If ∫ 0 x f t d t = x 2 + ∫ x 1 t 2 f t d t , then f ' 1 2 is

Differentiating both the sides w.r.t. x f x = 2 x - x 2 f x ⇒ f x = 2 x 1 + x 2 ⇒ f ' x = 1 + x 2 × 2 - 2 x × 2 x 1 + x 2 2 = 2 - 2 x 2 1 + x 2 2 Hence, f ' 1 2 = 24 25

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Mathematics - Definite Integration Question with Solution | TestHub

MathematicsDefinite IntegrationDerivatives (Newton- Leibnitz)Medium2 minPYQ_2019

MathematicsMediumsingle choice

If $\int_{0}^{x} f (t) d t = x^{2} + \int_{x}^{1} t^{2} f (t) d t,$ then $f^{'} (\frac{1}{2})$ is

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Answer:

Solution:

Differentiating both the sides w.r.t. $x$

$f (x) = 2 x - x^{2} f (x)$
$\Rightarrow f (x) = \frac{2 x}{1 + x^{2}}$
$\Rightarrow f^{'} (x) = \frac{(1 + x^{2}) \times 2 - 2 x \times 2 x}{{(1 + x^{2})}^{2}}$
$= \frac{2 - 2 x^{2}}{{(1 + x^{2})}^{2}}$

Hence, $f^{'} (\frac{1}{2}) = \frac{24}{25}$

Stream:JEESubject:MathematicsTopic:Definite IntegrationSubtopic:Derivatives (Newton- Leibnitz)

⏱ 2mℹ️ Source: PYQ_2019

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