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Polynomial Functions

If fx= 1+...

Question

If $f (x) =$ $∣ ∣ ∣ ∣ ∣ \begin{matrix} (1 + 3 x)^{2 a} & (1 + 2 x)^{3 b} & 1 1 & (1 + 3 x)^{2 a} & (1 + 2 x)^{3 b} (1 + 2 x)^{3 b} & 1 & (1 + 3 x)^{2 a} \end{matrix} ∣ ∣ ∣ ∣ ∣$ then

A

f (x)

has constant term

1

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B

constant term is

2 a - 3 b

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C

coefficient of

x

in

f (x)

is zero

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D

constant term is

2 a + 3 b

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Solution

The correct option is C coefficient of $x$ in $f (x)$ is zero
Let $f (x) =$

$∣ ∣ ∣ ∣ ∣ \begin{matrix} (1 + 3 x)^{2 a} & (1 + 2 x)^{3 b} & 1 1 & (1 + 3 x)^{2 a} & (1 + 2 x)^{3 b} (1 + 2 x)^{3 b} & 1 & (1 + 3 x)^{2 a} \end{matrix} ∣ ∣ ∣ ∣ ∣$ $= A + B x + C x^{2} + D x^{3} + . . . . . . . . . . . . . . . . . . . . . . . . .$ ----------(1)

To find the constant term put $x = 0$
Then equation (1) becomes
$∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 1 & 1 & 1 1 & 1 & 1 \end{matrix} ∣ ∣ ∣ ∣ = A$

$∴ A = 0$

Constant term of $f (x)$ is zero.

To find coeffcient of x, differentiate equation (1) with respect to x.

$f^{'} (x) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 2 a \times 3 (1 + 3 x)^{2 a - 1} & 3 b \times 2 (1 + 2 x)^{3 b - 1} & 0 1 & (1 + 3 x)^{2 a} & (1 + 2 x)^{3 b} (1 + 2 x)^{3 b} & 1 & (1 + 3 x)^{2 a} \end{matrix} ∣ ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ ∣ \begin{matrix} (1 + 3 x)^{2 a} & (1 + 2 x)^{3 b} & 1 0 & 2 a \times 3 (1 + 3 x)^{2 a - 1} & 3 b \times 2 (1 + 2 x)^{3 b - 1} (1 + 2 x)^{3 b} & 1 & (1 + 3 x)^{2 a} \end{matrix} ∣ ∣ ∣ ∣ ∣$

$+ ∣ ∣ ∣ ∣ ∣ \begin{matrix} (1 + 3 x)^{2 a} & (1 + 2 x)^{3 b} & 1 1 & (1 + 3 x)^{2 a} & (1 + 2 x)^{3 b} 3 b \times 2 (1 + 2 x)^{3 b - 1} & 0 & 2 a \times 3 (1 + 3 x)^{2 a - 1} \end{matrix} ∣ ∣ ∣ ∣ ∣ = B + 2 C x + 3 D x^{2} + . . . . . . . . . . . . . . . . . . . . . . .$ -------(2)

We need to find $B$ . For that substitute $x = 0.$

$f^{'} (0) = ∣ ∣ ∣ ∣ \begin{matrix} 6 a & 6 b & 0 1 & 1 & 1 1 & 1 & 1 \end{matrix} ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 0 & 6 a & 6 b 1 & 1 & 1 \end{matrix} ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 1 & 1 & 1 6 b & 0 & 6 a \end{matrix} ∣ ∣ ∣ ∣ = B$

If any two rows of a determinant are same, then the determinant is zero.
Hence $B = 0$ .
$∴$ Coefficient of $x$ in $f (x)$ is zero.

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Similar questions

f (x) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} (1 + x)^{a} & (1 + 2 x)^{b} & 1 1 & (1 + x)^{2} & (1 + 2 x)^{b} (1 + 2 x)^{b} & 1 & (1 + x)^{a} \end{matrix} ∣ ∣ ∣ ∣ ∣

, then find the sum of constant term and coefficient of

x

.

If f(x) = x - 1 ÷ x + 1 then f(2x) in terms of f(x) is?

Q.
The general term of

(2 a - 3 b)^{- 1 / 2}

Q. If the polynomial

f (x) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} {(1 + x)}^{a} & {(2 + x)}^{b} & 1 1 & {(1 + x)}^{a} & {(2 + x)}^{b} {(2 + x)}^{b} & 1 & {(1 + x)}^{a} \end{matrix} ∣ ∣ ∣ ∣ ∣

, then the constant term of

f (x)

f (x) = \frac{x - 1}{x + 1}

f (a x)

f (x)

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