:x=x_{\circ }\:\:is\:a\:point\:of\:local\:maximum\:if  f''(x)<0\:\:and\:local\:minimum\:if\:f''(x)>0

- wherein

Where\:\:y=f(x)

\frac{dy}{dx}=f'(x)

 

Range of x in s,

x^{2}-11x+30\leq 0

(x-5)(x-6)\leq 0

x\varepsilon [5,6]

Now, f(x)=3x^{3}-18x^{2}+27x-40

f{x}'=9x^{2}-36x+27

=9(x-1)(x-3)

\Rightarrow {f}'(x)>0 for x\varepsilon [5,6] so max f(x)

in s will be at x=6

f(6) = 648-648+162-40=122

 


Option 1)

 

-122

Option 2)

 

-222

Option 3)

 

222

Option 4)

 

122
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