The function f- R -1- e def ined by f x - X 2- e - X 2-1 isA- Neither one-one nor ontoB- One-one but not ontoC- Onto but not one-oneD- Both one-one and onto

The correct option is D Both one one and ontof(x)=x^2+e/x^2+1 f^'(x)=2x(x^2+1) 2x(x^2+e)/(x^2+1)^2=2x^3+2x 2x^3 2ex/(x^2+1)^2 =2x 2xe/(x^2+1)^2=2x(1 e)/(x^2+ ...

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Byju's Answer

Standard XI

Mathematics

Domain

The function ...

Question

The function $f : R + \to (1, e) d e f i n e d b y f (x) = \frac{X^{2} + e}{X^{2} + 1}$ is

One-one but not onto

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Onto but not one-one

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Both one-one and onto

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Neither one-one nor onto

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Solution

The correct option is C Both one-one and onto
$f (x) = \frac{x^{2} + e}{x^{2} + 1} f^{^{'}} (x) = \frac{2 x (x^{2} + 1) - 2 x (x^{2} + e)}{(x^{2} + 1)^{2}} = \frac{2 x^{3} + 2 x - 2 x^{3} - 2 e x}{(x^{2} + 1)^{2}} = \frac{2 x - 2 x e}{(x^{2} + 1)^{2}} = \frac{2 x (1 - e)}{(x^{2} + 1)^{2}} < 0 f^{^{'}} (x) < 0, f (x) i s d e c r e a s i n g$
Hence f is one-one function.
$x \to 0, f (x) \to e$
$x \to \infty, f (x) \to e$
Hence range = (1, e) = co-domain

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The function f:R+→(1,e) defined by f(x)=X2+eX2+1 is

The correct option is C Both one-one and ontof(x)=x2+ex2+1f′(x)=2x(x2+1)−2x(x2+e)(x2+1)2=2x3+2x−2x3−2ex(x2+1)2=2x−2xe(x2+1)2=2x(1−e)(x2+1)2<0f′(x)<0,f(x) is decreasing Hence f is one-one function. x→0,f(x)→e x→∞,f(x)→e Hence range = (1, e) = co-domain

The function $f : R + \to (1, e) d e f i n e d b y f (x) = \frac{X^{2} + e}{X^{2} + 1}$ is