Let f ( x ) = x 3 3 – x 2 d x . If 5 f ( 2 ) = – 4 , then f(1) is equal to [2025]

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Q.
Let $f (x) = \int x^{3} \sqrt{3 - x^{2}} d x$ . If $5 f (\sqrt{2}) = - 4$ , then f(1) is equal to [2025]

1 $- \frac{6 \sqrt{2}}{5}$

2 $- \frac{4 \sqrt{2}}{5}$

3 $- \frac{2 \sqrt{2}}{5}$

4 $- \frac{8 \sqrt{2}}{5}$

Ans.
(1)

We have, $f (x) = \int x^{3} \sqrt{3 - x^{2}} d x$

Put $3 - x^{2} = u^{2} \Rightarrow x d x = - u d u$

$∴ f (x) = \int (3 - u^{2}) \cdot u (- u d u)$

$= \int (u^{4} - 3 u^{2}) d u = \frac{u^{5}}{5} - u^{3} + C$

$\Rightarrow f (x) = \frac{{(3 - x^{2})}^{5 / 2}}{5} - {(3 - x^{2})}^{3 / 2} + C$ $[∵ u = \sqrt{3 - x^{2}}]$

$\Rightarrow f (\sqrt{2}) = \frac{1}{5} - 1 + C = \frac{- 4}{5} + C$

Now, $5 f (\sqrt{2}) = - 4$

$\Rightarrow - 4 + 5 C = - 4 \Rightarrow 5 C = 0 \Rightarrow C = 0$

Now, $f (1) = \frac{2^{5 / 2}}{5} - 2^{3 / 2} = 2^{1 / 2} (\frac{4}{5} - 2)$

$\Rightarrow f (1) = \frac{- 6 \sqrt{2}}{5}$ .

Solution

Q.
Let $f (x) = \int x^{3} \sqrt{3 - x^{2}} d x$ . If $5 f (\sqrt{2}) = - 4$ , then f(1) is equal to [2025]

1 $- \frac{6 \sqrt{2}}{5}$

2 $- \frac{4 \sqrt{2}}{5}$

3 $- \frac{2 \sqrt{2}}{5}$

4 $- \frac{8 \sqrt{2}}{5}$

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Solution

Q. Let f(x)=∫x33–x2dx. If 5f(2)=–4, then f(1) is equal to [2025]

1 –625

2 –425

3 –225

4 –825

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Q.
Let $f (x) = \int x^{3} \sqrt{3 - x^{2}} d x$ . If $5 f (\sqrt{2}) = - 4$ , then f(1) is equal to [2025]

1 $- \frac{6 \sqrt{2}}{5}$

2 $- \frac{4 \sqrt{2}}{5}$

3 $- \frac{2 \sqrt{2}}{5}$

4 $- \frac{8 \sqrt{2}}{5}$