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

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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

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