Let f be a continuous function satisfying 0 t 2 ( f ( x ) + x 2 ) ? d x = 4 3 t 3 , t 0 . Then f ( 2 4 ) is equal to [2023]

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Q.
Let $f$ be a continuous function satisfying $\int_{0}^{t^{2}} (f (x) + x^{2}) d x = \frac{4}{3} t^{3}, \forall t > 0 .$ Then $f (\frac{π^{2}}{4})$ is equal to [2023]

1 $π (1 - \frac{π^{3}}{16})$

2 $- π^{2} (1 + \frac{π^{3}}{16})$

3 $- π (1 + \frac{π^{3}}{16})$

4 $π^{2} (1 - \frac{π^{2}}{16})$

Ans.
(1)

$Given, \int_{0}^{t^{2}} (f (x) + x^{2}) d x = \frac{4}{3} t^{3}, \forall t > 0$

Using Leibnitz Rule, we get

$[f (t^{2}) + {(t^{2})}^{2}] \cdot 2 t = 4 t^{2} \Rightarrow f (t^{2}) + t^{4} = 2 t \Rightarrow f (t^{2}) = 2 t - t^{4}$

$Put t = \frac{π}{2}$

$f ({(\frac{π}{2})}^{2}) = 2 \frac{π}{2} - {(\frac{π}{2})}^{4} = π - \frac{π^{4}}{16} = π (1 - \frac{π^{3}}{16})$

Solution

Q.
Let $f$ be a continuous function satisfying $\int_{0}^{t^{2}} (f (x) + x^{2}) d x = \frac{4}{3} t^{3}, \forall t > 0 .$ Then $f (\frac{π^{2}}{4})$ is equal to [2023]

1 $π (1 - \frac{π^{3}}{16})$

2 $- π^{2} (1 + \frac{π^{3}}{16})$

3 $- π (1 + \frac{π^{3}}{16})$

4 $π^{2} (1 - \frac{π^{2}}{16})$

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Solution

Q. Let f be a continuous function satisfying ∫0t2(f(x)+x2) dx=43t3,∀t>0. Then f(π24) is equal to [2023]

1 π(1-π316)

2 -π2(1+π316)

3 -π(1+π316)

4 π2(1-π216)

Ans. (1) Given, ∫0t2(f(x)+x2) dx=43t3,∀ t>0 Using Leibnitz Rule, we get [f(t2)+(t2)2]·2t=4t2⇒f(t2)+t4=2t ⇒f(t2)=2t-t4 Put t=π2 f((π2)2)=2π2-(π2)4=π-π416=π(1-π316)

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Q.
Let $f$ be a continuous function satisfying $\int_{0}^{t^{2}} (f (x) + x^{2}) d x = \frac{4}{3} t^{3}, \forall t > 0 .$ Then $f (\frac{π^{2}}{4})$ is equal to [2023]

1 $π (1 - \frac{π^{3}}{16})$

2 $- π^{2} (1 + \frac{π^{3}}{16})$

3 $- π (1 + \frac{π^{3}}{16})$

4 $π^{2} (1 - \frac{π^{2}}{16})$