Question

If \(f(x)=\begin{vmatrix} x^{2} & \sin x & e^{x} \\ 2 & 1 & -1 \\ p & p^{2} & 1 \end{vmatrix}\) then \(f^{''}(0)=\)

• A. \(2p^{2}-p+1\)
• B. \(4p^{2}-p+1\)
• C. \(4p^{2}-p+2\)
• D. \(2p^{2}-p+2\)

Hint:

Expand the determinant along the first row, differentiate twice and put x equal to 0.

Answer 1

The Correct Option is C

Step 1: Differentiate the determinant
Only the first row has functions of $x$, so $f''(x)$ is the same determinant with the first row replaced by $2,\ -\sin x,\ e^x$.

Step 2: Substitute x = 0
The first row becomes $2,\ 0,\ 1$.

Step 3: Expand
$\begin{vmatrix}2&0&1\\2&1&-1\\p&p^2&1\end{vmatrix}=2(1+p^2)-0+1(2p^2-p)=4p^2-p+2$.

Final Answer:
$f''(0)=4p^2-p+2$. \[ \boxed{\text{(C) }4p^{2}-p+2} \]