Let begin array ll Wegin vmatrix x - 2 - x x -2- x - 6 x - x - 6lend vmatrix - A x -4- B x -3- C x -2- Dx - Elend array A- A - B - C - D -0B- A - B - C - D -1C- 5 A -4 B -3 C -2 D - E -11D- 5 A -4 B -3 C -2 D - E -11

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Byju's Answer

Standard XII

Mathematics

Summation by Sigma Method

Let begin arr...

Question

Let
$\begin{matrix} ∣ ∣ ∣ ∣ \begin{matrix} x & 2 & x x^{2} & x & 6 x & x & 6 \end{matrix} ∣ ∣ ∣ ∣ & = A x^{4} + B x^{3} + C x^{2} + D x + E \end{matrix}$

A+B+C+D = 0

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A+B+C+D = -1

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5A+4B+3C+2D+E = -11

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5A+4B+3C+2D+E = 11

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Solution

The correct options are
A
A+B+C+D = 0

C
5A+4B+3C+2D+E = -11

Expanding the determinant and equating:

A = 1, B= -1, C = -12, D = 12, E = 0

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Similar questions

Q. .Let

∣ ∣ ∣ ∣ \begin{matrix} x & 2 & x x^{2} & x & 6 x & x & 6 \end{matrix} ∣ ∣ ∣ ∣

a x^{4} + b x^{3} + c x^{2} + d x + e

,then the value of 5a + 4b + 3c + 2d + e is
equal to

Q. Let

∣ ∣ ∣ ∣ \begin{matrix} x & 2 & x x^{2} & x & 6 x & x & 6 \end{matrix} ∣ ∣ ∣ ∣ = A x^{4} + B x^{3} + C x^{2} + D x + E

, then the value of 5A + 4B + 3C + 2D + E is equal to

Q. Let

∣ ∣ ∣ ∣ \begin{matrix} x & 2 & x x^{2} & x & 6 x & x & 6 \end{matrix} ∣ ∣ ∣ ∣ = A x^{4} + B x^{3} + C x^{2} + D x + E

If $△= ∣ ∣ ∣ ∣ ∣ \begin{matrix} e^{x} & s i n x & 1 c o s x & l n (1 + x^{2}) & 1 x & x^{2} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = a + b x + c x^{2}$ then the value of b is

Let $f (x) = a x^{2} + b x + c$ . Then, match the following.
$\begin{matrix} a. Sum of roots of f(x) = 0 & 1. - \frac{b}{a} b. Product of roots of f(x) = 0 & 2. \frac{c}{a} c. Roots of f(x) = 0 are real and distinct & 3. b^{2} - 4 a c = 0 d. Roots of f(x) = 0 are real and identical. & 4. b^{2} - 4 a c > 0 \end{matrix}$

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Let ∣∣ ∣∣x2xx2x6xx6∣∣ ∣∣=Ax4+Bx3+Cx2+Dx+E

The correct options are A A+B+C+D = 0 C 5A+4B+3C+2D+E = -11 Expanding the determinant and equating: A = 1, B= -1, C = -12, D = 12, E = 0

Let
$\begin{matrix} ∣ ∣ ∣ ∣ \begin{matrix} x & 2 & x x^{2} & x & 6 x & x & 6 \end{matrix} ∣ ∣ ∣ ∣ & = A x^{4} + B x^{3} + C x^{2} + D x + E \end{matrix}$

The correct options are
A
A+B+C+D = 0

C
5A+4B+3C+2D+E = -11

Expanding the determinant and equating:

A = 1, B= -1, C = -12, D = 12, E = 0