If fx= x^3 2x^2+1 1+3x 3x^2+2 2x x^3+6 x^3−x 4 x^2−2 for all x∈ℝ, then 2f0+f^(')0 is equal to

Correct answer: (C) 42. Given: fx= x^3 2x^2+1 1+3x 3x^2+2 2x x^3+6 x^3-x 4 x^2-2 ⇒f0= 0 1 1 2 0 6 0 4 -2 ⇒f0=0+4+8 ⇒f0=12 ⇒f^(')x= x^3 2x^2+1 1+3x 3x^2+2 2x x^3+6 3x^2-1 0 2x + x^3 2x^2+1 1+3x 6x 2 3x^2 x^3-x 4 x^2-2 + 3x^2 4x 3 3x^2+2 2x x^3+6 x^3-x 4 x^2-2 ⇒f^(')0= 0 1 1 2 0 6 -1 0 0 + 0 1 1 0 2 0 0 4 -2 + 0 0 3 2 0 6 0 4 -2 ⇒f^(')0=0-0+6+0+0+0+0+38 ⇒f^(')0=-6+24=18 ⇒2f0+f^(')0=42

Q45 / 90 · JEE Main 2024 (31 Jan Shift 1) · Mathematics

If $f (x) = |\begin{matrix} x^{3} & 2 x^{2} + 1 & 1 + 3 x \\ 3 x^{2} + 2 & 2 x & x^{3} + 6 \\ x^{3} - x & 4 & x^{2} - 2 \end{matrix}|$ for all $x \in ℝ$ , then $2 f (0) + f^{'} (0)$ is equal to

A
$48$
B
$24$
Correct answer
C
$42$
D
$18$

Solution

Given: $f (x) = |\begin{matrix} x^{3} & 2 x^{2} + 1 & 1 + 3 x \\ 3 x^{2} + 2 & 2 x & x^{3} + 6 \\ x^{3} - x & 4 & x^{2} - 2 \end{matrix}|$

$\Rightarrow f (0) = |\begin{matrix} 0 & 1 & 1 \\ 2 & 0 & 6 \\ 0 & 4 & - 2 \end{matrix}|$

$\Rightarrow f (0) = 0 + 4 + 8$

$\Rightarrow f (0) = 12$

$\Rightarrow f^{'} (x) = |\begin{matrix} x^{3} & 2 x^{2} + 1 & 1 + 3 x \\ 3 x^{2} + 2 & 2 x & x^{3} + 6 \\ 3 x^{2} - 1 & 0 & 2 x \end{matrix}| + |\begin{matrix} x^{3} & 2 x^{2} + 1 & 1 + 3 x \\ 6 x & 2 & 3 x^{2} \\ x^{3} - x & 4 & x^{2} - 2 \end{matrix}| + |\begin{matrix} 3 x^{2} & 4 x & 3 \\ 3 x^{2} + 2 & 2 x & x^{3} + 6 \\ x^{3} - x & 4 & x^{2} - 2 \end{matrix}|$

$\Rightarrow f^{'} (0) = |\begin{matrix} 0 & 1 & 1 \\ 2 & 0 & 6 \\ - 1 & 0 & 0 \end{matrix}| + |\begin{matrix} 0 & 1 & 1 \\ 0 & 2 & 0 \\ 0 & 4 & - 2 \end{matrix}| + |\begin{matrix} 0 & 0 & 3 \\ 2 & 0 & 6 \\ 0 & 4 & - 2 \end{matrix}|$

$\Rightarrow f^{'} (0) = 0 - (0 + 6) + 0 + 0 + 0 + 0 + 3 (8)$

$\Rightarrow f^{'} (0) = - 6 + 24 = 18$

$\Rightarrow 2 f (0) + f^{'} (0) = 42$

Hence, the correct answer is option: 'C'.

If fx=x32x2+11+3x3x2+22xx3+6x3−x4x2−2 for all x∈ℝ, then 2f0+f'0 is equal to

Solution

More Determinants Questions

If $f (x) = |\begin{matrix} x^{3} & 2 x^{2} + 1 & 1 + 3 x \\ 3 x^{2} + 2 & 2 x & x^{3} + 6 \\ x^{3} - x & 4 & x^{2} - 2 \end{matrix}|$ for all $x \in ℝ$ , then $2 f (0) + f^{'} (0)$ is equal to