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JEE Mathematics 2024 Question with Solution

If f(x)=x3+2x2+1+3xf(x) = x^3 + 2x^2 + 1 + 3x, then 2f(0)+f′(0)2f(0) + f'(0) is equal to:

  • A

    4848

  • B

    2424

  • C

    4242

  • D

    1818

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: the solution defines f(x)f(x) as the determinant

∣x32x2+11+3x3x2+22xx3+6x3−x4x2−2∣\begin{vmatrix} x^3 & 2x^2 + 1 & 1 + 3x \\ 3x^2 + 2 & 2x & x^3 + 6 \\ x^3 - x & 4 & x^2 - 2 \end{vmatrix}

Find: 2f(0)+f′(0)2f(0) + f'(0).

First evaluate f(0)f(0):

∣01120604−2∣=12\begin{vmatrix} 0 & 1 & 1 \\ 2 & 0 & 6 \\ 0 & 4 & -2 \end{vmatrix} = 12

So, f(0)=12f(0) = 12.

Now differentiate the determinant row-wise as shown in the solution:

f′(x)=∣3x24x3x32x2+11+3xx3−x4x2−2∣+∣x32x2+11+3x6x2x3x2x3−x4x2−2∣+∣x32x2+11+3x3x2+22xx3+63x2−102x∣f'(x) = \begin{vmatrix} 3x^2 & 4x & 3 \\ x^3 & 2x^2 + 1 & 1 + 3x \\ x^3 - x & 4 & x^2 - 2 \end{vmatrix} + \begin{vmatrix} x^3 & 2x^2 + 1 & 1 + 3x \\ 6x & 2x & 3x^2 \\ x^3 - x & 4 & x^2 - 2 \end{vmatrix} + \begin{vmatrix} x^3 & 2x^2 + 1 & 1 + 3x \\ 3x^2 + 2 & 2x & x^3 + 6 \\ 3x^2 - 1 & 0 & 2x \end{vmatrix}

At x=0x = 0, this gives

∣00320604−2∣+∣01100004−2∣+∣011206−100∣\begin{vmatrix} 0 & 0 & 3 \\ 2 & 0 & 6 \\ 0 & 4 & -2 \end{vmatrix} + \begin{vmatrix} 0 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 4 & -2 \end{vmatrix} + \begin{vmatrix} 0 & 1 & 1 \\ 2 & 0 & 6 \\ -1 & 0 & 0 \end{vmatrix}

which evaluates, as stated in the solution, to

f′(0)=18f'(0) = 18

Therefore,

2f(0)+f′(0)=2⋅12+18=422f(0) + f'(0) = 2 \cdot 12 + 18 = 42

Hence, the correct option is C.

Using the displayed determinant evaluations

Given: The working in the solution uses a determinant form of f(x)f(x), not the plain polynomial written in the question statement. The answer has therefore been derived from the solution, which is the primary source.

From the solution:

f(0)=∣01120604−2∣=12f(0) = \begin{vmatrix} 0 & 1 & 1 \\ 2 & 0 & 6 \\ 0 & 4 & -2 \end{vmatrix} = 12

Also,

f′(0)=18f'(0) = 18

Now substitute these values:

2f(0)+f′(0)=2(12)+18=24+18=422f(0) + f'(0) = 2(12) + 18 = 24 + 18 = 42

Therefore, the required value is 4242, so the correct option is C.

Common mistakes

  • Using the polynomial f(x)=x3+2x2+1+3xf(x) = x^3 + 2x^2 + 1 + 3x directly gives f(0)=1f(0) = 1 and f′(0)=3f'(0) = 3, which conflicts with the solution. The solution actually treats f(x)f(x) as a determinant, so the determinant-based definition must be used here.

  • While differentiating a determinant, differentiating only one entry or one row is incorrect. The shown method adds determinants obtained by differentiating each row contribution separately, then evaluates at x=0x = 0.

  • Errors in substituting x=0x = 0 into the determinant entries can change signs and constants. Carefully convert the matrix entries to ∣01120604−2∣\begin{vmatrix} 0 & 1 & 1 \\ 2 & 0 & 6 \\ 0 & 4 & -2 \end{vmatrix} before computing f(0)f(0).

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