QJEEM21Maa14e: JEE Main 2021 Mathematics - Class 12 JEEM21Maa14e Solution: The function is such that . Consider two statements : Statement …

Question

Verified Solution • 2025–26 Session moderate

The function

f(x) = {x^3} - 6{x^2} + ax + b

is such that

f(2) = f(4) = 0

. Consider two statements :

Statement 1 : there exists x₁, x₂

\in

(2, 4), x₁ < x₂, such that f'(x₁) =

-

1 and f'(x₂) = 0.

Statement 2 : there exists x₃, x₄

\in

(2, 4), x₃ < x₄, such that f is decreasing in (2, x₄), increasing in (x₄, 4) and

2f'({x_3}) = \sqrt 3 f({x_4})

.

Then

Accepted Answer

f(x) = {x^3} - 6{x^2} + ax + b

f(2) = 8 - 24 + 2a + b = 0

2a + b = 16

.... (1)

f(4) = 64 - 96 + 4a + b = 0

4a + b = 32

.... (2)

Solving (1) and (2)

a = 8, b = 0

f(x) = {x^3} - 6{x^2} + 8x

f'(x) = 3{x^2} - 12x + 8

f''(x) = 6x - 12

\Rightarrow

f'(x) is

\uparrow

for x > 2, and f'(x) is

\downarrow

for x < 2

f'(2) = 12 - 24 + 8 = - 4

f'(4) = 48 - 48 + 8 = 8

f'(x) = 3{x^2} - 12x + 8

vertex (2,

-

4)

f'(2) =

-

4, f'(4) = 8, f'(3) = 27

-

36 + 8

f'(x₁) =

-

1, then x₁ = 3

f'(x₂) = 0

Again

f'(x) < 0 for x

\in

(2, x₄)

f'(x) > 0 for x

\in

(x₄, 4)

x₄

\in

(3, 4)

f(x) = x³

-

6x² + 8x

f(3) = 27

-

54 + 24 =

-

3

f(4) = 64

-

96 + 32 = 0

For x₄(3, 4)

f(x₄) <

-

3

\sqrt 3

and f'(x₃) >

-

4

2f'(x₃) >

-

8

So, 2f'(x₃) =

\sqrt 3

f(x₄)

Correct Ans. (a).

Step-by-Step Verified Solution