I- x2-4 x-4-0II- y 2-8 y -16-0A- x-y B- x - yC- x-y or no relation can be established between x and y D- x - yE- x-y

The correct option is E x lt; yEquation I ⇒ x^2+4x+4=0 ⇒ x^2+2x+2x+4=0 ⇒ x(x+2)+2(x+2)=0 ⇒ (x+2)^2=0 ⇒ x= 2 Equation II y^2 8y+16=0 ⇒ y^2 4y 4y+16=0 ⇒ y(y+4 ...

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Byju's Answer

Standard XII

Mathematics

Derivative of Standard Functions

I. x2+4 x+4=0...

Question

$I . x^{2} + 4 x + 4 = 0$
$I I . y^{2} - 8 y + 16 = 0$

\geq

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x > y

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\leq

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x < y

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x = y or no relation can be established between x and y

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Solution

The correct option is D x < y
Equation I
$\Rightarrow x^{2} + 4 x + 4 = 0$
$\Rightarrow x^{2} + 2 x + 2 x + 4 = 0$
$\Rightarrow x (x + 2) + 2 (x + 2) = 0$
$\Rightarrow (x + 2)^{2} = 0$
$\Rightarrow x = - 2$
Equation II
$y^{2} - 8 y + 16 = 0$
$\Rightarrow y^{2} - 4 y - 4 y + 16 = 0$
$\Rightarrow y (y + 4) - 4 (y - 4) = 0$
$\Rightarrow (y - 4) (y - 4) = 0$
$\Rightarrow (y - 4)^{2} = 0$
$∴ y = 4$
Hence, x < y

Suggest Corrections

I. x2+4x+4=0 II. y2−8y+16=0

The correct option is D x < yEquation I ⇒x2+4x+4=0 ⇒x2+2x+2x+4=0 ⇒x(x+2)+2(x+2)=0 ⇒(x+2)2=0 ⇒x=−2 Equation II y2−8y+16=0 ⇒y2−4y−4y+16=0 ⇒y(y+4)−4(y−4)=0 ⇒(y−4)(y−4)=0 ⇒(y−4)2=0 ∴y=4 Hence, x < y

$I . x^{2} + 4 x + 4 = 0$
$I I . y^{2} - 8 y + 16 = 0$