Let f- R - R be a continuous function satisfying fx-0x t ft d t-x2-0for all x - R-T henA- lim x - fx-2- lim x - fx-2C- f x has more than one point in common with the x-axisD- f x is an odd function

The correct option is B f'(x)+xf(x)+2x=0 dy/dx+xy= 2x I.F.=e^x^2/2 ⇒ y.e^x^2/2= ∫ e^x^2/2( 2x) dx+C e^x^2/2.y= 2e^x^2/2+C y= 2+Ce^ix^2/2 f(0)= 0 ⇒ C=2 f( ...

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Byju's Answer

Standard XII

Mathematics

Property 5

Let f: R → R ...

Question

Let $f : R \to R$ be a continuous function satisfying

$f (x) + \int_{0}^{x} t f (t) d t + x^{2} = 0$
$f o r a l l x ϵ R . T h e n$

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Solution

The correct option is B

$f^{'} (x) + x f (x) + 2 x = 0 \frac{d y}{d x} + x y = - 2 x I . F . = e^{\frac{x^{2}}{2}} \Rightarrow y . e^{\frac{x^{2}}{2}} = \int e^{\frac{x^{2}}{2}} (- 2 x) d x + C e^{\frac{x^{2}}{2}} . y = - 2 e^{\frac{x^{2}}{2}} + C y = - 2 + C e^{\frac{i x^{2}}{2}} f (0) = 0 \Rightarrow C = 2 f (x) = 2 ⌊ e^{\frac{- x^{2}}{2}} - 1 ⌋$

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Let f:R→R be a continuous function satisfying f(x)+∫x0tf(t)dt+x2=0 for all xϵR.Then

The correct option is B f′(x)+xf(x)+2x=0dydx+xy=−2xI.F.=ex22⇒y.ex22=∫ex22(−2x)dx+Cex22.y=−2ex22+Cy=−2+Ceix22f(0)=0⇒C=2f(x)=2⌊e−x22−1⌋

Let $f : R \to R$ be a continuous function satisfying

$f (x) + \int_{0}^{x} t f (t) d t + x^{2} = 0$
$f o r a l l x ϵ R . T h e n$