I- 4 x 2-4 x -3-0II- 4 y 2-12 y -5-0A- If x - yB- If x - yC- If x yE- If x - y

The correct option is E If x ≥ yEquation I 4x^2 4x 3=0 ⇒ 4x^2 6x+2x 3=0 ⇒ 2x(2x 3)+1(2x 3)=0 ⇒ (2x+1)(2x 3)=0 ⇒ x= 1/2, 3/2 Equation II 4y^2+12y+5=0 ⇒ 4y^2+1 ...

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Byju's Answer

Standard XII

Mathematics

Derivative of Standard Functions

I. 4 x 2-4 x ...

Question

I. $4 x^{2} - 4 x - 3 = 0$
II. $4 y^{2} + 12 y + 5 = 0$

x < y

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x \leq y

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x = y

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x > y

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x \geq y

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Solution

The correct option is E If $x \geq y$
Equation I
$4 x^{2} - 4 x - 3 = 0$
$\Rightarrow 4 x^{2} - 6 x + 2 x - 3 = 0$
$\Rightarrow 2 x (2 x - 3) + 1 (2 x - 3) = 0$
$\Rightarrow (2 x + 1) (2 x - 3) = 0$
$\Rightarrow x = - \frac{1}{2}, \frac{3}{2}$

Equation II
$4 y^{2} + 12 y + 5 = 0$
$\Rightarrow 4 y^{2} + 10 y + 2 y + 5 = 0$
$\Rightarrow 2 y (2 y + 5) + 1 (2 y + 5) = 0$
$\Rightarrow (2 y + 5) (2 y + 1) = 0$
$\Rightarrow y = - \frac{5}{2}, - \frac{1}{2}$
Hence $x \geq y$

Suggest Corrections

I. 4x2−4x−3=0 II. 4y2+12y+5=0

The correct option is E If x≥yEquation I 4x2−4x−3=0 ⇒4x2−6x+2x−3=0 ⇒2x(2x−3)+1(2x−3)=0 ⇒(2x+1)(2x−3)=0 ⇒x=−12,32 Equation II 4y2+12y+5=0 ⇒4y2+10y+2y+5=0 ⇒2y(2y+5)+1(2y+5)=0 ⇒(2y+5)(2y+1)=0 ⇒y=−52,−12 Hence x≥y

I. $4 x^{2} - 4 x - 3 = 0$
II. $4 y^{2} + 12 y + 5 = 0$