If the polynomial f(x) = left|begin{array}{cc | vedclass

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

Question

(A) To find the constant term of the polynomial $f(x)$, we set $x = 0$.Substituting $x = 0$ into the determinant, we get:$f(0) = \left|\begin{array}{c

Vedclass · Accepted Answer

$2 - 3 \cdot 2^{b} + 2^{3b}$

Vedclass · Answer

(A) To find the constant term of the polynomial $f(x)$, we set $x = 0$.Substituting $x = 0$ into the determinant, we get:$f(0) = \left|\begin{array}{ccc} (1+0)^{a} & (2+0)^{b} & 1 \ 1 & (1+0)^{a} & (2+0)^{b} \ (2+0)^{b} & 1 & (1+0)^{a} \end{array}ight| = \left|\begin{array}{ccc} 1 & 2^{b} & 1 \ 1 & 1 & 2^{b} \ 2^{b} & 1 & 1 \end{array}ight|$.Expanding the determinant along the first row:$f(0) = 1 \cdot \left|\begin{array}{cc} 1 & 2^{b} \ 1 & 1 \end{array}ight| - 2^{b} \cdot \left|\begin{array}{cc} 1 & 2^{b} \ 2^{b} & 1 \end{array}ight| + 1 \cdot \left|\begin{array}{cc} 1 & 1 \ 2^{b} & 1 \end{array}ight|$.Calculating the $2 	imes 2$ determinants:$f(0) = 1(1 - 2^{b}) - 2^{b}(1 - (2^{b})^{2}) + 1(1 - 2^{b})$.$f(0) = (1 - 2^{b}) - 2^{b}(1 - 2^{2b}) + (1 - 2^{b})$.$f(0) = 1 - 2^{b} - 2^{b} + 2^{3b} + 1 - 2^{b}$.$f(0) = 2 - 3 \cdot 2^{b} + 2^{3b}$.

List-$I$	List-$II$
a) $A$ matrix which is not a square matrix	i) Symmetric matrix
b) $A$ square matrix $A' = A$	ii) Null matrix
c) The diagonal elements of a diagonal matrix are same	iii) Rectangular matrix
d) $A$ matrix which is both symmetric and skew symmetric	iv) Scalar matrix

If the polynomial $f(x) = \left|\begin{array}{ccc} (1+x)^{a} & (2+x)^{b} & 1 \\ 1 & (1+x)^{a} & (2+x)^{b} \\ (2+x)^{b} & 1 & (1+x)^{a} \end{array}\right|$, then the constant term of $f(x)$ is ($a$ and $b$ are positive integers).

Similar Questions

Let $S$ be the set containing all $3 \times 3$ matrices with entries from $\{-1, 0, 1\}$. The total number of matrices $A \in S$ such that the sum of all the diagonal elements of $A^{T}A$ is $6$ is:

Let $a, b, c \in \mathbb{R}$ be all non-zero and satisfy $a^{3}+b^{3}+c^{3}=2$. If the matrix $A=\begin{bmatrix} a & b & c \\ b & c & a \\ c & a & b \end{bmatrix}$ satisfies $A^{T} A=I$,then a value of $abc$ can be

If $A=\left[\begin{array}{ccc}1 & 1 & 3 \\ 5 & 2 & 6 \\ -2 & -1 & -3\end{array}\right]$,then $A+A^3+A^4+A^5+3 I=$

If $z = \begin{bmatrix} 1 & 1+2i & -5i \\ 1-2i & -3 & 5+3i \\ 5i & 5-3i & 7 \end{bmatrix}$, then which of the following is true? (where $i = \sqrt{-1}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module