I- 3 x2-8 x-4-0II- 4 y 2-19 y -12-0A- If x - yB- If x-yC- If x - yD- If x - yE- x - y or the relationship cannot be established-

The correct option is C If x lt; yEquation I 3x^2+8x+4=0 ⇒ 3x^2+6x+2x+4=0 ⇒ 3x(x+2)+2(x+2)=0 ⇒ (x+2)(3x+2)=0 ⇒ x= 2, 2/3 Equation II 4y^2 19y + 12=0 ⇒ ...

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Byju's Answer

Standard XII

Mathematics

Derivative of Standard Functions

I. 3 x2+8 x+4...

Question

I. $3 x^{2} + 8 x + 4 = 0$
II. $4 y^{2} - 19 y + 12 = 0$

x > y

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x \geq y

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x < y

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x \leq y

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x = y

or the relationship cannot be established.

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Solution

The correct option is C If $x < y$
Equation I
$3 x^{2} + 8 x + 4 = 0$
$\Rightarrow 3 x^{2} + 6 x + 2 x + 4 = 0$
$\Rightarrow 3 x (x + 2) + 2 (x + 2) = 0$
$\Rightarrow (x + 2) (3 x + 2) = 0$
$\Rightarrow x = - 2, \frac{- 2}{3}$

Equation II
$4 y^{2} - 19 y + 12 = 0$
$\Rightarrow 4 y^{2} - 16 y - 3 y + 12 = 0$
$\Rightarrow 4 y (y - 4) - 3 (y - 4) = 0$
$\Rightarrow (4 y - 3) (y - 4) = 0$
$\Rightarrow y = 4, \frac{3}{4}$
Hence, $x < y$

Suggest Corrections

I. 3x2+8x+4=0 II. 4y2−19y+12=0

The correct option is C If x<yEquation I 3x2+8x+4=0 ⇒3x2+6x+2x+4=0 ⇒3x(x+2)+2(x+2)=0 ⇒(x+2)(3x+2)=0 ⇒x=−2,−23 Equation II 4y2−19y+12=0 ⇒4y2−16y−3y+12=0 ⇒4y(y−4)−3(y−4)=0 ⇒(4y−3)(y−4)=0 ⇒y=4,34 Hence, x<y

I. $3 x^{2} + 8 x + 4 = 0$
II. $4 y^{2} - 19 y + 12 = 0$