I- 2 x2-3 x-20-0II- 2 y 2-19 y -44-0A- x-yB- x - yC- x-y D- x - yE- x - y or no relationship can be established between x and y

The correct option is D x ≥ yEquation I 2x^2+3x 20=0 ⇒ 2x^2+8x 5x 20 ⇒ 2x(x+4) 5(x+4)=0 ⇒ (2x 5)(x+4)=0 ∴ x=5/2, 4 Equation II ⇒ 2y^2+19y+44=0 ⇒ 2y^2+11y+8y+4 ...

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Byju's Answer

Standard XII

Mathematics

Chain Rule of Differentiation

I. 2 x2+3 x-2...

Question

$I . 2 x^{2} + 3 x - 20 = 0$
$I I . 2 y^{2} + 19 y + 44 = 0$

x < y

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x > y

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x = y

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x \geq y

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x \leq y

or no relationship can be established between x and y

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Solution

The correct option is D $x \geq y$
Equation I
$2 x^{2} + 3 x - 20 = 0$
$\Rightarrow 2 x^{2} + 8 x - 5 x - 20$
$\Rightarrow 2 x (x + 4) - 5 (x + 4) = 0$
$\Rightarrow (2 x - 5) (x + 4) = 0$
$∴ x = \frac{5}{2}, - 4$
Equation II
$\Rightarrow 2 y^{2} + 19 y + 44 = 0$
$\Rightarrow 2 y^{2} + 11 y + 8 y + 44 = 0$
$\Rightarrow y (2 y + 11) + 4 (2 y + 11) = 0$
$\Rightarrow (y + 4) (2 y + 11) = 0$
$∴ y = - 4, y = - \frac{11}{2}$
Hence, $x \geq y$

Suggest Corrections

I. 2x2+3x−20=0 II. 2y2+19y+44=0

The correct option is D x≥yEquation I 2x2+3x−20=0 ⇒2x2+8x−5x−20 ⇒2x(x+4)−5(x+4)=0 ⇒(2x−5)(x+4)=0 ∴x=52,−4 Equation II ⇒2y2+19y+44=0 ⇒2y2+11y+8y+44=0 ⇒y(2y+11)+4(2y+11)=0 ⇒(y+4)(2y+11)=0 ∴y=−4,y=−112 Hence, x≥y

$I . 2 x^{2} + 3 x - 20 = 0$
$I I . 2 y^{2} + 19 y + 44 = 0$