Question

Let \( f(x) = \begin{vmatrix} x^3 & \sin x & \cos x \\ 6 & -1 & 0 \\ p & p^2 & p^3 \end{vmatrix} \) where \(p\) is a constant.
The value of \( \dfrac{d^3}{dx^3} f(x) \) at \(x = 0\) is

• A. \(6p^3\)
• B. \(p + p^2\)
• C. \(p + p^3\)
• D. independent of \(p\)

Hint:

Only the first row of the determinant depends on \(x\); differentiate that row alone, three times, then check what the row becomes at \(x=0\).

Answer 1

The Correct Option is D

Step 1: Set up the differentiation rule for the determinant.
The determinant \(f(x)\) has three rows. Only the first row, \((x^3, \sin x, \cos x)\), depends on \(x\). The second row \((6,-1,0)\) and the third row \((p,p^2,p^3)\) are made of constants.
A determinant is linear in each row taken separately. So when we differentiate \(f(x)\) with respect to \(x\), only the row that carries \(x\) changes. We replace that row with its derivative and keep the other two rows exactly as printed. This is much faster than expanding the whole determinant first.

Step 2: Differentiate the first row three times.
Let \(R_1(x) = (x^3, \sin x, \cos x)\).
\[ R_1'(x) = (3x^2, \cos x, -\sin x) \]
\[ R_1''(x) = (6x, -\sin x, -\cos x) \]
\[ R_1'''(x) = (6, -\cos x, \sin x) \]
So \(f'''(x)\) is the determinant with row 1 replaced by \(R_1'''(x)\), rows 2 and 3 unchanged:
\[ f'''(x) = \begin{vmatrix} 6 & -\cos x & \sin x \\ 6 & -1 & 0 \\ p & p^2 & p^3 \end{vmatrix} \]
Step 3: Put in \(x = 0\).
At \(x = 0\), \(\cos 0 = 1\) and \(\sin 0 = 0\). The first row becomes \((6, -1, 0)\).
\[ f'''(0) = \begin{vmatrix} 6 & -1 & 0 \\ 6 & -1 & 0 \\ p & p^2 & p^3 \end{vmatrix} \]
Step 4: Read off the value of the determinant.
Row 1 and row 2 are now exactly the same, \((6,-1,0)\). A determinant with two identical rows is always zero, no matter what sits in the third row. This is a basic property of determinants: swapping two equal rows leaves the value unchanged, but swapping any two rows must also flip the sign of the determinant, so the value has to equal its own negative, which forces it to be zero.
So \(f'''(0) = 0\) for every value of \(p\), including \(p=0\), \(p=1\), or any other number.

Step 5: Rule out the other options.
Option (A), \(6p^3\), and option (C), \(p+p^3\), both claim the answer changes with \(p\) in a specific nonzero way. We just showed the value is exactly zero for every \(p\), so neither expression matches (they are not identically zero). Option (B), \(p+p^2\), makes the same mistake.

Final Answer:
\(f'''(0) = 0\) for every value of \(p\), so the result does not change with \(p\).
\[ \boxed{\text{Independent of } p} \]