JEE Main 2024 — 4 April, Shift 2 — Question 23

$\text{Given:\; } f:\mathbb{R}\to\mathbb{R}\; \text{ is\; thrice\; differentiable\; with\; } f(0)=0, f(1)=1, f(2)=-1, f(3)=2, f(4)=-2.$

We are asked for the minimum number of zeros of $g(x) = 3 f'(x) f''(x) + f(x) f'''(x).$

$\text{Step\; 1:\; Rewrite\; } g(x)\; \text{ as\; a \; derivative:}$

$g(x) = 3 f' f'' + f f''' = \frac{d}{dx} \big( (f')^2 + f f'' \big).$

$\text{Let\; } h(x) = (f')^2 + f f'' \implies g(x) = h'(x).$

$\text{Step\; 2:\; Apply\; Rolle's\; theorem\; to } f:$

f(0)=0, f(1)=1 &\implies f'\text{ has at least 1 zero in } (0,1)? \text{(not guaranteed)}\\ f(1)=1, f(2)=-1 &\implies f'\text{ has at least 1 zero in } (1,2)\\ f(2)=-1, f(3)=2 &\implies f'\text{ has at least 1 zero in } (2,3)\\ f(3)=2, f(4)=-2 &\implies f'\text{ has at least 1 zero in } (3,4) \end{aligned}$$ $\text{Thus\; } f' \text{ has\; at\; least\; 3 \; zeros.}$ --- $\text{Step\; 3:\; Apply\; Rolle's\; theorem\; to\; } f' \text{ to\; get\; zeros\; of } f'':$ $f'' \text{ has\; at\; least\; 2\; zeros\; between\; consecutive\; zeros\; of } f'.$ --- $\text{Step \;4: \; Apply \;Rolle's\; theorem\; to\; } h(x) = (f')^2 + f f'':$ $g(x) = h'(x)\; \text{ has\; at\; least\; one\; lessv zero\; than\; number\; of\; extrema\; of\; } h(x).$ $\text{Given\; the\; pattern\; of\; } f(x),\; \text{ the\; minimum\; number\; of\; zeros\; of\; } g(x) = 4.$ --- $\boxed{4}$