F-2- - - -4- where fx-x4-xthen f-1xis

Y=4x x^2 x^2 4x+y=0 x=4±√(16 4y)/2 f^ 1(y)=2±√(4 y) f^ 1(x) = 2+√(4 x) 2 √(4 x) is not possible because image is not present in its co domain only 2+√(4 x) is p ...

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Byju's Answer

Standard XII

Mathematics

Inverse of a Function

f:[2,∞ → -∞ ,...

Question

$f : [2, \infty) \to (- \infty, 4], w h e r e f (x) = x (4 - x) t h e n f^{- 1} (x) i s$

2 - \sqrt{4 - x}

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2 + \sqrt{4 - x}

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- 2 + \sqrt{4 - x}

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- 2 - \sqrt{4 - x}

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Solution

The correct option is B $2 + \sqrt{4 - x}$
$y = 4 x - x^{2}$
$x^{2} - 4 x + y = 0$
$x = \frac{4 \pm \sqrt{16 - 4 y}}{2}$
$f^{- 1} (y) = 2 \pm \sqrt{4 - y}$
$f^{- 1} (x)$ = $2 + \sqrt{4 - x}$
$2 - \sqrt{4 - x}$ is not possible because image is not present in its co-domain only $2 + \sqrt{4 - x}$ is possible

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Similar questions

f : [2, \infty) \to (- \infty, 4], w h e r e f (x) = x (4 - x) t h e n f^{- 1} (x) i s

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f:[2,∞)→(−∞,4],where f(x)=x(4−x)then f−1(x)is

The correct option is B 2+√4−xy=4x−x2 x2−4x+y=0 x=4±√16−4y2 f−1(y)=2±√4−y f−1(x) = 2+√4−x 2−√4−x is not possible because image is not present in its co-domain only 2+√4−x is possible

$f : [2, \infty) \to (- \infty, 4], w h e r e f (x) = x (4 - x) t h e n f^{- 1} (x) i s$