The equation left(x^4+1right)=frac{1}{a}(x+1) | vedclass

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

Question

(B) Given equation is $\left(x^4+1\right)=\frac{1}{a}(x+1)^4$. Multiplying by $a$,we get $a(x^4+1) = (x+1)^4$. Expanding the right side: $a(x^4+1) = x

Vedclass · Accepted Answer

for all $a \in R-\{1\}$

Vedclass · Answer

(B) Given equation is $\left(x^4+1ight)=\frac{1}{a}(x+1)^4$. Multiplying by $a$,we get $a(x^4+1) = (x+1)^4$. Expanding the right side: $a(x^4+1) = x^4+4x^3+6x^2+4x+1$. Rearranging the terms: $(a-1)x^4 - 4x^3 - 6x^2 - 4x + (a-1) = 0$. For an equation to be a reciprocal equation,the coefficients of $x^k$ and $x^{n-k}$ must be equal (or proportional). Here,the coefficient of $x^4$ is $(a-1)$ and the constant term is $(a-1)$. For the equation to be of degree $4$,the coefficient of $x^4$ must not be zero,so $a-1 
eq 0$,which means $a 
eq 1$. Thus,the equation is a reciprocal equation for all $a \in R - \{1\}$.

The equation $\left(x^4+1\right)=\frac{1}{a}(x+1)^4$ is a reciprocal equation:

Similar Questions

If $\frac{5}{2}$ is the sum of two roots of the equation $6x^6-25x^5+31x^4-31x^2+25x-6=0$,then the sum of all non-real roots of the equation is

The diagram shows the graph of $y = ax^2 + bx + c$. Then:

If $\frac{x-P}{x^2-3x+2}$ takes all real values for $x \in \mathbb{R} \setminus \{1, 2\}$,then the range of $P$ is

Suppose,$\alpha$ is the minimum value of $x^2+bx+5$ and $\beta$ is the maximum value of $-x^2+ax+5$. If $[\alpha, \beta]$ is the interval of maximum length for $x$ in which $x^2-10x+24 \leq 0$,then $a^2b^2$ is equal to

On each face of a cuboid,the sum of its perimeter and its area is written. Among the six numbers so written,there are three distinct numbers and they are $16, 24$ and $31$. The volume of the cuboid lies between

Vedclass Test Series

Exam Paper Generator

Online Exam Module