Consider the polynomial function fx- 1-x a 1-2x b 1 1 1-x a 1-2x b 1-2x b 1 1-x a -a-b being positive integers-The constant term in fx is

The correct option is D 0Let f(x)=( 1+x ) nbsp;^ a amp; ( 1+2x ) nbsp;^ b amp; 1 1 amp; ( 1+x ) nbsp;^ a amp; ( 1+2x ) nbsp; ...

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Byju's Answer

Standard XII

Mathematics

Definition of a Determinant

Consider the ...

Question

Consider the polynomial function $f (x) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} {(1 + x)}^{a} & {(1 + 2 x)}^{b} & 1 1 & {(1 + x)}^{a} & {(1 + 2 x)}^{b} {(1 + 2 x)}^{b} & 1 & {(1 + x)}^{a} \end{matrix} ∣ ∣ ∣ ∣ ∣, a, b$ being positive integers.
The constant term in $f (x)$ is

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Solution

The correct option is D 0
Let $f (x) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} {(1 + x)}^{a} & {(1 + 2 x)}^{b} & 1 1 & {(1 + x)}^{a} & {(1 + 2 x)}^{b} {(1 + 2 x)}^{b} & 1 & {(1 + x)}^{a} \end{matrix} ∣ ∣ ∣ ∣ ∣ = a_{0} + a_{1} x + a_{2} x^{2} + . . .$
Substitute $x = 0$
$\Rightarrow ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 1 & 1 & 1 1 & 1 & 1 \end{matrix} ∣ ∣ ∣ ∣ = a_{0}$
$\Rightarrow a_{0} = 0$
Hence, the constant term in $f (x)$ is 0.

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Similar questions

Q. If

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

Q. If

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

, then find
(i) constant term
(ii) coefficient of

x

Prove that :

$(i) {(\frac{x^{a}}{x^{b}})}^{\frac{1}{a b}} {(\frac{x^{b}}{x^{c}})}^{\frac{1}{b c}} {(\frac{x^{c}}{x^{a}})}^{\frac{1}{c a}} = 1 (i i) \frac{1}{1 + x^{a - b}} + \frac{1}{1 + x^{b - a}} = 1$

Q. Prove that :

{(\frac{x^{a}}{x^{b}})}^{\frac{1}{a b}} . {(\frac{x^{b}}{x^{c}})}^{\frac{1}{b c}} . {(\frac{x^{c}}{x^{a}})}^{\frac{1}{c a}} = 1

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)

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Consider the polynomial function f(x)=∣∣ ∣ ∣∣(1+x)a(1+2x)b11(1+x)a(1+2x)b(1+2x)b1(1+x)a∣∣ ∣ ∣∣,a,b being positive integers.The constant term in f(x) is

The correct option is D 0Let f(x)=∣∣ ∣ ∣∣(1+x)a(1+2x)b11(1+x)a(1+2x)b(1+2x)b1(1+x)a∣∣ ∣ ∣∣=a0+a1x+a2x2+...Substitute x=0 ⇒∣∣ ∣∣111111111∣∣ ∣∣=a0⇒a0=0Hence, the constant term in f(x) is 0.

Consider the polynomial function $f (x) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} {(1 + x)}^{a} & {(1 + 2 x)}^{b} & 1 1 & {(1 + x)}^{a} & {(1 + 2 x)}^{b} {(1 + 2 x)}^{b} & 1 & {(1 + x)}^{a} \end{matrix} ∣ ∣ ∣ ∣ ∣, a, b$ being positive integers.
The constant term in $f (x)$ is