Let f : R R be a function defined by f ( x ) = x ( 1 + x 4 ) 1 / 4 , and g ( x ) = f ( f ( f ( f ( x ) ) ) ) . Then 18 0 2 5 x 2 g ( x ) ? d x is equal to [2024]

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Solution

Q.
Let $f : R \to R$ be a function defined by $f (x) = \frac{x}{{(1 + x^{4})}^{1 / 4}},$ and $g (x) = f (f (f (f (x)))) .$ Then $18 \int_{0}^{\sqrt{2 \sqrt{5}}} x^{2} g (x) d x$ is equal to [2024]

1 36

2 42

3 39

4 33

Ans.
(3)

We have, $f (x) = \frac{x}{{(1 + x^{4})}^{1 / 4}} and g (x) = f (f (f (x)))$

$∴ f (f (f (f (x)))) = \frac{x}{{(1 + 4 x^{4})}^{1 / 4}} = g (x)$

$Now, \int_{0}^{\sqrt{2 \sqrt{5}}} x^{2} g (x) d x = \int_{0}^{\sqrt{2 \sqrt{5}}} x^{2} \cdot \frac{x}{{(1 + 4 x^{4})}^{1 / 4}} d x$

$= \int_{0}^{\sqrt{2 \sqrt{5}}} \frac{x^{3}}{{(1 + 4 x^{4})}^{1 / 4}} d x$

$Put 1 + 4 x^{4} = t^{4} \Rightarrow 16 x^{3} d x = 4 t^{3} d t$

$= \frac{1}{4} \int_{1}^{3} t^{2} d t = \frac{1}{12} (27 - 1) = \frac{26}{12} = \frac{13}{6}$

$∴ 18 \int_{0}^{\sqrt{2 \sqrt{5}}} x^{2} g (x) d x = 18 \times \frac{13}{6} = 39$

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