If f(x)=left|begin{array}{ccc}x-3 & 2x^2-18 & | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given $f(x) = \left|\begin{array}{ccc} x-3 & 2(x^2-9) & 2x^3-81 \ x-5 & 2(x^2-25) & 4(x^3-125) \ 1 & 2 & 3 \end{array}ight|$.For $x=5$,the sec

Vedclass · Accepted Answer

$2183328$

Vedclass · Answer

(B) Given $f(x) = \left|\begin{array}{ccc} x-3 & 2(x^2-9) & 2x^3-81 \ x-5 & 2(x^2-25) & 4(x^3-125) \ 1 & 2 & 3 \end{array}ight|$.For $x=5$,the second row becomes $5-5=0$,$2(25-25)=0$,and $4(125-125)=0$. Since a row is zero,$f(5)=0$.Now calculate $f(1)$:$f(1) = \left|\begin{array}{ccc} -2 & -16 & -79 \ -4 & -48 & -496 \ 1 & 2 & 3 \end{array}ight| = -2888$.Now calculate $f(3)$:$f(3) = \left|\begin{array}{ccc} 0 & 0 & -27 \ -2 & -32 & -392 \ 1 & 2 & 3 \end{array}ight| = -27(0 - (-32 + 32)) = 756$ (Correction: Calculating determinant: $-27(-4 - (-32)) = -27(28) = -756$).Thus,$f(1) \cdot f(3) + f(3) \cdot f(5) + f(5) \cdot f(1) = (-2888 	imes -756) + 0 + 0 = 2183328$.

If $f(x)=\left|\begin{array}{ccc}x-3 & 2x^2-18 & 2x^3-81 \\ x-5 & 2x^2-50 & 4x^3-500 \\ 1 & 2 & 3\end{array}\right|$,then the value of $f(1) \cdot f(3)+f(3) \cdot f(5)+f(5) \cdot f(1)$ is:

Similar Questions

The roots of the equation $\left| \begin{matrix} x & 0 & 8 \\ 4 & 1 & 3 \\ 2 & 0 & x \end{matrix} \right| = 0$ are equal to

If $x^4+y^4+z^4=0$ then,$\left|\begin{array}{ccc}1 & xy & yz \\ zx & 1 & xy \\ yz & zx & 1\end{array}\right|=$ . . . . . . . $(\because x, y, z \in \mathbb{R})$

If $C = 2 \cos \theta$,then the value of the determinant $\Delta = \begin{vmatrix} C & 1 & 0 \\ 1 & C & 1 \\ 6 & 1 & C \end{vmatrix}$ is

$f(x)$ is an $n^{\text{th}}$ degree polynomial satisfying $f(x) = \frac{1}{2} \begin{vmatrix} f(x) & f(\frac{1}{x}) - f(x) \\ 1 & f(\frac{1}{x}) \end{vmatrix}$. If $f(2) = 33$,then the value of $f(3)$ is

If the system of equations $2x + 3y - z = 0$,$x + ky - 2z = 0$ and $2x - y + z = 0$ has a non-trivial solution $(x, y, z)$,then $\frac{x}{y} + \frac{y}{z} + \frac{z}{x} + k$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module