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4-2.Quadratic Equations and Inequations
normal
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
A
$4 t+3$
B
$8 t+5$
C
$10 t+3$
D
$6 t+5$
(KVPY-2016)
Solution
(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.
Standard 11
Mathematics
