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:$f(0) = 1(1 - 2^{b}) - 2^{b}(1 - (2^{b})^{2}) + 1(1 - 2^{b})$.$f(0) = (1 - 2^{b}) - 2^{b}(1 - 2^{2b}) + (1 - 2^{b})$.$f(0) = 1 - 2^{b} - 2^{b} + 2^{3b} + 1 - 2^{b}$.$f(0) = 2 - 3 \cdot 2^{b} + 2^{3b}$.

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

If both $\left( A - \frac{I}{2} \right)$ and $\left( A + \frac{I}{2} \right)$ are orthogonal matrices,then:

Let $A$ be a $3 \times 3$ real matrix such that $A^2(A-2I) - 4(A-I) = O$,where $I$ and $O$ are the identity and null matrices,respectively. If $A^5 = \alpha A^2 + \beta A + \gamma I$,where $\alpha, \beta$ and $\gamma$ are real constants,then $\alpha + \beta + \gamma$ is equal to:

If $A$ and $B$ are both $3 \times 3$ matrices,then which of the following statements are true?
$(i)$ $AB=0 \Rightarrow A=0$ or $B=0$
(ii) $AB=I_3 \Rightarrow A^{-1}=B$
(iii) $(A-B)^2=A^2-2AB+B^2$

Let $A$ and $B$ be two $3 \times 3$ real matrices such that $(A^{2}-B^{2})$ is an invertible matrix. If $A^{5}=B^{5}$ and $A^{3} B^{2}=A^{2} B^{3}$,then the value of the determinant of the matrix $A^{3}+B^{3}$ is equal to:

If $A = \begin{bmatrix} \cos \theta & i \sin \theta \\ i \sin \theta & \cos \theta \end{bmatrix}$,$\theta = \frac{\pi}{24}$ and $A^{5} = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$,where $i = \sqrt{-1}$,then which one of the following is not true?

Vedclass Test Series

Exam Paper Generator

Online Exam Module