Let t be real number such that t^2=a t+b for | vedclass

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

Question

(b)

Given,

$t^2=a t+b$, where $a, b$ are positive

integers. $t^3=a t^2+b t$

$\Rightarrow t^3=a(a t+b)+b t$

$\Rightarrow t^3=a^2 t+b t+a

Vedclass · Accepted Answer

$8 t+5$

Vedclass · Answer

(b)

Given,

$t^2=a t+b$, where $a, b$ are positive

integers. $t^3=a t^2+b t$

$\Rightarrow t^3=a(a t+b)+b t$

$\Rightarrow t^3=a^2 t+b t+a b$

$\Rightarrow t^3=\left(a^2+b\right) t+a b$

$(i)$ $4 t+3$ $a^2+b=4, a b=3$

$a=1, b=3$ it is possible

$(ii)$ $8 t+5$

$a^2+b=8, a b=5$

It is not possible

(iii) $10 t+3$

$a^2+b=10, a b=3$ $a=3, b=1$ it is possible

(iv) $6 t+5$

$a^2+b=6, a b=5$

$a=1, b=5$ it is also possible

Hence, option $(b)$ is correct.

Let $t$ be real number such that $t^2=a t+b$ for some positive integers $a$ and $b$. Then, for any choice of positive integers $a$ and $b, t^3$ is never equal to

Similar Questions

If $S$ is a set of $P(x)$ is polynomial of degree $ \le 2$ such that $P(0) = 0,$$P(1) = 1$,$P'(x) > 0{\rm{ }}\forall x \in (0,\,1)$, then

The equation $x^2-4 x+[x]+3=x[x]$, where $[x]$ denotes the greatest integer function, has:

The sum of all integral values of $\mathrm{k}(\mathrm{k} \neq 0$ ) for which the equation $\frac{2}{x-1}-\frac{1}{x-2}=\frac{2}{k}$ in $x$ has no real roots, is ..... .

The number of real solutions of the equation $e ^{4 x }+4 e ^{3 x }-58 e ^{2 x }+4 e ^{ x }+1=0$ is..........

The integer $'k'$, for which the inequality $x^{2}-2(3 k-1) x+8 k^{2}-7>0$ is valid for every $x$ in $R ,$ is