Consider the function $\mathrm{f}:\left[\frac{1}{2}, 1\right] \rightarrow \mathrm{R}$ defined by $f(x)=4 \sqrt{2} x^{3}-3 \sqrt{2} x-1$. Consider the statements (I) The curve $\mathrm{y}=\mathrm{f}(\mathrm{x})$ intersects the x-axis exactly at one point (II) The curve $y=f(x)$ intersects the $x$-axis at $\mathrm{x}=\cos \frac{\pi}{12}$ Then

(D) Both (I) and (II) are correct

(B) Both (I) and (II) are incorrect

JEE Main · 2024 · 29 Jan 2024 · Shift 1 · Application of Derivatives

Consider the function

f : [\frac{1}{2}, 1] \to R

defined by

f (x) = 42 x^{3} - 32 x - 1

. Consider the statements
(I) The curve

y = f (x)

intersects the x-axis exactly at one point
(II) The curve

y = f (x)

intersects the

x

-axis at

x = cos \frac{π}{12}

Then

Answer: (D)

f (x) = 42 x^{3} - 32 x - 1

[\frac{1}{2}, 1]

f^{'} (x) = 122 x^{2} - 32 = 32 (4 x^{2} - 1) = 32 (2 x - 1) (2 x + 1)

[\frac{1}{2}, 1]

this is

\geq 0

(zero only at

x = \frac{1}{2}

), so

f

is increasing there.

f (\frac{1}{2}) = \frac{2}{2} - \frac{3 2}{2} - 1 = - 2 - 1 < 0, f (1) = 42 - 32 - 1 = 2 - 1 > 0

Since

f

increases continuously from negative to positive, it crosses zero exactly once: Statement I is true.

For Statement II, write

x = cos θ

. Using

4 cos^{3} θ - 3 cos θ = cos 3 θ

f (cos θ) = 2 (4 cos^{3} θ - 3 cos θ) - 1 = 2 cos 3 θ - 1

Setting this to zero:

cos 3 θ = \frac{1}{2} ⟹ 3 θ = \frac{π}{4} ⟹ θ = \frac{π}{12}

, so

x = cos \frac{π}{12}

is indeed the (unique) root. Statement II is true.

Key confirmed: D (both correct).

JEE Main · 2024 · 29 Jan 2024 · Shift 1 · Single correct option