If the sum of the coefficients in the expansi | vedclass

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

Question

(C) The sum of the coefficients of a polynomial $P(x)$ is obtained by evaluating $P(1)$.Given the polynomial $P(x) = (\alpha ^2 x^2 - 2\alpha x + 1)^{

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) The sum of the coefficients of a polynomial $P(x)$ is obtained by evaluating $P(1)$.Given the polynomial $P(x) = (\alpha ^2 x^2 - 2\alpha x + 1)^{51}$.Setting $x = 1$,the sum of the coefficients is $(\alpha ^2(1)^2 - 2\alpha(1) + 1)^{51} = (\alpha ^2 - 2\alpha + 1)^{51}$.Since the sum of the coefficients vanishes,we have $(\alpha ^2 - 2\alpha + 1)^{51} = 0$.This implies $(\alpha - 1)^2 = 0$,which gives $\alpha = 1$.

If the sum of the coefficients in the expansion of $({\alpha ^2}{x^2} - 2\alpha x + 1)^{51}$ vanishes,then the value of $\alpha$ is

Similar Questions

Expand $\left(x^{2}+\frac{3}{x}\right)^{4}, x \neq 0$

In the expansion of the expression $1 + (1 + x) + (1 + x)^2 + \dots + (1 + x)^n$,the coefficient of $x^k$ $(0 \le k \le n)$ is

Using the Binomial Theorem,determine which number is larger: $(1.1)^{10000}$ or $1000$.

The expression $[x + (x^3 - 1)^{1/2}]^5 + [x - (x^3 - 1)^{1/2}]^5$ is a polynomial of degree

The expression $(2 + \sqrt{2})^4$ has a value lying between

Vedclass Test Series

Exam Paper Generator

Online Exam Module