I- x2-4-0II- y 2-6 y -9-0A- If x-yB- If x-yC- If x - yD- If x-yE- If x - y

The correct option is A If x gt; yEquation I x^2=4 ⇒ x = ± 2 Equation II y^2+6y+9=0 y^2+3y+3y+9=0 y(y+3)+3(y+3)=0 (y+3)(y+3)=0 ⇒ (y+3)^2 = 0 ⇒ y+3=0 ⇒ y= 3 ...

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Byju's Answer

Standard XII

Mathematics

Derivative of Standard Functions

I. x2-4=0II. ...

Question

I. $x^{2} - 4 = 0$
II. $y^{2} + 6 y + 9 = 0$

x < y

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x \leq y

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x = y

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x \geq y

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x > y

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Solution

The correct option is E If $x > y$
Equation I
$x^{2} = 4 \Rightarrow x = \pm 2$

Equation II
$y^{2} + 6 y + 9 = 0$
$y^{2} + 3 y + 3 y + 9 = 0$
$y (y + 3) + 3 (y + 3) = 0$
$(y + 3) (y + 3) = 0$
$\Rightarrow (y + 3)^{2} = 0$
$\Rightarrow y + 3 = 0$
$\Rightarrow y = - 3$
Hence, $x > y$

Suggest Corrections

I. x2−4=0 II. y2+6y+9=0

The correct option is E If x>yEquation I x2=4⇒x=±2 Equation II y2+6y+9=0 y2+3y+3y+9=0 y(y+3)+3(y+3)=0 (y+3)(y+3)=0 ⇒(y+3)2=0 ⇒y+3=0 ⇒y=−3 Hence, x>y

I. $x^{2} - 4 = 0$
II. $y^{2} + 6 y + 9 = 0$

The correct option is E If $x > y$
Equation I
$x^{2} = 4 \Rightarrow x = \pm 2$

Equation II
$y^{2} + 6 y + 9 = 0$
$y^{2} + 3 y + 3 y + 9 = 0$
$y (y + 3) + 3 (y + 3) = 0$
$(y + 3) (y + 3) = 0$
$\Rightarrow (y + 3)^{2} = 0$
$\Rightarrow y + 3 = 0$
$\Rightarrow y = - 3$
Hence, $x > y$