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  • Please help! - Limit , continuity and differentiability - JEE Main-17

If the Rolle’s theorem holds for the function f(x)=2x^{3}+ax^{2}+bx  in the interval [-1,1] for the point c=\frac{1}{2} ,  then the value of  2a+b  is :

  • Option 1)

    1

  • Option 2)

    -1

  • Option 3)

    2

  • Option 4)

    -2

 

Answers (1)

best_answer

As we have learned

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.

-

 

 

Geometrical interpretation of Rolle's theorem -

Let  f(x) be a function defined on [a, b] such that the curve  y = f(x) is continuous between points  {a, f(a)}  and  {b, f(b)}  at every points on the curve encept at the end point it is possible to draw a unique tangent and ordinates at  x = a  and  x = b are equal  f(a) = f(b).

- wherein

 

 We have 

f'(1/2)= \frac{f(1)-f(-1)}{2}= 0 \\\Rightarrow (6x^2+2ax+b)|_{x= 1/2}

\frac{2+a+b-(-2+a-b)}{2}= 0 \\ \Rightarrow 3/2 +a+b = \frac{4+2b}{2}=0 \\\Rightarrow 3+ 2a +2b = 4+2b = 0 \\ \Rightarrow a = 1/2 \: \: and \: \: b = -2\\\therefore 2a+b = -1

 

 

 

 

 

 


Option 1)

1

Option 2)

-1

Option 3)

2

Option 4)

-2

Posted by

Himanshu

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