JEE Main 2021 — Application Of Derivatives Question with Solution

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

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

$$2a+b=16$$ .... (1)

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

$$4a+b=32$$ .... (2)

Solving (1) and (2)

a = 8, b = 0

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

$$f^{\prime }(x)=3x^2-12x+8$$

$$f^{\prime \prime }(x)=6x-12$$

$$\Rightarrow$$ f'(x) is $$\uparrow$$ for x > 2, and f'(x) is $$\downarrow$$ for x < 2

$$f^{\prime }(2)=12-24+8=-4$$

$$f^{\prime }(4)=48-48+8=8$$

$$f^{\prime }(x)=3x^2-12x+8$$

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).