Suppose a is a positive real number such that | vedclass

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Question

(c)

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, \f

Vedclass · Accepted Answer

$3 < a^6 < 4$

Vedclass · Answer

(c) Given, $a^5-a^3+a=2$ $\Rightarrow \quad a^5-a^3+a-2=0$ Let $f(a)=a^5-a^3+a-2$ $f^{\prime}(a)=5 a^4-3 a^2+1$ $f^{\prime}(a)>0, \forall a \in R$ $ herefore a^5-a^3+a-2=0$ has only one roots. for $a^6=3 \Rightarrow a=(3)^{1 / 6}=12$ [by calculation] $f\left(4^{1 / 6} ight) > 0$ and at $a^6=4, a=(4)^{1 / 6}$ So one root lies in $(3,4)$. $ herefore 3 < a^6 < 4$

Suppose $a$ is a positive real number such that $a^5-a^3+a=2$. Then,

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