If Rolle's theorem holds for the function

f(x)= 2x^{3}+bx^{2}+cx,x\epsilon \left [ -1,1 \right ]  at the point

x= \frac{1}{2},\: \: then\: \: 2b+cequals to:

  • Option 1)

    1

  • Option 2)

    -1

  • Option 3)

    2

  • Option 4)

    -3

 

Answers (1)

As we learnt in 

Rolle's Theorems -

Let f(x) be a function of x subject to the following conditions.

1.  f(x) is continuous function of    x:x\epsilon [a,b]

2.  f'(x) is exists for every point :  x\epsilon [a,b]

3.  f(a)=f(b)\:\:\:then\:\:f'(c)=0\:\:such \:that\:\:a<c<b.

-

 

 f(x)= 2x^{3}+bx^{2}+cx  

f(1)= 2+b+c

f(-1)= -2+b-c

\therefore f(1) =f(-1)

=>b+c+2=b-2-c

=>c+2=0

\therefore c=-2

now

f'(x)=6x^{2}+2bx+c

0=6\left ( \frac{1}{2} \right )^{2}+b\times 2\times \frac{1}{2}-2

0=\frac{6}{4}+b-2

0=b+\left ( \frac{-1}{2} \right )

=>b=\frac{1}{2}

2b+c=-1

 


Option 1)

1

Incorrect option    

Option 2)

-1

Correct option

Option 3)

2

Incorrect option    

Option 4)

-3

Incorrect option    

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