JEE Main20257 Apr 2025Evening ShiftMathematicsApplication of DerivativesActual
Let f: R R be a polynomial function of degree four having extreme values at x=4 and x=5 . If _ x 0 f(x) x^2 =5 , then f(2) is equal to :
Options
- A12
- B10
- C8
- D14
Correct answer
B. 10
Step-by-step solution
aligned & _ x 0 f(x) x^2 =5 & _ x 0 .a x^4+b x^3+c x^2+d x+e ) x^2 =5 & c=5 and d=e=0 & f(x)=a x^4+b x^3+5 x^2 & f^ (x)=4 a x^3+3 b x^2+10 x & =x (4 a x^2+3 b x+10 ) aligned has extremes at 4 and so f^ (4)=0 & f^ (5)=0 so a = 1 8 & ~b = -3 2 so f(2)= 1 8 2^4- 3 2 2^3+5 2^2 =2-12+20=10