JEE Main20211 Sep 2021Evening ShiftMathematicsApplication of DerivativesActual
The function f x = x 3 - 6 x 2 + a x + b is such that f 2 = f 4 = 0 . Consider two statements: S 1 there exists x 1 , x 2 ∈ 2 , 4 , x 1 < x 2 , such that f ' x 1 = - 1 and f ' x 2 = 0 . S 2 there exists x 3 , x 4 ∈ 2 , 4 , x 3 < x 4 , such that f is decreasing in 2 , x 4 , increasing in x 4 , 4 and 2 f ' x 3 = 3 f x 4 then
Options
- AS 1 is true and S 2 is false
- Bboth S 1 and S 2 are false
- Cboth S 1 and S 2 are true
- DS 1 is false and S 2 is true
Correct answer
C. both S 1 and S 2 are true
Step-by-step solution
Given: f x = x 3 - 6 x 2 + a x + b Given that: f 2 = 0 ⇒ 2 a + b = 16         . . . . i f 4 = 0 ⇒ 4 a + b = 32         . . . . ii Solving both equations, we get a = 8 ,   b = 0 ∴     f x = x 3 - 6 x 2 + 8 x ⇒     f x = x x - 2 x - 4 Now, f ' x = 3 x 2 - 12 x + 8 Now, If f ' x = - 1 ⇒ 3 x 2 - 12 x + 8 = - 1 ⇒ 3 x 2 - 12 x + 9 = 0 ⇒ x 2 - 4 x + 3 = 0 ⇒ x 2 - 3 x - x + 3 = 0 ⇒ x - 1 x - 3 = 0 ⇒