$$\text { If } f(x)=\left|\begin{array}{ccc} x^3 & 2 x^2+1 & 1+3 x \\ 3 x^2+2 & 2 x & x^3+6 \\ x^3-x & 4 & x^2-2 \end{array}\right| \text { for all } x \in \mathbb{R} \text {, then } 2 f(0)+f^{\prime}(0) \text { is equal to }$$
Solution
<p>$$\begin{aligned}
& f(0)=\left|\begin{array}{ccc}
0 & 1 & 1 \\
2 & 0 & 6 \\
0 & 4 & -2
\end{array}\right|=12 \\
& f^{\prime}(x)=\left|\begin{array}{ccc}
3 x^2 & 4 x & 3 \\
3 x^2+2 & 2 x & x^3+6 \\
x^3-x & 4 & x^2-2
\end{array}\right|+
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \left|\begin{array}{ccc}
x^3 & 2 x^2+1 & 1+3 x \\
6 x & 2 & 3 x^2 \\
x^3-x & 4 & x^2-2
\end{array}\right|+ \\
& \left|\begin{array}{ccc}
x^3 & 2 x^2+1 & 1+3 x \\
3 x^2+2 & 2 x & x^3+6 \\
3 x^2-1 & 0 & 2 x
\end{array}\right| \\
& \therefore f^{\prime}(0)=\left|\begin{array}{ccc}
0 & 0 & 3 \\
2 & 0 & 6 \\
0 & 4 & -2
\end{array}\right|+\left|\begin{array}{ccc}
0 & 1 & 1 \\
0 & 2 & 0 \\
0 & 4 & -2
\end{array}\right|+\left|\begin{array}{ccc}
0 & 1 & 1 \\
2 & 0 & 6 \\
-1 & 0 & 0
\end{array}\right| \\
& =24-6=18 \\
& \therefore 2 f(0)+f^{\prime}(0)=42
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Application of Derivatives · Topic: Tangents and Normals
This question is part of PrepWiser's free JEE Main question bank. 99 more solved questions on Application of Derivatives are available — start with the harder ones if your accuracy is >70%.