I - - 6 X - 2-5 X -1-0 1-II- 15 Y- 2-8 Y -1-0 A- X - YB- X - YC- X - YD- X - YE- Relationship between X and Y cannot be established-

The correct option is A X ≤ YEquation I 6X^2+5X+1=0 ⇒ 6X^2+3X+2X+1=0 ⇒ 3X(2X+1)+1(2X+1)=0 ⇒ (3X+1)(2X+1)=0 ∴ X= 1/3 or X= 1/2 Equation II 15Y^2+8Y+1=0 ⇒ 15Y^2+ ...

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Byju's Answer

Standard XII

Mathematics

Derivative of Standard Functions

I . ∼ 6 X ∧ 2...

Question

$I . 6 X^{2} + 5 X + 1 = 0 I I . 15 Y^{2} + 8 Y + 1 = 0$

X > Y

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X \leq Y

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X < Y

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X \geq Y

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Relationship between X and Y cannot be established.

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Solution

The correct option is B $X \leq Y$
Equation I
$6 X^{2} + 5 X + 1 = 0$
$\Rightarrow 6 X^{2} + 3 X + 2 X + 1 = 0$
$\Rightarrow 3 X (2 X + 1) + 1 (2 X + 1) = 0$
$\Rightarrow (3 X + 1) (2 X + 1) = 0$
$∴ X = \frac{- 1}{3} o r X = - \frac{1}{2}$
Equation II
$15 Y^{2} + 8 Y + 1 = 0$
$\Rightarrow 15 Y^{2} + 5 Y + 3 Y + 1 = 0$
$\Rightarrow 5 Y (3 Y + 1) + 1 (3 Y + 1) = 0$
$\Rightarrow (5 Y + 1) (3 Y + 1) = 0$
$∴ Y X = - \frac{1}{5} o r - \frac{1}{3}$
Hence, $X \leq Y$

Suggest Corrections

I. 6X2+5X+1=0II. 15Y2+8Y+1=0

The correct option is B X≤YEquation I 6X2+5X+1=0 ⇒6X2+3X+2X+1=0 ⇒3X(2X+1)+1(2X+1)=0 ⇒(3X+1)(2X+1)=0 ∴X=−13 or X=−12 Equation II 15Y2+8Y+1=0 ⇒15Y2+5Y+3Y+1=0 ⇒5Y(3Y+1)+1(3Y+1)=0 ⇒(5Y+1)(3Y+1)=0 ∴YX=−15 or −13 Hence, X≤Y

$I . 6 X^{2} + 5 X + 1 = 0 I I . 15 Y^{2} + 8 Y + 1 = 0$