MHT CET202313 May 2023Evening ShiftMathematicsApplication of DerivativesActual
If Rolle's theorem holds for the function f(x)=x^3+b x^2+a x+5 on [1,3] with c=2+ 1 3 , then the values of a and b respectively are
Options
- A-11,-6
- B11,6
- C11,-6
- D6,11
Correct answer
C. 11,-6
Step-by-step solution
Since f (x) satisfies the Rolle's theorem, f(1)=f(3) 1+b+a+5=27+9 b+3 a+5 2 a+8 b=-26 a+4 b=-13 ...(i) f (x)=x^3+b x^2+a x+5 f ^ (x)=3 x^2+2 ~b x+ a Now, f ^ ( c )=0 aligned & f ^ (2+ 1 3 )=0 & 3 (2+ 1 3 )^2+2 ~b (2+ 1 3 )+ a =0 & 3 (4+ 4 3 + 1 3 )+4 ~b + 2 ~b 3 + a =0 & a +4 ~b + 2 ~b +12 3 +13=0 & -13+ 2 ~b +12 3 +13=0 & 2 ~b +12 3 =0 & b =-6 aligned Substituting b=-6 in (i), we get a=11