JEE Main 2021 (Online) 1st September Evening Shift
MCQ (Single Correct Answer)
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
A
both Statement 1 and Statement 2 are true
B
Statement 1 is false and Statement 2 is true
C
both Statement 1 and Statement 2 are false
D
Statement 1 is true and Statement 2 is false
Explanation
$$f(x) = {x^3} - 6{x^2} + ax + b$$
$$f(2) = 8 - 24 + 2a + b = 0$$
$$2a + b = 16$$ .... (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