II 24 x- 2-38 x-15-0 011II- 12 y- 2-28 y-15-0

The correct option is C x≤ yEquation I 24x^2+38x+15=0 ⇒ 24x^2+20x+18x+15=0 ⇒ 4x(6x+5)+3(6x+5)=0 ⇒ (4x+3)(6x+5)=0 ∴ x= 3/4, 5/6 Equation II 12y^2+28y+15=0 ⇒ 12 ...

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Byju's Answer

Standard XII

Mathematics

Derivative of Standard Functions

II 24 x∧ 2+38...

Question

$I . 24 x^{2} + 38 x + 15 = 0 I I . 12 y^{2} + 28 y + 15 = 0$

x \geq y

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x > y

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x \leq y

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x < y

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x = y

or relationship between x and y cannot be established

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Solution

The correct option is C $x \leq y$
Equation I
$24 x^{2} + 38 x + 15 = 0$
$\Rightarrow 24 x^{2} + 20 x + 18 x + 15 = 0$
$\Rightarrow 4 x (6 x + 5) + 3 (6 x + 5) = 0$
$\Rightarrow (4 x + 3) (6 x + 5) = 0$
$∴ x = - \frac{3}{4}, - \frac{5}{6}$
Equation II
$12 y^{2} + 28 y + 15 = 0$
$\Rightarrow 12 y^{2} + 18 y + 10 y + 15 = 0$
$\Rightarrow 6 y (2 y + 3) + 5 (2 y + 3) = 0$
$\Rightarrow (6 y + 5) (2 y + 3) = 0$
$∴ y = - \frac{3}{2}, - \frac{5}{6}$
Hence, $x \leq y$

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I. 24x2+38x+15=0II. 12y2+28y+15=0

The correct option is C x≤yEquation I 24x2+38x+15=0 ⇒24x2+20x+18x+15=0 ⇒4x(6x+5)+3(6x+5)=0 ⇒(4x+3)(6x+5)=0 ∴x=−34,−56 Equation II 12y2+28y+15=0 ⇒12y2+18y+10y+15=0 ⇒6y(2y+3)+5(2y+3)=0 ⇒(6y+5)(2y+3)=0 ∴y=−32,−56 Hence, x≤y

$I . 24 x^{2} + 38 x + 15 = 0 I I . 12 y^{2} + 28 y + 15 = 0$