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Standard XII

Mathematics

Polynomial Functions

The function ...

Question

The function $f : [2, \infty) \to Y$ defined by $f (x) = x^{2} - 4 x + 5$ is one-one if (more than one option can be correct)

Y = R

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Y = [1, \infty)

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Y = [0, \infty)

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Y = [- 1, \infty)

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Solution

The correct option is A $Y = [1, \infty)$
$f (x) = (x - 2)^{2} + 1$ which for domain $[2, \infty)$ is one one and has minimum value at 1
$∴ Y = [1, \infty)$

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Similar questions

Q. The function

f : [2, \infty) \to Y

defined by

f (x) = x^{2} - 4 x + 5

is both one-one & onto if:

Q. Let

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

and

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

be functions defined by

f (x) = [x^{2} - 3]

and

g (x) = | x | f (x) + | 4 x - 7 | f (x)

, where

[y]

denotes the greatest integer less than or equal to y for

y \in R

. Then

Let $f : [- \frac{1}{2}, 2] \to R$ and $g : [- \frac{1}{2}, 2] \to R$ be functions defined by
$f (x) = [x^{2} - 3] and g (x) = | x | f (x) + | 4 x - 7 | f (x)$ ,where [y] denotes the greatest integer less than or equal to y for $y ϵ R$ . Then

Q. Let

f

be any function defined on

R

and let it satisfy the condition :

| f (x) - f (y) | \leq | (x - y)^{2} |, \forall x, y \in R

f (0) = 1,

then

Q. Let f:

X \to Y

be a function defined by

f (x) = a sin (x + \frac{π}{4}) + b cos x + c

. If f is both one-one and onto, then find the sets

X

and

Y

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The function f:[2,∞)→Y defined by f(x)=x2−4x+5 is one-one if (more than one option can be correct)

The correct option is A Y=[1,∞)f(x)=(x−2)2+1 which for domain [2,∞) is one one and has minimum value at 1∴Y=[1,∞)

The function $f : [2, \infty) \to Y$ defined by $f (x) = x^{2} - 4 x + 5$ is one-one if (more than one option can be correct)

The correct option is A $Y = [1, \infty)$
$f (x) = (x - 2)^{2} + 1$ which for domain $[2, \infty)$ is one one and has minimum value at 1
$∴ Y = [1, \infty)$