JEE MAIN - Mathematics (2021 - 1st September Evening Shift - No. 6)
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
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
both Statement 1 and Statement 2 are true
Statement 1 is false and Statement 2 is true
both Statement 1 and Statement 2 are false
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
$$f'(2) = 12 - 24 + 8 = - 4$$
$$f'(4) = 48 - 48 + 8 = 8$$
$$f'(x) = 3{x^2} - 12x + 8$$
vertex (2, $$-$$4)
f'(2) = $$-$$4, f'(4) = 8, f'(3) = 27 $$-$$ 36 + 8
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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).
$$f(2) = 8 - 24 + 2a + b = 0$$
$$2a + b = 16$$ .... (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).
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