4.2 Quadratic equations and inequations — MATHS STD 11 Science — Question

Given, $a^5-a^3+a=2$

$\Rightarrow \quad a^5-a^3+a-2=0$

Let $f(a)=a^5-a^3+a-2$

$f^{\prime}(a)=5 a^4-3 a^2+1$

$f^{\prime}(a)>0, \forall a \in R$

$\therefore a^5-a^3+a-2=0$ has only one roots.

for $a^6=3 \Rightarrow a=(3)^{1 / 6}=12$ [by calculation]

$f\left(4^{1 / 6}\right) > 0$ and at $a^6=4, a=(4)^{1 / 6}$

So one root lies in $(3,4)$.

$\therefore 3 < a^6 < 4$