Suppose a, b, c are three distinct real numbe | vedclass

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

Question

(A) Given the expression $P(x) = \frac{(x-b)(x-c)}{(a-b)(a-c)} + \frac{(x-c)(x-a)}{(b-c)(b-a)} + \frac{(x-a)(x-b)}{(c-a)(c-b)}$.Evaluate $P(x)$ at $x=

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Given the expression $P(x) = \frac{(x-b)(x-c)}{(a-b)(a-c)} + \frac{(x-c)(x-a)}{(b-c)(b-a)} + \frac{(x-a)(x-b)}{(c-a)(c-b)}$.Evaluate $P(x)$ at $x=a$:$P(a) = \frac{(a-b)(a-c)}{(a-b)(a-c)} + 0 + 0 = 1$.Evaluate $P(x)$ at $x=b$:$P(b) = 0 + \frac{(b-c)(b-a)}{(b-c)(b-a)} + 0 = 1$.Evaluate $P(x)$ at $x=c$:$P(c) = 0 + 0 + \frac{(c-a)(c-b)}{(c-a)(c-b)} = 1$.Since $P(x)$ is a polynomial of degree at most $2$ and it takes the value $1$ for three distinct values $x = a, b, c$,it must be the constant polynomial $P(x) = 1$ for all real $x$.Therefore,the simplified value is $1$.

Suppose $a, b, c$ are three distinct real numbers. Let $P(x) = \frac{(x-b)(x-c)}{(a-b)(a-c)} + \frac{(x-c)(x-a)}{(b-c)(b-a)} + \frac{(x-a)(x-b)}{(c-a)(c-b)}$. When simplified,$P(x)$ becomes:

Similar Questions

For $x \in R$, the least value of $\frac{x^2-6x+5}{x^2+2x+1}$ is

The sum of the roots of the equation $e^{4t} - 10e^{3t} + 29e^{2t} - 20e^t + 4 = 0$ is

If $x$ is real,then the difference between the greatest and least values of $\frac{x^2-x+1}{x^2+x+1}$ is

Let $a_1, a_2, a_3, a_4$ be real numbers such that $a_1+a_2+a_3+a_4=0$ and $a_1^2+a_2^2+a_3^2+a_4^2=1$. Then,the smallest possible value of the expression $(a_1-a_2)^2+(a_2-a_3)^2+(a_3-a_4)^2+(a_4-a_1)^2$ lies in the interval

The number of solutions of the equations $x+y+z=12$,$x^2+y^2+z^2=50$,and $x^3+y^3+z^3=216$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module