Differentiable function f - R - R satisfying the equation f x -1- x 2-1-0 x f t dt -1- t 2- isA- fx-k ex1-x2B- fx-k ex1-x3C- fx-k ex1-x2D- fx-k ex1-x3

The correct option is C f(x) = ke^x (1+x^2)f(x)/1+x^2 = 1 + ∫^x _0f(t)dt/1+t^2 Differentiating both sides nbsp; f'(x)(1+x^2) f(x).2x/(1+x^2)^2 = f(x)/1+x^2⇒ ...

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Standard XII

Mathematics

Fundamental Laws of Logarithms

Differentiabl...

Question

Differentiable function $f : R \to R$ satisfying the equation $f (x) = (1 + x^{2}) [1 + x \int 0 \frac{f (t) d t}{1 + t^{2}}]$ is -

f (x) = k e^{x} (1 - x^{2})

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f (x) = k e^{x} (1 + x^{3})

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f (x) = k e^{x} (1 + x^{2})

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f (x) = k e^{x} (1 - x^{3})

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Solution

The correct option is C $f (x) = k e^{x} (1 + x^{2})$
$\frac{f (x)}{1 + x^{2}} = 1 + x \int 0 \frac{f (t) d t}{1 + t^{2}}$
Differentiating both sides
$\frac{f^{'} (x) (1 + x^{2}) - f (x) .2 x}{(1 + x^{2})^{2}} = \frac{f (x)}{1 + x^{2}} \Rightarrow f^{'} (x) = \frac{2 x f (x)}{1 + x^{2}} + f (x)$
$\Rightarrow \frac{f^{'} (x)}{f (x)} = 1 + \frac{2 x}{x^{2} + 1} \Rightarrow ln f (x) = x + ln (1 + x^{2}) + ln k$
$\Rightarrow f (x) = k e^{x} (1 + x^{2})$

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Differentiable function f:R→R satisfying the equation f(x)=(1+x2)[1+x∫0f(t)dt1+t2] is -

The correct option is C f(x)=kex(1+x2)f(x)1+x2=1+x∫0f(t)dt1+t2 Differentiating both sides f′(x)(1+x2)−f(x).2x(1+x2)2=f(x)1+x2⇒f′(x)=2xf(x)1+x2+f(x) ⇒f′(x)f(x)=1+2xx2+1⇒lnf(x)=x+ln(1+x2)+lnk ⇒f(x)=kex(1+x2)

Differentiable function $f : R \to R$ satisfying the equation $f (x) = (1 + x^{2}) [1 + x \int 0 \frac{f (t) d t}{1 + t^{2}}]$ is -