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: Understanding the Concept
Expand \(f(x)\) along the first row and then differentiate twice.

Step 2: Expand
\(f(x)=x^2(1+p^2)-\sin x\,(2+p)+e^x(2p^2-p)\).

Step 3: Differentiate twice
\(f'(x)=2x(1+p^2)-\cos x\,(2+p)+e^x(2p^2-p)\)
\(f''(x)=2(1+p^2)+\sin x\,(2+p)+e^x(2p^2-p)\).

Step 4: Put x = 0
\(f''(0)=2+2p^2+0+2p^2-p=4p^2-p+2\).
This matches the third option.

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