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MathematicsSingle correct+4 / -1Question 73 of 90

JEE Main 2014 · 9 Apr · Shift 1, Mathematics question 73

If the Rolle's theorem holds for the function f(x)=2x3+ax2+bxf(x)=2x^3+ax^2+bx in the interval [−1,1][-1,1] for the point c=12c=\dfrac12, then the value of 2a+b2a+b is :

Choose one option
Answer and worked solution

Official answer: B

  • f(−1)=f(1)f(-1)=f(1): −2+a−b=2+a+b-2+a-b=2+a+b, so b=−2b=-2.
  • f′(12)=32+a+b=0f'\left(\frac12\right)=\frac32+a+b=0, so a=12a=\frac12.

2a+b=1−2=−12a+b=1-2=-1.

Topics:Rolle's and mean value theorems
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