Sets, Relations & Functions
Zeros of a Function on a Closed Interval
nta_pyq_2024_jan
Grade 11
Question:
Consider the function $f:\left[\dfrac{1}{2},1\right]\to\mathbb{R}$ defined by $f(x)=4\sqrt{2}x^3-3\sqrt{2}x-1$. 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 $x=\cos\dfrac{\pi}{12}$
Then
Only (II) is correct
Both (I) and (II) are incorrect
Only (I) is correct
Both (I) and (II) are correct
Step-by-Step Solution
Key Concept: $f'(x)=12\sqrt{2}x^2-3\sqrt{2}\geq0$ on $[1/2,1]$ (since $x\geq1/2\Rightarrow12x^2\geq3$). So $f$ is increasing on $[1/2,1]$. Check endpoints. Use the substitution $x=\cos\alpha$, recognize $4\cos^3\alpha-3\cos\alpha=\cos3\alpha$ pattern.
$f'\geq0$ on $[1/2,1]$: exactly one root. $4x^3-3x=1/\sqrt{2}=\cos(\pi/4)$. $x=\cos(\pi/12)$ satisfies this. Both statements correct.
Correct Answer: 4