I- x2-7 x-12-0II- y 2- y -12-0A- x-y or the relationship cannot be established-B- x-y C- x - yD- x-y E- x - y

The correct option is C x ≥ yEquation I x^2 7x + 12 = 0 ⇒ x^2 3x 4x + 12=0 ⇒ x(x 3) 4(x 3)=0 ⇒ (x 3)(x 4)=0 ∴ x=3, 4 Equation II y^2 + y 12=0 ⇒ y^2 +4y ...

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Byju's Answer

Standard XII

Mathematics

Derivative of Standard Functions

I. x2-7 x+12=...

Question

I. $x^{2} - 7 x + 12 = 0$
II. $y^{2} + y - 12 = 0$

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 B $x \geq y$
Equation I
$x^{2} - 7 x + 12 = 0$
$\Rightarrow x^{2} - 3 x - 4 x + 12 = 0$
$\Rightarrow x (x - 3) - 4 (x - 3) = 0$
$\Rightarrow (x - 3) (x - 4) = 0$
$∴ x = 3, 4$

Equation II
$y^{2} + y - 12 = 0$
$\Rightarrow y^{2} + 4 y - 3 y - 12 = 0$
$\Rightarrow y (y - 4) - 3 (y + 4) = 0$
$\Rightarrow (y - 3) (y + 4) = 0$
$∴ y = 3, - 4$
Hence, $x \geq y$

Suggest Corrections

I. x2−7x+12=0 II. y2+y−12=0

The correct option is B x≥yEquation I x2−7x+12=0 ⇒x2−3x−4x+12=0 ⇒x(x−3)−4(x−3)=0 ⇒(x−3)(x−4)=0 ∴x=3,4 Equation II y2+y−12=0 ⇒y2+4y−3y−12=0 ⇒y(y−4)−3(y+4)=0 ⇒(y−3)(y+4)=0 ∴y=3,−4 Hence, x≥y

I. $x^{2} - 7 x + 12 = 0$
II. $y^{2} + y - 12 = 0$