Q56 / 90 · JEE Main 2024 (04 Apr Shift 2) · Mathematics
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a thrice differentiable function such that $f(0)=0, f(1)=1, f(2)=-1, f(3)=2$ and $f(4)=-2$. Then, the minimum number of zeros of $\left(3 f^{\prime} f^{\prime \prime}+f f^{\prime \prime \prime}\right)(x)$ is _______
Correct answer: 5
Solution
$\begin{aligned} & \left(3 f^{\prime} f^{\prime \prime}+f f^{\prime \prime \prime}\right)(x)=\left(\left(f f^{\prime \prime}+\left(f^{\prime}\right)^2\right)(x)\right)^{\prime} \\ & \left(f f f^{\prime \prime}+\left(f^{\prime}\right)^2\right)(x)=\left(\left(f f^{\prime}\right)(x)\right)^{\prime} \\ & \therefore\left(3 f^{\prime} f^{\prime \prime}+f^{\prime \prime \prime}\right)(x)=\left(f(x) \cdot f^{\prime}(x)\right)^{\prime \prime}\end{aligned}$
$\begin{aligned} & \min \text { roots of } \mathrm{f}(\mathrm{x}) \rightarrow 4 \\ & \therefore \min \text {. roots of } \mathrm{f}^{\prime}(\mathrm{x}) \rightarrow 3 \\ & \therefore \min \text { roots of }\left(\mathrm{f}(\mathrm{x}) \cdot \mathrm{f}^{\prime}(\mathrm{x})\right) \rightarrow 7 \\ & \therefore \min \text { roots of }\left(\mathrm{f}(\mathrm{x}) \cdot \mathrm{f}^{\prime}(\mathrm{x})\right)^{\prime \prime} \rightarrow 5\end{aligned}$
$\begin{aligned} & \min \text { roots of } \mathrm{f}(\mathrm{x}) \rightarrow 4 \\ & \therefore \min \text {. roots of } \mathrm{f}^{\prime}(\mathrm{x}) \rightarrow 3 \\ & \therefore \min \text { roots of }\left(\mathrm{f}(\mathrm{x}) \cdot \mathrm{f}^{\prime}(\mathrm{x})\right) \rightarrow 7 \\ & \therefore \min \text { roots of }\left(\mathrm{f}(\mathrm{x}) \cdot \mathrm{f}^{\prime}(\mathrm{x})\right)^{\prime \prime} \rightarrow 5\end{aligned}$
Hence, the correct answer is 5.