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$\frac{x^{2}}{f (4 a)} + \frac{y^{2}}{f (a^{2} - 5)} = 1 represents an ellipse with major axis as y -axis and ' f' is decreasing function, positive for all ‘a’,$
$then a belongs to$

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$(5, \infty)$

(1, 4)

(-1, 1)

(-1, 5)

answer is C.

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Detailed Solution

$\begin{array}{l} ' f' is decreasing, major axis is along y -axis \\ f (4 a) < f (a^{2} - 5) \\ \Rightarrow 4 a > a^{2} - 5 \\ \Rightarrow a^{2} - 4 a - 5 < 0 \end{array}$

$\Rightarrow a \in (- 1, 5)$

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