If f(x) = left|begin{array}{ccc} 1 & x & x+1 | vedclass

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

Question

(A) Given,$f(x) = \left|\begin{array}{ccc} 1 & x & x+1 \ 2x & x(x-1) & x(x+1) \ 3x(x-1) & x(x-1)(x-2) & (x-1)x(x+1) \end{array}ight|$.Taking $x$ c

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) Given,$f(x) = \left|\begin{array}{ccc} 1 & x & x+1 \ 2x & x(x-1) & x(x+1) \ 3x(x-1) & x(x-1)(x-2) & (x-1)x(x+1) \end{array}ight|$.Taking $x$ common from $R_2$,$x(x-1)$ common from $R_3$,and $(x+1)$ common from $C_3$,we get:$f(x) = x \cdot x(x-1) \cdot (x+1) \left|\begin{array}{ccc} 1 & x & 1 \ 2 & x-1 & 1 \ 3 & x-2 & 1 \end{array}ight|$.Since $C_1$ and $C_3$ are not identical,let us simplify the determinant by performing row operations.Applying $R_2 ightarrow R_2 - R_1$ and $R_3 ightarrow R_3 - R_1$:$f(x) = x^2(x^2-1) \left|\begin{array}{ccc} 1 & x & 1 \ 1 & -1 & 0 \ 2 & -2 & 0 \end{array}ight|$.Expanding along $C_3$:$f(x) = x^2(x^2-1) \cdot [1 \cdot (1(-2) - 2(-1))] = x^2(x^2-1) \cdot [-2 + 2] = 0$.Thus,$f(x) = 0$ for all $x$.Therefore,$f(2012) = 0$.

If $f(x) = \left|\begin{array}{ccc} 1 & x & x+1 \\ 2x & x(x-1) & x(x+1) \\ 3x(x-1) & x(x-1)(x-2) & (x-1)x(x+1) \end{array}\right|$,then $f(2012)$ is equal to:

Similar Questions

The line $\frac{x + 1}{2} = \frac{y + 1}{3} = \frac{z + 1}{4}$ meets the plane $x + 2y + 3z = 14$ at the point:

If a bimetallic strip is heated,it will

If the angles of a triangle are in the ratio $4:1:1$,then the ratio of the longest side to the perimeter is

If $z = x + iy$ is a complex number satisfying $\left| z + \frac{i}{2} \right|^2 = \left| z - \frac{i}{2} \right|^2$,then the locus of $z$ is

$2 \ mol$ of zinc is dissolved in $HCl$ at $25 \ ^\circ C$. The work done in an open vessel is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module