JEE MAIN - Mathematics (2024 - 27th January Morning Shift - No. 23)
Let $f(x)=x^3+x^2 f^{\prime}(1)+x f^{\prime \prime}(2)+f^{\prime \prime \prime}(3), x \in \mathbf{R}$. Then $f^{\prime}(10)$ is equal to ____________.
Answer
202
Explanation
$$\begin{aligned}
& f(x)=x^3+x^2 \cdot f^{\prime}(1)+x \cdot f^{\prime \prime}(2)+f^{\prime \prime \prime}(3) \\
& f^{\prime}(x)=3 x^2+2 x f^{\prime}(1)+f^{\prime \prime}(2) \\
& f^{\prime \prime}(x)=6 x+2 f^{\prime}(1) \\
& f^{\prime \prime \prime}(x)=6 \\
& f^{\prime}(1)=-5, f^{\prime \prime}(2)=2, f^{\prime \prime \prime}(3)=6 \\
& f(x)=x^3+x^2 \cdot(-5)+x \cdot(2)+6 \\
& f^{\prime}(x)=3 x^2-10 x+2 \\
& f^{\prime}(10)=300-100+2=202
\end{aligned}$$
Comments (0)
