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(C) Given,$a^5-a^3+a=2$.Let $f(a) = a^5-a^3+a-2$.The derivative is $f'(a) = 5a^4-3a^2+1$.For the quadratic in $a^2$,the discriminant $D = (-3)^2 - 4(5

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$3 < a^6 < 4$

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(C) Given,$a^5-a^3+a=2$.Let $f(a) = a^5-a^3+a-2$.The derivative is $f'(a) = 5a^4-3a^2+1$.For the quadratic in $a^2$,the discriminant $D = (-3)^2 - 4(5)(1) = 9 - 20 = -11 0$ for all $a \in \mathbb{R}$.Thus,$f(a)$ is strictly increasing and has exactly one real root.We test values for $f(a)$:$f(1) = 1^5 - 1^3 + 1 - 2 = -1 $f(2) = 2^5 - 2^3 + 2 - 2 = 32 - 8 = 24 > 0$.So the root $a$ lies in $(1, 2)$.Now,check $a^6$ values:If $a^6 = 3$,$a = 3^{1/6} \approx 1.2009$.$f(3^{1/6}) = (3^{1/6})^5 - (3^{1/6})^3 + 3^{1/6} - 2 = 3^{5/6} - 3^{1/2} + 3^{1/6} - 2 \approx 2.44 - 1.73 + 1.20 - 2 = -0.09 If $a^6 = 4$,$a = 4^{1/6} = 2^{1/3} \approx 1.2599$.$f(2^{1/3}) = (2^{1/3})^5 - (2^{1/3})^3 + 2^{1/3} - 2 = 2^{5/3} - 2 + 2^{1/3} - 2 = 2^{5/3} + 2^{1/3} - 4 \approx 3.17 + 1.26 - 4 = 0.43 > 0$.Since $f(3^{1/6}) 0$,the root $a$ satisfies $3^{1/6} < a < 4^{1/6}$,which implies $3 < a^6 < 4$.

List-$I$	List-$II$
$a$. If $y=\|x\|+\|x-2\|$, then at $x=2, \frac{dy}{dx}=$	$i$. $2$
$b$. If $f(x)=\|\cos 2x\|$, then $f^{\prime}(\frac{\pi}{4}+)=$	$ii$. $0$
$c$. If $f(x)=\sin(\pi[x])$, where $[\cdot]$ denotes the greatest integer function, then $f^{\prime}(1-)=$	$iii$. $-2$
$d$. If $f(x)=\log\|x-1\|, x \neq 1$, then $f^{\prime}(\frac{1}{2})=$	$iv$. Does not exist

$(a)$ $x\|x\|$	$(i)$ continuous in $(-1, 1)$
$(b)$ $\sqrt{\|x\|}$	$(ii)$ differentiable in $(-1, 1)$
$(c)$ $x+[x]$	$(iii)$ strictly increasing in $(-1, 1)$
$(d)$ $\|x-1\|+\|x+1\|$	$(iv)$ not differentiable at,at least one point in $(-1, 1)$

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

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