I- 12 x 2-11 x -12-10 x 2-22 xII- 13 y 2-18 y -3-9 y 2-10 yA- x - yB- x - yC- xyE- x-y or the relationship cannot be established

The correct option is A x ≥ yEquation I 12x^2+11x+12=10x^2+22x ⇒ 2x^2 11x+12=0 ⇒ 2x^2 8x 3x+12=0 ⇒ 2x(x 4) 3(x 4)=0 Then (2x 3)=0 ⇒ x=3/2 and (x 4)=0 ...

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Byju's Answer

Standard XII

Mathematics

Derivative of Standard Functions

I. 12 x 2+11 ...

Question

$I . 12 x^{2} + 11 x + 12 = 10 x^{2} + 22 x$
$I I . 13 y^{2} - 18 y + 3 = 9 y^{2} - 10 y$

x > y

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\geq

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x < y

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\leq

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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
$12 x^{2} + 11 x + 12 = 10 x^{2} + 22 x$
$\Rightarrow 2 x^{2} - 11 x + 12 = 0$
$\Rightarrow 2 x^{2} - 8 x - 3 x + 12 = 0$
$\Rightarrow 2 x (x - 4) - 3 (x - 4) = 0$
Then (2x-3)=0
$\Rightarrow x = \frac{3}{2}$
and (x-4)=0
$\Rightarrow x = 4$
Equation II
$13 y^{2} - 18 y + 3 = 9 y^{2} - 10 y$
$\Rightarrow 4 y^{2} - 8 y + 3 = 0$
$\Rightarrow 4 y^{2} - 6 y - 2 y + 3 = 0$
$\Rightarrow 2 y (2 y - 3) - 1 (2 y - 3) = 0$
$\Rightarrow (2 y - 1) (2 y - 3) = 0$
Then, $(2 y - 1) = 0 \Rightarrow y = \frac{1}{2}$
and $(2 y - 3) = 0 \Rightarrow y = \frac{3}{2}$
Hence, x $\geq$ y

Suggest Corrections

I. 12x2+11x+12=10x2+22x II. 13y2−18y+3=9y2−10y

$I . 12 x^{2} + 11 x + 12 = 10 x^{2} + 22 x$
$I I . 13 y^{2} - 18 y + 3 = 9 y^{2} - 10 y$