Which of the following function satisfy given mapping diagram.

Given that $x \in Domain & y \in Range$

y = \sqrt{x}

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y = x^{2}

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y = x^{3}

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y + x^{2} = 0

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Solution

The correct option is A $y = \sqrt{x}$
Given that
Domain: ${4, 25, 36, 49}$
Range : ${2, 6, 5, 7}$

On checking through options,
$O p t i o n (a) - --------- - : y = \sqrt{x}$

Here, on putting values of domain and range in the function,
$2 = \sqrt{4}, 5 = \sqrt{25}, 6 = \sqrt{36}, 7 = \sqrt{49}$

hence, $y = \sqrt{x}$ satisfies given mapping.

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Similar questions

y = {3x}^{2} - 5 x + 12

y = x^{2}

{1, 2, - 1, - 2, 3, - 3}

{1, 4, 9}

x \in Domain & y \in Range

X

Y

y = 2 x^{2} - 2 x + 2

X : {\frac{3}{2}, \sqrt{2}, - 1, 1, 0}

Y : {2, 6, \frac{7}{2}, 6 - 2 \sqrt{2}}

x \in X & y \in Y

R

A = {1, 2, 3, 5}

B = {1, 4, 9, 15}

R = {(x, y) : x = \sqrt{y}; x \in A, y \in B},

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