Consider the function fx - x2-x2 - 1-If f is defined from R -1- 1- R then f is

The correct option is D neither injective nor surjectiveSince nbsp; f( x ) = f( x) , function is not injective.Also there does not exist nbsp;any pre ...

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Standard XII

Mathematics

Bijective Function

Consider the ...

Question

Consider the function $f (x) = \frac{x^{2}}{x^{2} - 1}$ .If $f$ is defined from $R - (- 1, 1) \to R$ then $f$ is

injective but not surjective

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surjective but not injective

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injective as well as surjective

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neither injective nor surjective

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Solution

The correct option is D neither injective nor surjective
Since $f (x) = f (- x)$ , function is not injective.
Also there does not exist any pre-image for $y = 0$ as $x = 0$ is not the element of domain set.
Hence the function is neither injective nor surjective.

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Similar questions

Q. The function

f : R \to R, f (x) = x^{2}

is
(a) injective but not surjective
(b) surjective but not injective
(c) injective as well as surjective
(d) neither injective nor surjective

Q. The function :

R \to [- \frac{1}{2}, \frac{1}{2}]

defined as

f (x) = \frac{x}{1 + x^{2}}

Q. Let

f : R \to R

be a function defined by

f (x) = x^{3} + x^{2} + 3 x + sin x .

Then

f

Q. The function

f : R \to [- \frac{1}{2}, \frac{1}{2}]

defined as

f (x) = \frac{x}{1 + x^{2}},

Q. Let a function

f : (0, \infty) \to (0, \infty)

be defined by

f (x) = ∣ ∣ ∣ 1 - \frac{1}{x} ∣ ∣ ∣

. Then

f

is :

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Consider the function f(x)=x2x2−1.If f is defined from R−(−1,1)→R then f is

The correct option is D neither injective nor surjectiveSince f(x)=f(−x), function is not injective.Also there does not exist any pre-image for y=0 as x=0 is not the element of domain set. Hence the function is neither injective nor surjective.

Consider the function $f (x) = \frac{x^{2}}{x^{2} - 1}$ .If $f$ is defined from $R - (- 1, 1) \to R$ then $f$ is

The correct option is D neither injective nor surjective
Since $f (x) = f (- x)$ , function is not injective.
Also there does not exist any pre-image for $y = 0$ as $x = 0$ is not the element of domain set.
Hence the function is neither injective nor surjective.