Medium MCQ +4 / -1 PYQ · JEE Mains 2024

$$\text { If } f(x)=\left|\begin{array}{ccc} x^3 & 2 x^2+1 & 1+3 x \\ 3 x^2+2 & 2 x & x^3+6 \\ x^3-x & 4 & x^2-2 \end{array}\right| \text { for all } x \in \mathbb{R} \text {, then } 2 f(0)+f^{\prime}(0) \text { is equal to }$$

  1. A 24
  2. B 18
  3. C 42 Correct answer
  4. D 48

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

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