I- 4 x2-8 x-3-0II- 2 y 2-7 y -6-0A- If x - yB- If xy

The correct option is A If x ≤ yEquation I 4x^2 8x+3=0 ⇒ 4x^2 6x 2x+3=0 ⇒ 2x(2x 3) 1(2x 3)=0 ⇒ (2x 1)(2x 3)=0 ∴ x = 1/2 or 3/2 Equation II 2y^2 7y+6=0 ⇒ 2y^2 ...

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Byju's Answer

Standard XII

Mathematics

Derivative of Standard Functions

I. 4 x2-8 x+3...

Question

I. $4 x^{2} - 8 x + 3 = 0$
II. $2 y^{2} - 7 y + 6 = 0$

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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x > y

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Solution

The correct option is B If $x \leq y$
Equation I
$4 x^{2} - 8 x + 3 = 0$
$\Rightarrow 4 x^{2} - 6 x - 2 x + 3 = 0$
$\Rightarrow 2 x (2 x - 3) - 1 (2 x - 3) = 0$
$\Rightarrow (2 x - 1) (2 x - 3) = 0$
$∴ x = \frac{1}{2}$ or $\frac{3}{2}$

Equation II
$2 y^{2} - 7 y + 6 = 0$
$\Rightarrow 2 y^{2} - 4 y - 3 y + 6 = 0$
$\Rightarrow 2 y (y - 2) - 3 (y - 2) = 0$
$\Rightarrow (2 y - 3) (y - 2) = 0$
$\Rightarrow y = \frac{3}{2}$ or $2$
Hence, $x \leq y$

Suggest Corrections

I. 4x2−8x+3=0 II. 2y2−7y+6=0

The correct option is B If x≤yEquation I 4x2−8x+3=0 ⇒4x2−6x−2x+3=0 ⇒2x(2x−3)−1(2x−3)=0 ⇒(2x−1)(2x−3)=0 ∴x=12 or 32 Equation II 2y2−7y+6=0 ⇒2y2−4y−3y+6=0 ⇒2y(y−2)−3(y−2)=0 ⇒(2y−3)(y−2)=0 ⇒y=32 or 2 Hence, x≤y

I. $4 x^{2} - 8 x + 3 = 0$
II. $2 y^{2} - 7 y + 6 = 0$